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  • How do you pass user credentials from WebClient to a WCF REST service?

    - by Alex
    I am trying to expose a WCT REST service and only users with valid username and password would be able to access it. The username and password are stored in a SQL database. Here is the service contract: public interface IDataService { [OperationContract] [WebGet(ResponseFormat = WebMessageFormat.Json)] byte[] GetData(double startTime, double endTime); } Here is the WCF configuration: <bindings> <webHttpBinding> <binding name="SecureBinding"> <security mode="Transport"> <transport clientCredentialType="Basic"/> </security> </binding> </webHttpBinding> </bindings> <behaviors> <serviceBehaviors> <behavior name="DataServiceBehavior"> <serviceMetadata httpGetEnabled="true"/> <serviceCredentials> <userNameAuthentication userNamePasswordValidationMode="Custom" customUserNamePasswordValidatorType= "CustomValidator, WCFHost" /> </serviceCredentials> </behavior> </serviceBehaviors> </behaviors> <services> <service behaviorConfiguration="DataServiceBehavior" name="DataService"> <endpoint address="" binding="webHttpBinding" bindingConfiguration="SecureBinding" contract="IDataService" /> </service> </services> I am accessing the service via the WebClient class within a Silverlight application. However, I have not been able to figure out how to pass the user credentials to the service. I tried various values for client.Credentials but none of them seems to trigger the code in my custom validator. I am getting the following error: The underlying connection was closed: An unexpected error occurred on a send. Here is some sample code I have tried: WebClient client = new WebClient(); client.Credentials = new NetworkCredential("name", "password", "domain"); client.OpenReadCompleted += new OpenReadCompletedEventHandler(GetData); client.OpenReadAsync(new Uri(uriString)); If I set the security mode to None, the whole thing works. I also tried other clientCredentialType values and none of them worked. I also self-hosted the WCF service to eliminate the issues related to IIS trying to authenticate a user before the service gets a chance. Any comment on what the underlying issues may be would be much appreciated. Thanks. Update: Thanks to Mehmet's excellent suggestions. Here is the tracing configuration I had: <system.diagnostics> <sources> <source name="System.ServiceModel" switchValue="Information, ActivityTracing" propagateActivity="true"> <listeners> <add name="xml" /> </listeners> </source> <source name="System.IdentityModel" switchValue="Information, ActivityTracing" propagateActivity="true"> <listeners> <add name="xml" /> </listeners> </source> </sources> <sharedListeners> <add name="xml" type="System.Diagnostics.XmlWriterTraceListener" initializeData="c:\Traces.svclog" /> </sharedListeners> </system.diagnostics> But I did not see any message coming from my Silverlight client. As for https vs http, I used https as follows: string baseAddress = "https://localhost:6600/"; _webServiceHost = new WebServiceHost(typeof(DataServices), new Uri(baseAddress)); _webServiceHost.Open(); However, I did not configure any SSL certificate. Is this the problem?

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  • Android - Saving an object in onSaveInstanceState?

    - by Donal Rafferty
    I have created a small XML parsing application for Android that displays information in a listview and then allows a user to click on the list view and a dialog with further info will pop up. The problem is that when the screen orientation is changed when a dialog screen is open I get a null pointer error. The null pointer occurs on the following line: if(setting.getAddForPublicUserNames() == 1){ This line is part of my dialogPrepare method: @Override public void onPrepareDialog(int id, Dialog dialog) { switch(id) { case (SETTINGS_DIALOG) : afpunText = ""; if(setting.getAddForPublicUserNames() == 1){ afpunText = "Yes"; } else{ afpunText = "No"; } String Text = "Login Settings: " + "\n" + "Password: " + setting.getPassword() + "\n" + "Server: " + setting.getServerAddress() + "\n" + "Register: " + setting.getRegistrarAddress() + "\n" + "Realm: " + setting.getRealm() + "\n" + "Public UserNames: " + afpunText + "\n" + "Preference Settings: " + "\n" + "Request VDN: " + setting.getRequestVDN() + "\n" + "Handover Settings: " + "\n" + "Enable Handover: " + setting.getEnableHandover() + "\n" + "Hand Over Number: " + setting.getHandoverNum() + "\n"; AlertDialog settingsDialog = (AlertDialog)dialog; settingsDialog.setTitle("Auth ID: " + setting.getUserName()); tv = (TextView)settingsDialog.findViewById(R.id.detailsTextView); if (tv != null) tv.setText(Text); break; } } So the error is that my variable setting is null after the screen orientation changes. I have tried to use the onSaveInstance state methods to fix that as follows: @Override public void onSaveInstanceState(Bundle savedInstanceState) { for(int i = 0; i < settings.size(); i++){ savedInstanceState.putString("Username"+i, settings.get(i).getUserName()); savedInstanceState.putString("Password"+i, settings.get(i).getPassword()); savedInstanceState.putString("Server"+i, settings.get(i).getServerAddress()); savedInstanceState.putString("Registrar"+i, settings.get(i).getRegistrarAddress()); savedInstanceState.putString("Realm"+i, settings.get(i).getRealm()); savedInstanceState.putInt("PUserNames"+i, settings.get(i).getAddForPublicUserNames()); savedInstanceState.putString("RequestVDN"+i, settings.get(i).getRequestVDN()); savedInstanceState.putString("EnableHandOver"+i, settings.get(i).getEnableHandover()); savedInstanceState.putString("HandOverNum"+i, settings.get(i).getHandoverNum()); } super.onSaveInstanceState(savedInstanceState); } and @Override public void onRestoreInstanceState(Bundle savedInstanceState) { super.onRestoreInstanceState(savedInstanceState); //Check to see if this is required // Restore UI state from the savedInstanceState. // This bundle has also been passed to onCreate. for(int i = 0; i<settings.size(); i++){ settings.get(i).setUserName(savedInstanceState.getString("Username"+i)); settings.get(i).setPassword(savedInstanceState.getString("Password"+i)) ; settings.get(i).setServerAddress(savedInstanceState.getString("Server"+i)); settings.get(i).setRegistrarAddress(savedInstanceState.getString("Registrar"+i)); settings.get(i).setRealm(savedInstanceState.getString("Realm"+i)); settings.get(i).setAddForPublicUserNames(savedInstanceState.getInt("PUserNames"+i)); settings.get(i).setRequestVDN(savedInstanceState.getString("RequestVDN"+i)); settings.get(i).setEnableHandover(savedInstanceState.getString("EnableHandOver"+i)); settings.get(i).setHandoverNum(savedInstanceState.getString("HandOverNum"+i)); } } However the error still remains, I think I have to save the selected setting from what was selected from the ListView? But how do I save a setting object in onSavedInstance?

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  • nhibernate says 'mapping exception was unhandled' no persister for: MyNH.Domain.User

    - by mrblah
    Hi, I am using nHibernate and fluent. I created a User.cs: public class User { public virtual int Id { get; set; } public virtual string Username { get; set; } public virtual string Password { get; set; } public virtual string Email { get; set; } public virtual DateTime DateCreated { get; set; } public virtual DateTime DateModified { get; set; } } Then in my mappinds folder: public class UserMapping : ClassMap<User> { public UserMapping() { WithTable("ay_users"); Not.LazyLoad(); Id(x => x.Id).GeneratedBy.Identity(); Map(x => x.Username).Not.Nullable().WithLengthOf(256); Map(x => x.Password).Not.Nullable().WithLengthOf(256); Map(x => x.Email).Not.Nullable().WithLengthOf(100); Map(x => x.DateCreated).Not.Nullable(); Map(x => x.DateModified).Not.Nullable(); } } Using the repository pattern for the nhibernate blog: public class UserRepository : Repository<User> { } public class Repository<T> : IRepository<T> { public ISession Session { get { return SessionProvider.GetSession(); } } public T GetById(int id) { return Session.Get<T>(id); } public ICollection<T> FindAll() { return Session.CreateCriteria(typeof(T)).List<T>(); } public void Add(T product) { Session.Save(product); } public void Remove(T product) { Session.Delete(product); } } public interface IRepository<T> { T GetById(int id); ICollection<T> FindAll(); void Add(T entity); void Remove(T entity); } public class SessionProvider { private static Configuration configuration; private static ISessionFactory sessionFactory; public static Configuration Configuration { get { if (configuration == null) { configuration = new Configuration(); configuration.Configure(); configuration.AddAssembly(typeof(User).Assembly); } return configuration; } } public static ISessionFactory SessionFactory { get { if (sessionFactory == null) sessionFactory = Configuration.BuildSessionFactory(); return sessionFactory; } } private SessionProvider() { } public static ISession GetSession() { return SessionFactory.OpenSession(); } } My config: <?xml version="1.0" encoding="utf-8" ?> <hibernate-configuration xmlns="urn:nhibernate-configuration-2.2"> <session-factory> <property name="connection.provider">NHibernate.Connection.DriverConnectionProvider</property> <property name="dialect">NHibernate.Dialect.MsSql2005Dialect</property> <property name="connection.driver_class">NHibernate.Driver.SqlClientDriver</property> <property name="connection.connection_string">Server=.\SqlExpress;Initial Catalog=TestNH;User Id=dev;Password=123</property> <property name="show_sql">true</property> </session-factory> </hibernate-configuration> I created a console application to test the output: static void Main(string[] args) { Console.WriteLine("starting..."); UserRepository users = new UserRepository(); User user = users.GetById(1); Console.WriteLine("user is null: " + (null == user)); if(null != user) Console.WriteLine("User: " + user.Username); Console.WriteLine("ending..."); Console.ReadLine(); } Error: nhibernate says 'mapping exception was unhandled' no persister for: MyNH.Domain.User What could be the issue, I did do the mapping?

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  • Simple Observation in Django: How Can I Correctly Modify The `attrs` sent to __new__ of a Django Mod

    - by DGGenuine
    Hello, I'm a strong proponent of the observer pattern, and this is what I'd like to be able to do in my Django models.py: class AModel(Model): __metaclass__ = SomethingMagical @post_save(AnotherModel) @classmethod def observe_another_model_saved(klass, sender, instance, created, **kwargs): pass @pre_init('YetAnotherModel') @classmethod def observe_yet_another_model_initializing(klass, sender, *args, **kwargs): pass @post_delete('DifferentApp.SomeModel') @classmethod def observe_some_model_deleted(klass, sender, **kwargs): pass This would connect a signal with sender = the decorator's argument and receiver = the decorated method. Right now my signal connection code all exists in __init__.py which is okay, but a little unmaintainable. I want this code all in one place, the models.py file. Thanks to helpful feedback from the community I'm very close (I think.) (I'm using a metaclass solution instead of the class decorator solution in the previous question/answer because you can't set attributes on classmethods, which I need.) I am having a strange error I don't understand. At the end of my post are the contents of a models.py that you can pop into a fresh project/application to see the error. Set your database to sqlite and add the application to installed apps. This is the error: Validating models... Unhandled exception in thread started by Traceback (most recent call last): File "/Library/Python/2.6/site-packages//lib/python2.6/site-packages/django/core/management/commands/runserver.py", line 48, in inner_run File "/Library/Python/2.6/site-packages/django/core/management/base.py", line 253, in validate raise CommandError("One or more models did not validate:\n%s" % error_text) django.core.management.base.CommandError: One or more models did not validate: local.myothermodel: 'my_model' has a relation with model MyModel, which has either not been installed or is abstract. I've indicated a few different things you can comment in/out to fix the error. First, if you don't modify the attrs sent to the metaclass's __new__, then the error does not arise. (Note even if you copy the dictionary element by element into a new dictionary, it still fails; only using the exact attrs dictionary works.) Second, if you reference the first model by class rather than by string, the error also doesn't arise regardless of what you do in __new__. I appreciate your help. I'll be githubbing the solution if and when it works. Maybe other people would enjoy a simplified way to use Django signals to observe application happenings. #models.py from django.db import models from django.db.models.base import ModelBase from django.db.models import signals import pdb class UnconnectedMethodWrapper(object): sender = None method = None signal = None def __init__(self, signal, sender, method): self.signal = signal self.sender = sender self.method = method def post_save(sender): return _make_decorator(signals.post_save, sender) def _make_decorator(signal, sender): def decorator(view): return UnconnectedMethodWrapper(signal, sender, view) return decorator class ConnectableModel(ModelBase): """ A meta class for any class that will have static or class methods that need to be connected to signals. """ def __new__(cls, name, bases, attrs): unconnecteds = {} ## NO WORK newattrs = {} for name, attr in attrs.iteritems(): if isinstance(attr, UnconnectedMethodWrapper): unconnecteds[name] = attr newattrs[name] = attr.method #replace the UnconnectedMethodWrapper with the method it wrapped. else: newattrs[name] = attr ## NO WORK # newattrs = {} # for name, attr in attrs.iteritems(): # newattrs[name] = attr ## WORKS # newattrs = attrs new = super(ConnectableModel, cls).__new__(cls, name, bases, newattrs) for name, unconnected in unconnecteds.iteritems(): _connect_signal(unconnected.signal, unconnected.sender, getattr(new, name), new._meta.app_label) return new def _connect_signal(signal, sender, receiver, default_app_label): # full implementation also accepts basestring as sender and will look up model accordingly signal.connect(sender=sender, receiver=receiver) class MyModel(models.Model): __metaclass__ = ConnectableModel @post_save('In my application this string matters') @classmethod def observe_it(klass, sender, instance, created, **kwargs): pass @classmethod def normal_class_method(klass): pass class MyOtherModel(models.Model): ## WORKS # my_model = models.ForeignKey(MyModel) ## NO WORK my_model = models.ForeignKey('MyModel')

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  • ASP.net roles and Projects

    - by Zyphrax
    EDIT - Rewrote my original question to give a bit more information Background info At my work I'm working on a ASP.Net web application for our customers. In our implementation we use technologies like Forms authentication with MembershipProviders and RoleProviders. All went well until I ran into some difficulties with configuring the roles, because the roles aren't system-wide, but related to the customer accounts and projects. I can't name our exact setup/formula, because I think our company wouldn't approve that... What's a customer / project? Our company provides management information for our customers on a yearly (or other interval) basis. In our systems a customer/contract consists of: one Account: information about the Company per Account, one or more Products: the bundle of management information we'll provide per Product, one or more Measurements: a period of time, in which we gather and report the data Extranet site setup Eventually we want all customers to be able to access their management information with our online system. The extranet consists of two sites: Company site: provides an overview of Account information and the Products Measurement site: after selecting a Measurement, detailed information on that period of time The measurement site is the most interesting part of the extranet. We will create submodules for new overviews, reports, managing and maintaining resources that are important for the research. Our Visual Studio solution consists of a number of projects. One web application named Portal for the basis. The sites and modules are virtual directories within that application (makes it easier to share MasterPages among things). What kind of roles? The following users (read: roles) will be using the system: Admins: development users :) (not customer related, full access) Employees: employees of our company (not customer related, full access) Customer SuperUser: top level managers (full access to their account/measurement) Customer ContactPerson: primary contact (full access to their measurement(s)) Customer Manager: a department manager (limited access, specific data of a measurement) What about ASP.Net users? The system will have many ASP.Net users, let's focus on the customer users: Users are not shared between Accounts SuperUser X automatically has access to all (and new) measurements User Y could be Primary contact for Measurement 1, but have no role for Measurement 2 User Y could be Primary contact for Measurement 1, but have a Manager role for Measurement 2 The department managers are many individual users (per Measurement), if Manager Z had a login for Measurement 1, we would like to use that login again if he participates in Measurement 2. URL structure These are typical urls in our application: http://host/login - the login screen http://host/project - the account/product overview screen (measurement selection) http://host/project/1000 - measurement (id:1000) details http://host/project/1000/planning - planning overview (for primary contact/superuser) http://host/project/1000/reports - report downloads (manager department X can only access report X) We will also create a document url, where you can request a specific document by it's GUID. The system will have to check if the user has rights to the document. The document is related to a Measurement, the User or specific roles have specific rights to the document. What's the problem? (finally ;)) Roles aren't enough to determine what a user is allowed to see/access/download a specific item. It's not enough to say that a certain navigation item is accessible to Managers. When the user requests Measurement 1000, we have to check that the user not only has a Manager role, but a Manager role for Measurement 1000. Summarized: How can we limit users to their accounts/measurements? (remember superusers see all measurements, some managers only specific measurements) How can we apply roles at a product/measurement level? (user X could be primarycontact for measurement 1, but just a manager for measurement 2) How can we limit manager access to the reports screen and only to their department's reports? All with the magic of asp.net classes, perhaps with a custom roleprovider implementation. Similar Stackoverflow question/problem http://stackoverflow.com/questions/1367483/asp-net-how-to-manage-users-with-different-types-of-roles

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  • System.UnsupportedException using WCF on Windows Phone 7

    - by Igor Zevaka
    Has anyone been able to communicate using WCF on Windows Phone Series 7 emulator? I've been trying for the past two days and it's just happening for me. I can get a normal Silverlight control to work in both Silverlight 3 and Silverlight 4, but not the phone version. Here are two versions that I've tried: Version 1 - Using Async Pattern BasicHttpBinding basicHttpBinding = new BasicHttpBinding(); EndpointAddress endpointAddress = new EndpointAddress("http://localhost/wcf/Authentication.svc"); Wcf.IAuthentication auth1 = new ChannelFactory<Wcf.IAuthentication>(basicHttpBinding, endpointAddress).CreateChannel(endpointAddress); AsyncCallback callback = (result) => { Action<string> write = (str) => { this.Dispatcher.BeginInvoke(delegate { //Display something }); }; try { Wcf.IAuthentication auth = result.AsyncState as Wcf.IAuthentication; Wcf.AuthenticationResponse response = auth.EndLogin(result); write(response.Success.ToString()); } catch (Exception ex) { write(ex.Message); System.Diagnostics.Debug.WriteLine(ex.Message); } }; auth1.BeginLogin("user0", "test0", callback, auth1); This version breaks on this line: Wcf.IAuthentication auth1 = new ChannelFactory<Wcf.IAuthentication>(basicHttpBinding, endpointAddress).CreateChannel(endpointAddress); Throwing System.NotSupportedException. The exception is not very descriptive and the callstack is equally not very helpful: at System.ServiceModel.DiagnosticUtility.ExceptionUtility.BuildMessage(Exception x) at System.ServiceModel.DiagnosticUtility.ExceptionUtility.LogException(Exception x) at System.ServiceModel.DiagnosticUtility.ExceptionUtility.ThrowHelperError(Exception e) at System.ServiceModel.ChannelFactory`1.CreateChannel(EndpointAddress address) at WindowsPhoneApplication2.MainPage.DoLogin() .... Version 2 - Blocking WCF call Here is the version that doesn't use the async pattern. [System.ServiceModel.ServiceContract] public interface IAuthentication { [System.ServiceModel.OperationContract] AuthenticationResponse Login(string user, string password); } public class WcfClientBase<TChannel> : System.ServiceModel.ClientBase<TChannel> where TChannel : class { public WcfClientBase(string name, bool streaming) : base(GetBinding(streaming), GetEndpoint(name)) { ClientCredentials.UserName.UserName = WcfConfig.UserName; ClientCredentials.UserName.Password = WcfConfig.Password; } public WcfClientBase(string name) : this(name, false) {} private static System.ServiceModel.Channels.Binding GetBinding(bool streaming) { System.ServiceModel.BasicHttpBinding binding = new System.ServiceModel.BasicHttpBinding(); binding.MaxReceivedMessageSize = 1073741824; if(streaming) { //binding.TransferMode = System.ServiceModel.TransferMode.Streamed; } /*if(XXXURLXXX.StartsWith("https")) { binding.Security.Mode = BasicHttpSecurityMode.Transport; binding.Security.Transport.ClientCredentialType = HttpClientCredentialType.None; }*/ return binding; } private static System.ServiceModel.EndpointAddress GetEndpoint(string name) { return new System.ServiceModel.EndpointAddress(WcfConfig.Endpoint + name + ".svc"); } protected override TChannel CreateChannel() { throw new System.NotImplementedException(); } } auth.Login("test0", "password0"); This version crashes in System.ServiceModel.ClientBase<TChannel> constructor. The call stack is a bit different: at System.Reflection.MethodInfo.get_ReturnParameter() at System.ServiceModel.Description.ServiceReflector.HasNoDisposableParameters(MethodInfo methodInfo) at System.ServiceModel.Description.TypeLoader.CreateOperationDescription(ContractDescription contractDescription, MethodInfo methodInfo, MessageDirection direction, ContractReflectionInfo reflectionInfo, ContractDescription declaringContract) at System.ServiceModel.Description.TypeLoader.CreateOperationDescriptions(ContractDescription contractDescription, ContractReflectionInfo reflectionInfo, Type contractToGetMethodsFrom, ContractDescription declaringContract, MessageDirection direction) at System.ServiceModel.Description.TypeLoader.CreateContractDescription(ServiceContractAttribute contractAttr, Type contractType, Type serviceType, ContractReflectionInfo& reflectionInfo, Object serviceImplementation) at System.ServiceModel.Description.TypeLoader.LoadContractDescriptionHelper(Type contractType, Type serviceType, Object serviceImplementation) at System.ServiceModel.Description.TypeLoader.LoadContractDescription(Type contractType) at System.ServiceModel.ChannelFactory1.CreateDescription() at System.ServiceModel.ChannelFactory.InitializeEndpoint(Binding binding, EndpointAddress address) at System.ServiceModel.ChannelFactory1..ctor(Binding binding, EndpointAddress remoteAddress) at System.ServiceModel.ClientBase1..ctor(Binding binding, EndpointAddress remoteAddress) at Wcf.WcfClientBase1..ctor(String name, Boolean streaming) at Wcf.WcfClientBase`1..ctor(String name) at Wcf.AuthenticationClient..ctor() at WindowsPhoneApplication2.MainPage.DoLogin() ... Any ideas?

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  • getting Error while set up the connection pool in jboss

    - by Yashwant Chavan
    Hi as per following Connection pool configuration facing some issue. Place a copy of mysql-connector-java-[version]-bin.jar in $JBOSS_HOME/server/all/lib. Then, follow the example configuration file named mysql-ds.xml in the $JBOSS_HOME/docs/examples/jca directory that comes with a JBoss binary installation. To activate your DataSource, place an xml file that follows the format of mysql-ds.xml in the deploy subdirectory in either $JBOSS_HOME/server/all, $JBOSS_HOME/server/default, or $JBOSS_HOME/server/[yourconfig] as appropriate. I am getting following error resource-ref: jdbc/buinessCaliberDb has no valid JNDI binding. Check the jboss-web/resource-ref. This is my mysql-ds.xml <datasources> <local-tx-datasource> <jndi-name>jdbc/buinessCaliberDb</jndi-name> <connection-url>jdbc:mysql:///BUSINESS</connection-url> <driver-class>com.mysql.jdbc.Driver</driver-class> <user-name>root</user-name> <password>password</password> <exception-sorter-class-name>org.jboss.resource.adapter.jdbc.vendor.MySQLExceptionSorter</exception-sorter-class-name> <!-- should only be used on drivers after 3.22.1 with "ping" support <valid-connection-checker-class-name>org.jboss.resource.adapter.jdbc.vendor.MySQLValidConnectionChecker</valid-connection-checker-class-name> --> <!-- sql to call when connection is created <new-connection-sql>some arbitrary sql</new-connection-sql> --> <!-- sql to call on an existing pooled connection when it is obtained from pool - MySQLValidConnectionChecker is preferred for newer drivers <check-valid-connection-sql>some arbitrary sql</check-valid-connection-sql> --> <!-- corresponding type-mapping in the standardjbosscmp-jdbc.xml (optional) --> <metadata> <type-mapping>mySQL</type-mapping> </metadata> </local-tx-datasource> </datasources> and this my web.xml entry <resource-ref> <description>DB Connection</description> <res-ref-name>jdbc/buinessCaliberDb</res-ref-name> <res-type>javax.sql.DataSource</res-type> <res-auth>Container</res-auth> </resource-ref> and this jboss-web.xml entry <jboss-web> <resource-ref> <description>DB Connection</description> <res-ref-name>jdbc/buinessCaliberDb</res-ref-name> <res-type>javax.sql.DataSource</res-type> <res-auth>Container</res-auth> </resource-ref> </jboss-web> Please help

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  • FaceBook Login Problem

    - by toman
    Hi All, I am for an application which extracts information from facebook search, hence i require to login facebook. i have registered my application in facebook developers site and have got api key and secret key. in my code i am getting an exception when i am trying to login. Here is my code for login to facebook: import com.facebook.api.FacebookRestClient; import org.apache.commons.httpclient.HttpClient; import org.apache.commons.httpclient.HttpState; import org.apache.commons.httpclient.NameValuePair; import org.apache.commons.httpclient.methods.GetMethod; import org.apache.commons.httpclient.methods.PostMethod; import org.apache.commons.httpclient.params.HttpClientParams; public class FaceLogin { public FaceLogin(){ getUserID("username", "password"); } private static void getUserID(String email, String password) { String session = null; try { HttpClient http = new HttpClient(); http.setParams(new HttpClientParams()); //http.getHostConfiguration().setHost("http://www.facebook.com/"); http.setState(new HttpState()); String api_key = "****some key****"; String secret = "****some key****"; FacebookRestClient client = new FacebookRestClient(api_key, secret); client.setIsDesktop(true); String token = client.auth_createToken(); final String loginId = "http://www.facebook.com/login.php"; GetMethod get = new GetMethod(loginId + "?api_key=" + api_key + "&v=1.0&auth_token=" +token); System.out.println("Get="+get); http.executeMethod(get); PostMethod post = new PostMethod(loginId); post.addParameter(new NameValuePair("api_key", api_key)); post.addParameter(new NameValuePair("v", "1.0")); post.addParameter(new NameValuePair("auth_token", token)); post.addParameter(new NameValuePair("fbconnect","true")); post.addParameter(new NameValuePair("return_session","true")); post.addParameter(new NameValuePair("session_key_only","true")); post.addParameter(new NameValuePair("req_perms","read_stream,publish_stream")); post.addParameter(new NameValuePair("lsd","8HYdi")); post.addParameter(new NameValuePair("locale","en_US")); post.addParameter(new NameValuePair("persistent","1")); post.addParameter(new NameValuePair("email", email)); post.addParameter(new NameValuePair("pass", password)); System.out.println("Token ="+token); int postStatus = http.executeMethod(post); System.out.println("Response : " + postStatus); session = client.auth_getSession(token); // Here I am getting error System.out.println("Session string: " + session); long userid = client.users_getLoggedInUser(); System.out.println("User Id is : " + userid); } catch (Exception e) { e.printStackTrace(); } } public static void main(String k[]) { FaceLogin facebookLoginObj=new FaceLogin(); } } I am getting the following exception: org.apache.commons.httpclient.HttpMethodBase processResponseHeaders WARNING: Cookie rejected: "$Version=0; $Domain=deleted; $Path=/; $Domain=.facebook.com". Cookie name may not start with $ Response : 200 Jun 8, 2010 2:07:36 PM org.apache.commons.httpclient.HttpMethodBase processResponseHeaders WARNING: Cookie rejected: "$Version=0; $Path=deleted; $Path=/; $Domain=.facebook.com". Cookie name may not start with $ Facebook returns error code 100 com.facebook.api.FacebookException: Invalid parameter - v - 1.0 at com.facebook.api.FacebookRestClient.callMethod(FacebookRestClient.java:828) - auth_token - 004e90dc8818d5f0921d1065d24508d3 at com.facebook.api.FacebookRestClient.callMethod(FacebookRestClient.java:606) - method - facebook.auth.getSession - call_id - 1275986256796 - api_key - f7cb1e48c383ef599da9021fc4dec322 at com.facebook.api.FacebookRestClient.auth_getSession(FacebookRestClient.java:1891) at facebookcrawler.FacebookLogin.getUserID(FacebookLogin.java:81) at facebookcrawler.FacebookLogin.( - sig - 9344ec75b74a0a87bcae645046d45da8 FacebookLogin.java:24) at facebookcrawler.FaceLogin.main(FaceLogin.java:80) Here may be the problem is for creating session, i searched for all the solutions on net but could not helped me to get login. Please help me if you can suggest me some way to resolve this problem. i thanks to all your valuable suggestion.

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  • How to configure the roles in my tomcat application to work with JNDI(WIN AUTH)

    - by Itay Levin
    Hi, I'm trying to change the authentication mode of my application from JDBC-REALM to JNDI-REALM. I configured the following section inside the Server.xml <Realm className="org.apache.catalina.realm.JNDIRealm" debug="99" connectionURL="ldap://****:389/DC=onsetinc,DC=com??sAMccountName?sub?(objectClass=*)" connectionName="[email protected]" connectionPassword="password" userBase="CN=Users" referrals="follow" userSearch="(sAMAccountName={0})" userSubtree="true" roleBase="CN=Users" roleName="name" roleSubtree="true" roleSearch="(member={1})"/> I have also configured the web.xml under my appfolder to contain the following: <security-role> <role-name>Admin</role-name> </security-role> <security-role> <role-name>WaterlooUsers</role-name> </security-role> <security-constraint> <web-resource-collection> <web-resource-name>Tube</web-resource-name> <url-pattern>/ComposeMessage.jsp</url-pattern> <url-pattern>/PageStatus.jsp</url-pattern> <url-pattern>/UserStatus.jsp</url-pattern> <url-pattern>/SearchEC.jsp</url-pattern> <url-pattern>/SearchEC2.jsp</url-pattern> <url-pattern>/SearchMessageStatisticsEC.jsp</url-pattern> <url-pattern>/SearchMessageStatus.jsp</url-pattern> <url-pattern>/SearchMessageStatisticsPager.jsp</url-pattern> <url-pattern>/SearchPageStatus.jsp</url-pattern> </web-resource-collection> <auth-constraint> <role-name>WaterlooUsers</role-name> </auth-constraint> </security-constraint> In my Active directory i have created a new group called WaterlooUsers It's distinguish name is : distinguishedName: CN=WaterlooUsers,CN=Users,DC=onsetinc,DC=com It has a property member which contains the following user: member: CN=Itay Levin,CN=Users,DC=onsetinc,DC=com (which is my user) My record on the active directory looks like that: sAMAccountName: itayL distinguishedName: CN=Itay Levin,CN=Users,DC=onsetinc,DC=com memberOf: CN=WaterlooUsers,CN=Users,DC=onsetinc,DC=com and when i get the popup for user/password i enter the username "ItayL" in the authentication message box (and my password) I have 2 questions: How do i configure correctly the roles parameters correctly in the Realm section in the server.xml to enable me to both authenticate and authorize both this group of users WaterlooUsers and also assign them to the appropriate role so that they can see all the relevant pages in my website. - currently it seems that all the Users in my domain are authenticated to the site but get the http-403 Error and can't access any of the pages in the site. I also want to be able to create 2 different set of roles in my site - which can both have access to the same pages - but will see different things on the page. (for instance adding some administrative ability to the admin) Hope it was clear enough and not too long. Thanks in advance, Itay

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  • Send smtp mail in php with HTML page attach as a text

    - by Nirmal
    Hello All.... I have a requirement of sending mail using smtp server in php. Now I am able to send the mail using smtp for a plain text. but I have a requirement where I need to attach an HTML page, which includes set of images. Now for that I am trying the following code : <?php require_once "Mail.php"; $to = '[email protected]'; $from = '[email protected]'; $subject = $_POST['subject']; $body = $_POST['message']; $fileatt = $_FILES['fileatt']['tmp_name']; $fileatt_type = $_FILES['fileatt']['type']; $fileatt_name = $_FILES['fileatt']['name']; $headers = array ('From' => $from, 'To' => $to, 'Subject' => $subject); if (is_uploaded_file($fileatt)) { echo("<p>Inside 1</p>"); $file = fopen($fileatt,'rb'); $data = fread($file,filesize($fileatt)); fclose($file); // Generate a boundary string $semi_rand = md5(time()); $mime_boundary = "==Multipart_Boundary_x{$semi_rand}x"; array_push(&$headers, 'MIME-Version: 1.0'); array_push(&$headers, 'Content-Type: multipart/mixed;'); array_push(&$headers, " boundary=\"{$mime_boundary}\""); echo("<p>Inside 2</p>"); $body = "This is a multi-part message in MIME format.\n\n" . "--{$mime_boundary}\n" . "Content-Type: text/plain; charset=\"iso-8859-1\"\n" . "Content-Transfer-Encoding: 7bit\n\n" . $body . "\n\n"; echo("<p>Inside 3</p>"); $data = chunk_split(base64_encode($data)); echo("<p>Inside 4</p>"); $body .= "--{$mime_boundary}\n" . "Content-Type: {$fileatt_type};\n" . " name=\"{$fileatt_name}\"\n" . "Content-Transfer-Encoding: base64\n\n" . $data . "\n\n" . "--{$mime_boundary}--\n"; echo("<p>Inside 5</p>"); } $host = "[email protected]"; $username = "[email protected]"; $password = "user"; $smtp = Mail::factory('smtp', array ('host' => $host, 'auth' => true, 'username' => $username, 'password' => $password)); $mail = $smtp->send($to, $headers, $body); if (PEAR::isError($mail)) { echo("<p>" . $mail->getMessage() . "</p>"); } else { echo("<p>Message successfully sent!</p>"); } ?> Now this code works fine for me, and it's sending the mail to the target email address. But when I open this email in the inbox, it's showing me the following text in the mailbox: This is a multi-part message in MIME format. --==Multipart_Boundary_x368d72fe1ff44518e90537abdb4bf029x Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit test 1011 --==Multipart_Boundary_x368d72fe1ff44518e90537abdb4bf029x Content-Type: text/html; name="mailing.html" Content-Transfer-Encoding: base64 PCFET0NUWVBFIGh0bWwgUFVCTElDICItLy9XM0MvL0RURCBYSFRNTCAxLjAgVHJhbnNpdGlvbmFs Ly9FTiIgImh0dHA6Ly93d3cudzMub3JnL1RSL3hodG1sMS9EVEQveGh0bWwxLXRyYW5zaXRpb25h ................ So, it's clearly showing me the encoded data. So, what should modify to send the proper html page that should be visible in targeted email's inbox? Thanks in advance...

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  • Fatal error on a non-object

    - by Chris Leah
    Hey, so I have created a function to check the DB for unique entries, but when I call the function it doesn't seem to work and gives me a fatal error any ideas ? Thanks :) //Check for unique entries function checkUnique($table, $field, $compared) { $query = $mysqli->query('SELECT '.$mysqli->real_escape_string($field).' FROM '.$mysqli->real_escape_string($table).' WHERE "'.$mysqli->real_escape_string($field).'" = "'.$mysqli->real_escape_string($compared).'"'); if(!$query){ return TRUE; } else { return FALSE; } } The page calling it..... //Start session session_start(); //Check if the session is already set, if so re-direct to the game if(isset($_SESSION['id'], $_SESSION['logged_in'])){ Header('Location: ../main/index.php'); }; //Require database connection require_once('../global/includes/db.php'); require_once('../global/functions/functions.php'); //Check if the form has been submitted if (isset($_POST['signup'])){ //Validate input if (!empty($_POST['username']) && !empty($_POST['password']) && $_POST['password']==$_POST['password_confirm'] && !empty($_POST['email']) && validateEmail($_POST['email']) == TRUE && checkUnique('users', 'email', $_POST['email']) == TRUE && checkUnique('users', 'username', $_POST['username']) == TRUE) { //Insert user to the database $insert_user = $mysqli->query('INSERT INTO (`username, `password`, `email`, `verification_key`) VALUES ("'.$mysqli->real_escape_string($_POST['username']).'", "'.$mysqli-real_escape_string(md5($_POST['password'])).'", "'.$mysqli->real_escape_string($_POST['email']).'", "'.randomString('alnum', 32). '"') or die($mysqli->error()); //Get user information $getUser = $mysqli->query('SELECT id, username, email, verification_key FROM users WHERE username = "'.$mysqli->real_escape_string($_POST['username']).'"' or die($mysqli->error())); //Check if the $getUser returns true if ($getUser->num_rows == 1) { //Fetch associated fields to this user $row = $getUser->fetch_assoc(); //Set mail() variables $headers = 'From: [email protected]'."\r\n". 'Reply-To: [email protected]'."\r\n". 'X-Mailer: PHP/'.phpversion(); $subject = 'Activate your account (Music Battles.net)'; //Set verification email message $message = 'Dear '.$row['username'].', I would like to welcome you to Music Battles. Although in order to enjoy the gmae you must first activate your account. \n\n Click the following link: http://www.musicbattles.net/home/confirm.php?id='.$row['id'].'key='.$row['verification_key'].'\n Thanks for signing up, enjoy the game! \n Music Battles Team'; //Attempts to send the email if (mail($row['email'], $subject, $message, $headers)) { $msg = '<p class="success">Accound has been created, please go activate it from your email.</p>'; } else { $error = '<p class="error">The account was created but your email was not sent.</p>'; } } else { $error = '<p class="error">Your account was not created.</p>'; } } else { $error = '<p class="error">One or more fields contain non or invalid data.</p>'; } } Erorr.... Fatal error: Call to a member function query() on a non-object in /home/mbattles/public_html/global/functions/functions.php on line 5

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  • AES Cipher not picking up IV

    - by timothyjc
    I am trying to use an IV with AES so that the encrypted text is unpredictable. However, the encrypted hex string is always the same. I have actually tried a few methods of attempting to add some randomness by passing some additional parameters to the cipher init call: 1) Manual IV generation byte[] iv = generateIv(); IvParameterSpec ivspec = new IvParameterSpec(iv); 2) Asking cipher to generate IV AlgorithmParameters params = cipher.getParameters(); params.getParameterSpec(IvParameterSpec.class); 3) Using a PBEParameterSpec byte[] encryptionSalt = generateSalt(); PBEParameterSpec pbeParamSpec = new PBEParameterSpec(encryptionSalt, 1000); All of these seem to have no influence on the encrypted text.... help!!! My code: package com.citc.testencryption; import java.security.NoSuchAlgorithmException; import java.security.SecureRandom; import javax.crypto.Cipher; import javax.crypto.SecretKey; import javax.crypto.SecretKeyFactory; import javax.crypto.spec.IvParameterSpec; import javax.crypto.spec.PBEKeySpec; import android.app.Activity; import android.os.Bundle; import android.util.Log; public class Main extends Activity { public static final int SALT_LENGTH = 20; public static final int PBE_ITERATION_COUNT = 1000; private static final String RANDOM_ALGORITHM = "SHA1PRNG"; private static final String PBE_ALGORITHM = "PBEWithSHA256And256BitAES-CBC-BC"; private static final String CIPHER_ALGORITHM = "PBEWithSHA256And256BitAES-CBC-BC"; private static final String TAG = Main.class.getSimpleName(); @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.main); try { String password = "password"; String plainText = "plaintext message to be encrypted"; // byte[] salt = generateSalt(); byte[] salt = "dfghjklpoiuytgftgyhj".getBytes(); Log.i(TAG, "Salt: " + salt.length + " " + HexEncoder.toHex(salt)); PBEKeySpec pbeKeySpec = new PBEKeySpec(password.toCharArray(), salt, PBE_ITERATION_COUNT); SecretKeyFactory keyFac = SecretKeyFactory.getInstance(PBE_ALGORITHM); SecretKey secretKey = keyFac.generateSecret(pbeKeySpec); byte[] key = secretKey.getEncoded(); Log.i(TAG, "Key: " + HexEncoder.toHex(key)); // PBEParameterSpec pbeParamSpec = new PBEParameterSpec(salt, ITERATION_COUNT); Cipher encryptionCipher = Cipher.getInstance(CIPHER_ALGORITHM); // byte[] encryptionSalt = generateSalt(); // Log.i(TAG, "Encrypted Salt: " + encryptionSalt.length + " " + HexEncoder.toHex(encryptionSalt)); // PBEParameterSpec pbeParamSpec = new PBEParameterSpec(encryptionSalt, 1000); // byte[] iv = params.getParameterSpec(IvParameterSpec.class).getIV(); // Log.i(TAG, encryptionCipher.getParameters() + " "); byte[] iv = generateIv(); IvParameterSpec ivspec = new IvParameterSpec(iv); encryptionCipher.init(Cipher.ENCRYPT_MODE, secretKey, ivspec); byte[] encryptedText = encryptionCipher.doFinal(plainText.getBytes()); Log.i(TAG, "Encrypted: " + HexEncoder.toHex(encryptedText)); // <== Why is this always the same :( Cipher decryptionCipher = Cipher.getInstance(CIPHER_ALGORITHM); decryptionCipher.init(Cipher.DECRYPT_MODE, secretKey, ivspec); byte[] decryptedText = decryptionCipher.doFinal(encryptedText); Log.i(TAG, "Decrypted: " + new String(decryptedText)); } catch (Exception e) { e.printStackTrace(); } } private byte[] generateSalt() throws NoSuchAlgorithmException { SecureRandom random = SecureRandom.getInstance(RANDOM_ALGORITHM); byte[] salt = new byte[SALT_LENGTH]; random.nextBytes(salt); return salt; } private byte[] generateIv() throws NoSuchAlgorithmException { SecureRandom random = SecureRandom.getInstance(RANDOM_ALGORITHM); byte[] iv = new byte[16]; random.nextBytes(iv); return iv; } }

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  • Eager/Lazy loaded member always empty with JPA one-to-many relationship

    - by Kaleb Pederson
    I have two entities, a User and Role with a one-to-many relationship from user to role. Here's what the tables look like: mysql> select * from User; +----+-------+----------+ | id | name | password | +----+-------+----------+ | 1 | admin | admin | +----+-------+----------+ 1 row in set (0.00 sec) mysql> select * from Role; +----+----------------------+---------------+----------------+ | id | description | name | summary | +----+----------------------+---------------+----------------+ | 1 | administrator's role | administrator | Administration | | 2 | editor's role | editor | Editing | +----+----------------------+---------------+----------------+ 2 rows in set (0.00 sec) And here's the join table that was created: mysql> select * from User_Role; +---------+----------+ | User_id | roles_id | +---------+----------+ | 1 | 1 | | 1 | 2 | +---------+----------+ 2 rows in set (0.00 sec) And here's the subset of orm.xml that defines the tables and relationships: <entity class="User" name="User"> <table name="User" /> <attributes> <id name="id"> <generated-value strategy="AUTO" /> </id> <basic name="name"> <column name="name" length="100" unique="true" nullable="false"/> </basic> <basic name="password"> <column length="255" nullable="false" /> </basic> <one-to-many name="roles" fetch="EAGER" target-entity="Role" /> </attributes> </entity> <entity class="Role" name="Role"> <table name="Role" /> <attributes> <id name="id"> <generated-value strategy="AUTO"/> </id> <basic name="name"> <column name="name" length="40" unique="true" nullable="false"/> </basic> <basic name="summary"> <column name="summary" length="100" nullable="false"/> </basic> <basic name="description"> <column name="description" length="255"/> </basic> </attributes> </entity> Yet, despite that, when I retrieve the admin user, I get back an empty collection. I'm using Hibernate as my JPA provider and it shows the following debug SQL: select user0_.id as id8_, user0_.name as name8_, user0_.password as password8_ from User user0_ where user0_.name=? limit ? When the one-to-many mapping is lazy loaded, that's the only query that's made. This correctly retrieves the one admin user. I changed the relationship to use eager loading and then the following query is made in addition to the above: select roles0_.User_id as User1_1_, roles0_.roles_id as roles2_1_, role1_.id as id9_0_, role1_.description as descript2_9_0_, role1_.name as name9_0_, role1_.summary as summary9_0_ from User_Role roles0_ left outer join Role role1_ on roles0_.roles_id=role1_.id where roles0_.User_id=? Which results in the following results: +----------+-----------+--------+----------------------+---------------+----------------+ | User1_1_ | roles2_1_ | id9_0_ | descript2_9_0_ | name9_0_ | summary9_0_ | +----------+-----------+--------+----------------------+---------------+----------------+ | 1 | 1 | 1 | administrator's role | administrator | Administration | | 1 | 2 | 2 | editor's role | editor | Editing | +----------+-----------+--------+----------------------+---------------+----------------+ 2 rows in set (0.00 sec) Hibernate obviously knows about the roles, yet getRoles() still returns an empty collection. Hibernate also recognized the relationship sufficiently to put the data in the first place. What problems can cause these symptoms?

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  • Input string was not in the correct format using int.Parse

    - by JDWebs
    I have recently been making a login 'representation' which is not secure. So before answering, please note I am aware of security risks etc., and this will not be on a live site. Also note I am a beginner :P. For my login representation, I am using LINQ to compare values of a DDL to select a username and a Textbox to enter a password, when a login button is clicked. However, an error is thrown 'Input string was not in the correct format', when using int.Parse. Front End: <%@ Page Language="C#" AutoEventWireup="true" CodeFile="Login_Test.aspx.cs" Inherits="Login_Login_Test" %> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head runat="server"> <title>Login Test</title> </head> <body> <form id="LoginTest" runat="server"> <div> <asp:DropDownList ID="DDL_Username" runat="server" Height="20px" DataTextField="txt"> </asp:DropDownList> <br /> <asp:TextBox ID="TB_Password" runat="server" TextMode="Password"></asp:TextBox> <br /> <asp:Button ID="B_Login" runat="server" onclick="B_Login_Click" Text="Login" /> <br /> <asp:Literal ID="LI_Result" runat="server"></asp:Literal> </div> </form> </body> </html> Back End: using System; using System.Collections.Generic; using System.Linq; using System.Web; using System.Web.UI; using System.Web.UI.WebControls; public partial class Login_Login_Test : System.Web.UI.Page { protected void Page_Load(object sender, EventArgs e) { if (!Page.IsPostBack) { Binder(); } } private void Binder() { using (DataClassesDataContext db = new DataClassesDataContext()) { DDL_Username.DataSource = from x in db.DT_Honeys select new { id = x.UsernameID, txt = x.Username }; DDL_Username.DataValueField = "id"; DDL_Username.DataTextField = "txt"; DDL_Username.DataBind(); } } protected void B_Login_Click(object sender, EventArgs e) { if (TB_Password.Text != "") { using (DataClassesDataContext db = new DataClassesDataContext()) { DT_Honey blah = new DT_Honey(); blah = db.DT_Honeys.SingleOrDefault(x => x.UsernameID == int.Parse(DDL_Username.SelectedValue.ToString())); if (blah == null) { LI_Result.Text = "Something went wrong :/"; } if (blah.Password == TB_Password.Text) { LI_Result.Text = "Credentials recognised :-)"; } else { LI_Result.Text = "Error with credentials :-("; } } } } } I am aware this problem is very common, but none of the help I have found online is useful/relevant. Any help/suggestions appreciated; thank you for your time :-).

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  • python Requests login to website returns 403

    - by Jeff
    I'm trying to use requests to login to a website but as you can guess I'm having a problem here's the the code that I'm using import requests EMAIL = '***' PASSWORD = '***' URL = 'https://portal.bitcasa.com/login' client = requests.session(config={'verbose': sys.stderr}) login_data = {'username': EMAIL, 'password': PASSWORD,} r = client.post(URL, data=login_data, headers={"Referer": "foo"}) print r and if I print out r.text I get <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html lang="en"> <head><script type="text/javascript">var NREUMQ=NREUMQ||[];NREUMQ.push(["mark","firstbyte",new Date().getTime()])</script> <meta http-equiv="content-type" content="text/html; charset=utf-8"> <meta name="robots" content="NONE,NOARCHIVE"> <title>403 Forbidden</title> <style type="text/css"> html * { padding:0; margin:0; } body * { padding:10px 20px; } body * * { padding:0; } body { font:small sans-serif; background:#eee; } body>div { border-bottom:1px solid #ddd; } h1 { font-weight:normal; margin-bottom:.4em; } h1 span { font-size:60%; color:#666; font-weight:normal; } #info { background:#f6f6f6; } #info ul { margin: 0.5em 4em; } #info p, #summary p { padding-top:10px; } #summary { background: #ffc; } #explanation { background:#eee; border-bottom: 0px none; } </style> </head> <body> <div id="summary"> <h1>Forbidden <span>(403)</span></h1> <p>CSRF verification failed. Request aborted.</p> </div> <div id="explanation"> <p><small>More information is available with DEBUG=True.</small></p> </div> <script type="text/javascript">if(!NREUMQ.f){NREUMQ.f=function(){NREUMQ.push(["load",new Date().getTime()]);var e=document.createElement("script");e.type="text/javascript";e.src=(("http:"===document.location.protocol)?"http:":"https:")+"//"+"d1ros97qkrwjf5.cloudfront.net/42/eum/rum.js";document.body.appendChild(e);if(NREUMQ.a)NREUMQ.a();};NREUMQ.a=window.onload;window.onload=NREUMQ.f;};NREUMQ.push(["nrfj","beacon-1.newrelic.com","0e859e0620",778660,"ZAZRbUcHWBAHURFYX11MdUxbBUIKCVxKVVpSDVRWGwtfBwJeAEZRQQYdWkYUUFklQRdXZloGRHRcAlIPA0UEQ1UdE0FWVgNFEDlEDFRH",0,7,new Date().getTime(),"","","","",""])</script></body> </html> They're using a combination of django and pyramid. I've been playing around with this for about two days now but, obviously, have gotten nowhere. Thanks for your help.

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  • asp .net MVC 2.0 xval Validation element

    - by ANDyW
    I got one problem with showing error message to element. Is there any option to turn on messages on place where is Html.ValidationMessageFor(model = model.ConfirmPassword). Becsoue for me it isn’t show up. I would like to have summary and near field information too not only red border. Any one know how to do it? using (Ajax.BeginForm("CreateValidForm", "Test", new AjaxOptions { HttpMethod = "Post" })) {%> <div id="validationSummary1"> <%= Html.ValidationSummary(true)%> </div> <fieldset> <legend>Fields</legend> <div class="editor-label"> <%= Html.LabelFor(model => model.Name)%> </div> <div class="editor-field"> <%= Html.TextBoxFor(model => model.Name)%> <%= Html.ValidationMessageFor(model => model.Name)%> </div> <div class="editor-label"> <%= Html.LabelFor(model => model.Email)%> </div> <div class="editor-field"> <%= Html.TextBoxFor(model => model.Email)%> <%= Html.ValidationMessageFor(model => model.Email)%> </div> <div class="editor-label"> <%= Html.LabelFor(model => model.Password)%> </div> <div class="editor-field"> <%= Html.TextBoxFor(model => model.Password)%> <%= Html.ValidationMessageFor(model => model.Password)%> </div> <div class="editor-label"> <%= Html.LabelFor(model => model.ConfirmPassword)%> </div> <div class="editor-field"> <%= Html.TextBoxFor(model => model.ConfirmPassword)%> <%= Html.ValidationMessageFor(model => model.ConfirmPassword)%> </div> <p> <input type="submit" value="Create" /> </p> </fieldset> <% } %> <%= Html.ClientSideValidation<ValidModel>() .UseValidationSummary("validationSummary1", "Please fix the following problems:") %> Here is link for sample project http://www.sendspace.com/file/m9gl54 .

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  • java.sql.SQLException: No suitable driver found for jdbc:db2:

    - by Celia
    Im using hibernate to connect to my DB2 database. I got java.sql.SQLException: No suitable driver found for jdbc:db2://ldild4268.mycompany.com:55000/myDB. I have db2jcc.jar, db2jcc_javax.jar, db2jcc_license_cu.jar, db2policy.jar, db2ggjava.jar and db2umplugin.jar added into my Java Build Path. I am able to connect to my database through SQuirrel. database.properties: jdbc.driverClassName=com.ibm.db2.jcc.DB2Driver jdbc.url=jdbc:db2://ldild4268.mycompany.com:55000/myDB jdbc.username=uname jdbc.password=pwd datasource.xml: <bean class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"> <property name="location"> <value>/WEB-INF/database.properties</value> </property> </bean> <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"> <property name="driverClassName" value="${jdbc.driverClassName}" /> <property name="url" value="${jdbc.url}" /> <property name="username" value="${jdbc.username}" /> <property name="password" value="${jdbc.password}" /> </bean> hibernate.xml: <bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean"> <property name="dataSource"> <ref bean="dataSource" /> </property> <property name="hibernateProperties"> <props> <prop key="hibernate.dialect">org.hibernate.dialect.DB2Dialect</prop> <prop key="hibernate.show_sql">true</prop> </props> </property> <property name="mappingResources"> <list> <value>/myModel.hbm.xml</value> </list> </property> </bean> myModel.hbm.xml: <hibernate-mapping> <class name="com.myCompany.model.myModel" table="table1" catalog=""> <composite-id> <key-property name="key1" column="key1" length="10"/> <key-property name="key2" column="key2" length="19"/> </composite-id> <property name="name" type="string"> <column name="Name" length="50"/> </property> </class> </hibernate-mapping> myModelDaoImpl: @Repository("myModelDao") public class myModelDaoImpl extends PortfolioHibernateDaoSupport implements myModelDao{ private SessionFactory sessionFactory; public List<Date> getKey1() { return this.sessionFactory.getCurrentSession() .createQuery("select pn.key1 from com.myCompany.model.myModel pn") .list(); } public String getPs() { String query = "select pn.name from com.myCompany.model.myModel pn where pn.key1='2011-09-30' and pn.key2=1049764"; List list = getHibernateTemplate().find(query); } } also, the method getKey1 throws nullPointer exception. How can I use createquery instead of hibernateTemplate? Thanks in advance!

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  • Why does sending post data with WebRequest take so long?

    - by Paramiliar
    I am currently creating a C# application to tie into a php / MySQL online system. The application needs to send post data to scripts and get the response. When I send the following data username=test&password=test I get the following responses... Starting request at 22/04/2010 12:15:42 Finished creating request : took 00:00:00.0570057 Transmitting data at 22/04/2010 12:15:42 Transmitted the data : took 00:00:06.9316931 <<-- Getting the response at 22/04/2010 12:15:49 Getting response 00:00:00.0360036 Finished response 00:00:00.0360036 Entire call took 00:00:07.0247024 As you can see it is taking 6 seconds to actually send the data to the script, I have done further testing bye sending data from telnet and by sending post data from a local file to the url and they dont even take a second so this is not a problem with the hosted script on the site. Why is it taking 6 seconds to transmit the data when it is two simple strings? I use a custom class to send the data class httppostdata { WebRequest request; WebResponse response; public string senddata(string url, string postdata) { var start = DateTime.Now; Console.WriteLine("Starting request at " + start.ToString()); // create the request to the url passed in the paramaters request = (WebRequest)WebRequest.Create(url); // set the method to post request.Method = "POST"; // set the content type and the content length request.ContentType = "application/x-www-form-urlencoded"; request.ContentLength = postdata.Length; // convert the post data into a byte array byte[] byteData = Encoding.UTF8.GetBytes(postdata); var end1 = DateTime.Now; Console.WriteLine("Finished creating request : took " + (end1 - start)); var start2 = DateTime.Now; Console.WriteLine("Transmitting data at " + start2.ToString()); // get the request stream and write the data to it Stream dataStream = request.GetRequestStream(); dataStream.Write(byteData, 0, byteData.Length); dataStream.Close(); var end2 = DateTime.Now; Console.WriteLine("Transmitted the data : took " + (end2 - start2)); // get the response var start3 = DateTime.Now; Console.WriteLine("Getting the response at " + start3.ToString()); response = request.GetResponse(); //Console.WriteLine(((WebResponse)response).StatusDescription); dataStream = response.GetResponseStream(); StreamReader reader = new StreamReader(dataStream); var end3 = DateTime.Now; Console.WriteLine("Getting response " + (end3 - start3)); // read the response string serverresponse = reader.ReadToEnd(); var end3a = DateTime.Now; Console.WriteLine("Finished response " + (end3a - start3)); Console.WriteLine("Entire call took " + (end3a - start)); //Console.WriteLine(serverresponse); reader.Close(); dataStream.Close(); response.Close(); return serverresponse; } } And to call it I use private void btnLogin_Click(object sender, EventArgs e) { // string postdata; if (txtUsername.Text.Length < 3 || txtPassword.Text.Length < 3) { MessageBox.Show("Missing your username or password."); } else { string postdata = "username=" + txtUsername.Text + "&password=" + txtPassword.Text; httppostdata myPost = new httppostdata(); string response = myPost.senddata("http://www.domainname.com/scriptname.php", postdata); MessageBox.Show(response); } }

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  • web application with secured sections, sessions and related trouble

    - by spirytus
    I would like to create web application with admin/checkout sections being secured. Assuming I have SSL set up for subdomain.mydomain.com I would like to make sure that all that top-secret stuff ;) like checkout pages and admin section is transferred securely. Would it be ok to structure my application as below? subdomain.mydomain.com adminSectionFolder adminPage1.php adminPage2.php checkoutPagesFolder checkoutPage1.php checkoutPage2.php checkoutPage3.php homepage.php loginPage.php someOtherPage.php someNonSecureFolder nonSecurePage1.php nonSecurePage2.php nonSecurePage3.php imagesFolder image1.jpg image2.jpg image3.jpg Users would access my web application via http as there is no need for SSL for homepage and similar. Checkout/admin pages would have to be accessed via https though (that I would ensure via .htaccess redirects). I would also like to have login form on every page of the site, including non-secure pages. Now my questions are: if I have form on non-secure page e.g http://subdomain.mydomain.com/homepage.php and that form sends data to http://subdomain.mydomain.com/loginPage.php, is data being send encrypted as if it were sent from https://subdomain.mydomain.com/homepage.php? I do realize users will not see padlock, but browser still should encrypt it, is it right? If on secure page loginPage.php (or any other accessed via https for that instance) I created session, session ID would be assigned, and in case of my web app. something like username of the logged in user. Would I be able to access these session variable from http://subdomain.mydomain.com/homepage.php to for example display greeting message? If session ID is stored in cookies then it would be trouble I assume, but could someone clarify how it should be done? It seems important to have username and password send over SSL. Related to above question I think.. would it actually make any sense to have login secured via SSL so usenrame/password would be transferred securely, and then session ID being transferred with no SSL? I mean wouldnt it be the same really if someone caught username and password being transferred, or caught session ID? Please let me know if I make sense here cause it feels like I'm missing something important. EDIT: I came up with idea but again please let me know if that would work. Having above, so assuming that sharing session between http and https is as secure as login in user via plain http (not https), I guess on all non secure pages, like homepage etc. I could check if user is already logged in, and if so from php redirect to https version of same page. So user fills in login form from homepage.php, over ssl details are send to backend so probably https://.../homepage.php. Trying to access http://.../someOtherPage.php script would always check if session is created and if so redirect user to https version of this page so https://.../someOtherPage.php. Would that work? 4.To avoid browser popping message "this page contains non secure items..." my links to css, images and all assets, e.g. in case of http://subdomain.mydomain.com/checkoutPage1.php should be absolute so "/images/image1.jpg" or relative so "../images/image1.jpg"? I guess one of those would have to work :) wow that's long post, thanks for your patience if you got that far and any answers :) oh yeh and I use php/apache on shared hosting

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  • Facebook iFrame APP not working in IE, works on every other browser

    - by Sean Ashmore
    So im getting a blank page when loading this page within an iFrame on Internet explorer, every other browser works fine.. I have also tried using p3p headers as other people have suggested, but to no avail. <?php require ("connect.php"); require ("config.php"); require ("fb_config.php"); ?> <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <title>Login handler</title> <meta http-equiv="content-type" content="text/html; charset=iso-8859-1"> <link rel="stylesheet" href="css/login.css" type="text/css"> </head> <body> <?//=$user?> <?php if($user == 0) { echo "You are not logged into facebook. Nice try."; }else{ $query = "SELECT id,fb_id,login_ip,login_count,activated,sitestate FROM login WHERE fb_id='".mysql_real_escape_string($user)."'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result); if (mysql_num_rows($result) == 0) { $sql = "INSERT INTO login SET id = '', fb_id ='" .mysql_real_escape_string($user). "', name = '" .rand(10000000000000000,99999999999999999999). "', signup =NOW() , password = '" .mysql_real_escape_string($pass). "', state = '0', mail = '" .mysql_real_escape_string($_POST['mail']). "',location='".mysql_real_escape_string($randomlocation)."',location_start='".mysql_real_escape_string($randomlocation)."', signup_ip='".mysql_real_escape_string($_SERVER['REMOTE_ADDR'])."',ref='".mysql_real_escape_string($_POST['ref'])."', activation_id = '" .mysql_real_escape_string($activation_link). "',activated='2', killprotection = '$twodayprot',gender='" .mysql_real_escape_string($_POST["gender"]). "'"; $res = mysql_query($sql); } //if($row['fb_id'] != $user){ //echo "Your facebook ID: $user is NOT in the MW DB."; //exit(); //}else{ if(empty($row['login_ip'])){ $row['login_ip'] = $_SERVER['REMOTE_ADDR']; }else{ $ip_information = explode("-", $row['login_ip']); if (in_array($_SERVER['REMOTE_ADDR'], $ip_information)) { $row['login_ip'] = $row['login_ip']; }else{ $row['login_ip'] = $row['login_ip']."-".$_SERVER['REMOTE_ADDR']; } } $update_login = mysql_query("UPDATE login SET login_count=login_count+'1' WHERE name='".mysql_real_escape_string($_POST['username'])."'") or die(mysql_error()); $_SESSION['user_id'] = $row['id']; $result = mysql_query("UPDATE login SET userip='".mysql_real_escape_string($_SERVER['REMOTE_ADDR'])."',login_ip='".mysql_real_escape_string($row['login_ip'])."',login_count='0' WHERE id='".mysql_real_escape_string($_SESSION['user_id'])."'") or die(mysql_error()); if ($row['sitestate'] == 0){ header("location: home.php"); } elseif ($row['sitestate'] == 2) { header("location: killed.php?id={$row['id']}&encrypted={$row['password']}"); } else { header("location: banned.php?id={$row['id']}&encrypted={$row['password']}"); } }// id check. ?> </body> </html>

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  • Mbean registered but not found in mbean Server

    - by Prisco
    I have a problem about the mbeans. I have created a simple mbean and I have registered it on the default mBeanServer that is run (Via eclipse or java -jar mbean.jar) and in the same process if I try to fouund the mbean registered with a simple query: for (ObjectInstance instance : mbs.queryMBeans(ObjectNameMbean, null)) { System.out.println(instance.toString()); } the query retuerns my mbean, but if I start another process and try to search this mbean registered the mbeas is not found! why? The approch is : (Process that is running) public static void main(String[] args) throws Exception { MBeanServer mbeanServer =ManagementFactory.getPlatformMBeanServer(); ObjectName objectName = new ObjectName(ObjectNameMbean); Simple simple = new Simple (1, 0); mbeanServer.registerMBean(simple, objectName); while (true) { wait (Is this necessary?) } } So this is the first process that is running (that has the only pourpose to registry the mbean, because there is another process that want to read these informations. So I start another process to search this mbean but nothing. I 'm not using jboss but the local Java virtual Machine but my scope is to deploy this simple application in one ejb (autostart) and another ejb will read all informations. All suggestions are really apprecciated. This example should be more useful : Object Hello: public class Hello implements HelloMBean { public void sayHello() { System.out.println("hello, world"); } public int add(int x, int y) { return x + y; } public String getName() { return this.name; } public int getCacheSize() { return this.cacheSize; } public synchronized void setCacheSize(int size) { this.cacheSize = size; System.out.println("Cache size now " + this.cacheSize); } private final String name = "Reginald"; private int cacheSize = DEFAULT_CACHE_SIZE; private static final int DEFAULT_CACHE_SIZE = 200; } Interface HelloBean (implemented by Hello) public interface HelloMBean { public void sayHello(); public int add(int x, int y); public String getName(); public int getCacheSize(); public void setCacheSize(int size); } Simple Main import java.lang.management.ManagementFactory; import java.util.logging.Logger; import javax.management.MBeanServer; import javax.management.ObjectName; public class Main { static Logger aLog = Logger.getLogger("MBeanTest"); public static void main(String[] args) { try{ MBeanServer mbs = ManagementFactory.getPlatformMBeanServer(); ObjectName name = new ObjectName("ApplicationDomain:type=Hello"); Hello mbean = new Hello(); mbs.registerMBean(mbean, name); // System.out.println(mbs.getAttribute(name, "Name")); aLog.info("Waiting forever..."); Thread.sleep(Long.MAX_VALUE); } catch(Exception x){ x.printStackTrace(); aLog.info("exception"); } } } So now I have exported this project as jar file and run it as "java -jar helloBean.jar" and by eclipse I have modified the main class to read informations of this read (Example "Name" attribute) by using the same objectname used to registry it . Main to read : public static void main(String[] args) { try{ MBeanServer mbs = ManagementFactory.getPlatformMBeanServer(); ObjectName name = new ObjectName("ApplicationDomain:type=Hello"); System.out.println(mbs.getAttribute(name, "Name")); } catch(Exception x){ x.printStackTrace(); aLog.info("exception"); } } But nothing, the bean is not found.

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  • Why my onsubmitg is not firing - Spring

    - by GigaPr
    Hi, i have a controller public class EditUserController extends BaseController { public EditUserController() { setCommandClass(User.class); setCommandName("editaUser"); } public ModelAndView handleRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { ModelAndView modelAndView = new ModelAndView("editUser"); String id = request.getParameter("id"); if(!id.isEmpty()) { User user = this.userServiceImplementation.get(Integer.parseInt(id)); modelAndView.addObject("editaUser", user); } return modelAndView; } } and the view <form:form method="POST" commandName="editaUser" cssClass="addUserForm"> <div class="floatL"> <div class="padding5"> <div class="fieldContainer"> <strong>First Name:</strong>&nbsp; </div> <form:errors path="firstName" cssClass="error"/> <form:input path="firstName" cssClass="textArea" /> </div> <div class="padding5"> <div class="fieldContainer"> <strong>Last Name:</strong>&nbsp; </div> <form:errors path="lastName" cssClass="error"/> <form:input path="lastName" cssClass="textArea" /> </div> </div> <div class="floatR"> <div class="padding5"> <div class="fieldContainer"> <strong>Username:</strong>&nbsp; </div> <form:errors path="username" cssClass="error"/> <form:input path="username" cssClass="textArea" /> </div> <div class="padding5"> <div class="fieldContainer"> <strong>Password</strong>&nbsp; </div> <form:errors path="password" cssClass="error"/> <form:input path="password" cssClass="textArea"/> </div> </div> <input type="submit" class="floatR" value="Save" > </form:form> and the bean definition looks like <bean name="/editUser.htm" class="com.rssFeed.mvc.EditUserController"> <property name="userServiceImplementation" ref="userServiceImplementation"/> <property name="commandName" value="editaUser" /> <property name="successView" value="users"/> <property name="sessionForm" value="true"/> </bean> I populate the view using the querystring but i would lke to update the record in the database on click of the submit button. i tried to insert a on submit method protected ModelAndView onSubmit(Object command, BindException bindException) throws Exception { return new ModelAndView(getSuccessView()); } but it never fires What is the problem i do not get it?? thanks

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  • Doesn't get the output in Java Database Connectivity

    - by Dooree
    I'm working on Java Database Connectivity through Eclipse IDE. I built a database through Ubuntu Terminal, and I need to connect and work with it. However, when I tried to run the following code, I don't get any error, but the following output is showed, anybody knows why I don't get the output from the code ? //STEP 1. Import required packages import java.sql.*; public class FirstExample { // JDBC driver name and database URL static final String JDBC_DRIVER = "com.mysql.jdbc.Driver"; static final String DB_URL = "jdbc:mysql://localhost/EMP"; // Database credentials static final String USER = "username"; static final String PASS = "password"; public static void main(String[] args) { Connection conn = null; Statement stmt = null; try{ //STEP 2: Register JDBC driver Class.forName("com.mysql.jdbc.Driver"); //STEP 3: Open a connection System.out.println("Connecting to database..."); conn = DriverManager.getConnection(DB_URL,USER,PASS); //STEP 4: Execute a query System.out.println("Creating statement..."); stmt = conn.createStatement(); String sql; sql = "SELECT id, first, last, age FROM Employees"; ResultSet rs = stmt.executeQuery(sql); //STEP 5: Extract data from result set while(rs.next()){ //Retrieve by column name int id = rs.getInt("id"); int age = rs.getInt("age"); String first = rs.getString("first"); String last = rs.getString("last"); //Display values System.out.print("ID: " + id); System.out.print(", Age: " + age); System.out.print(", First: " + first); System.out.println(", Last: " + last); } //STEP 6: Clean-up environment rs.close(); stmt.close(); conn.close(); }catch(SQLException se){ //Handle errors for JDBC se.printStackTrace(); }catch(Exception e){ //Handle errors for Class.forName e.printStackTrace(); }finally{ //finally block used to close resources try{ if(stmt!=null) stmt.close(); }catch(SQLException se2){ }// nothing we can do try{ if(conn!=null) conn.close(); }catch(SQLException se){ se.printStackTrace(); }//end finally try }//end try System.out.println("Goodbye!"); }//end main }//end FirstExample <ConnectionProperties> <PropertyCategory name="Connection/Authentication"> <Property name="user" required="No" default="" sortOrder="-2147483647" since="all"> The user to connect as </Property> <Property name="password" required="No" default="" sortOrder="-2147483646" since="all"> The password to use when connecting </Property> <Property name="socketFactory" required="No" default="com.mysql.jdbc.StandardSocketFactory" sortOrder="4" since="3.0.3"> The name of the class that the driver should use for creating socket connections to the server. This class must implement the interface 'com.mysql.jdbc.SocketFactory' and have public no-args constructor. </Property> <Property name="connectTimeout" required="No" default="0" sortOrder="9" since="3.0.1"> Timeout for socket connect (in milliseconds), with 0 being no timeout. Only works on JDK-1.4 or newer. Defaults to '0'. </Property> ...

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  • Run Windows in Ubuntu with VMware Player

    - by Matthew Guay
    Are you an enthusiast who loves their Ubuntu Linux experience but still needs to use Windows programs?  Here’s how you can get the full Windows experience on Ubuntu with the free VMware Player. Linux has become increasingly consumer friendly, but still, the wide majority of commercial software is only available for Windows and Macs.  Dual-booting between Windows and Linux has been a popular option for years, but this is a frustrating solution since you have to reboot into the other operating system each time you want to run a specific application.  With virtualization, you’ll never have to make this tradeoff.  VMware Player makes it quick and easy to install any edition of Windows in a virtual machine.  With VMware’s great integration tools, you can copy and paste between your Linux and Windows programs and even run native Windows applications side-by-side with Linux ones. Getting Started Download the latest version of VMware Player for Linux, and select either the 32-bit or 64-bit version, depending on your system.  VMware Player is a free download, but requires registration.  Sign in with your VMware account, or create a new one if you don’t already have one. VMware Player is fairly easy to install on Linux, but you will need to start out the installation from the terminal.  First, enter the following to make sure the installer is marked as executable, substituting version/build_number for the version number on the end of the file you downloaded. chmod +x ./VMware-Player-version/build_number.bundle Then, enter the following to start the install, again substituting your version number: gksudo bash ./VMware-Player-version/build_number.bundle You may have to enter your administrator password to start the installation, and then the VMware Player graphical installer will open.  Choose whether you want to check for product updates and submit usage data to VMware, and then proceed with the install as normal. VMware Player installed in only a few minutes in our tests, and was immediately ready to run, no reboot required.  You can now launch it from your Ubuntu menu: click Applications \ System Tools \ VMware Player. You’ll need to accept the license agreement the first time you run it. Welcome to VMware Player!  Now you can create new virtual machines and run pre-built ones on your Ubuntu desktop. Install Windows in VMware Player on Ubuntu Now that you’ve got VMware setup, it’s time to put it to work.  Click the Create a New Virtual Machine as above to start making a Windows virtual machine. In the dialog that opens, select your installer disk or ISO image file that you want to install Windows from.  In this example, we’re select a Windows 7 ISO.  VMware will automatically detect the operating system on the disk or image.  Click Next to continue. Enter your Windows product key, select the edition of Windows to install, and enter your name and password. You can leave the product key field blank and enter it later.  VMware will ask if you want to continue without a product key, so just click Yes to continue. Now enter a name for your virtual machine and select where you want to save it.  Note: This will take up at least 15Gb of space on your hard drive during the install, so make sure to save it on a drive with sufficient storage space. You can choose how large you want your virtual hard drive to be; the default is 40Gb, but you can choose a different size if you wish.  The entire amount will not be used up on your hard drive initially, but the virtual drive will increase in size up to your maximum as you add files.  Additionally, you can choose if you want the virtual disk stored as a single file or as multiple files.  You will see the best performance by keeping the virtual disk as one file, but the virtual machine will be more portable if it is broken into smaller files, so choose the option that will work best for your needs. Finally, review your settings, and if everything looks good, click Finish to create the virtual machine. VMware will take over now, and install Windows without any further input using its Easy Install.  This is one of VMware’s best features, and is the main reason we find it the easiest desktop virtualization solution to use.   Installing VMware Tools VMware Player doesn’t include the VMware Tools by default; instead, it automatically downloads them for the operating system you’re installing.  Once you’ve downloaded them, it will use those tools anytime you install that OS.  If this is your first Windows virtual machine to install, you may be prompted to download and install them while Windows is installing.  Click Download and Install so your Easy Install will finish successfully. VMware will then download and install the tools.  You may need to enter your administrative password to complete the install. Other than this, you can leave your Windows install unattended; VMware will get everything installed and running on its own. Our test setup took about 30 minutes, and when it was done we were greeted with the Windows desktop ready to use, complete with drivers and the VMware tools.  The only thing missing was the Aero glass feature.  VMware Player is supposed to support the Aero glass effects in virtual machines, and although this works every time when we use VMware Player on Windows, we could not get it to work in Linux.  Other than that, Windows is fully ready to use.  You can copy and paste text, images, or files between Ubuntu and Windows, or simply drag-and-drop files between the two. Unity Mode Using Windows in a window is awkward, and makes your Windows programs feel out of place and hard to use.  This is where Unity mode comes in.  Click Virtual Machine in VMware’s menu, and select Enter Unity. Your Windows desktop will now disappear, and you’ll see a new Windows menu underneath your Ubuntu menu.  This works the same as your Windows Start Menu, and you can open your Windows applications and files directly from it. By default, programs from Windows will have a colored border and a VMware badge in the corner.  You can turn this off from the VMware settings pane.  Click Virtual Machine in VMware’s menu and select Virtual Machine Settings.  Select Unity under the Options tab, and uncheck the Show borders and Show badges boxes if you don’t want them. Unity makes your Windows programs feel at home in Ubuntu.  Here we have Word 2010 and IE8 open beside the Ubuntu Help application.  Notice that the Windows applications show up in the taskbar on the bottom just like the Linux programs.  If you’re using the Compiz graphics effects in Ubuntu, your Windows programs will use them too, including the popular wobbly windows effect. You can switch back to running Windows inside VMware Player’s window by clicking the Exit Unity button in the VMware window. Now, whenever you want to run Windows applications in Linux, you can quickly launch it from VMware Player. Conclusion VMware Player is a great way to run Windows on your Linux computer.  It makes it extremely easy to get Windows installed and running, lets you run your Windows programs seamlessly alongside your Linux ones.  VMware products work great in our experience, and VMware Player on Linux was no exception. If you’re a Windows user and you’d like to run Ubuntu on Windows, check out our article on how to Run Ubuntu in Windows with VMware Player. Link Download VMware Player 3 (Registration required) Download Windows 7 Enterprise 90-day trial Similar Articles Productive Geek Tips Enable Copy and Paste from Ubuntu VMware GuestInstall VMware Tools on Ubuntu Edgy EftRestart the Ubuntu Gnome User Interface QuicklyHow to Add a Program to the Ubuntu Startup List (After Login)How To Run Ubuntu in Windows 7 with VMware Player TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips Xobni Plus for Outlook All My Movies 5.9 CloudBerry Online Backup 1.5 for Windows Home Server Snagit 10 Get a free copy of WinUtilities Pro 2010 World Cup Schedule Boot Snooze – Reboot and then Standby or Hibernate Customize Everything Related to Dates, Times, Currency and Measurement in Windows 7 Google Earth replacement Icon (Icons we like) Build Great Charts in Excel with Chart Advisor

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  • Azure - Part 4 - Table Storage Service in Windows Azure

    - by Shaun
    In Windows Azure platform there are 3 storage we can use to save our data on the cloud. They are the Table, Blob and Queue. Before the Chinese New Year Microsoft announced that Azure SDK 1.1 had been released and it supports a new type of storage – Drive, which allows us to operate NTFS files on the cloud. I will cover it in the coming few posts but now I would like to talk a bit about the Table Storage.   Concept of Table Storage Service The most common development scenario is to retrieve, create, update and remove data from the data storage. In the normal way we communicate with database. When we attempt to move our application over to the cloud the most common requirement should be have a storage service. Windows Azure provides a in-build service that allow us to storage the structured data, which is called Windows Azure Table Storage Service. The data stored in the table service are like the collection of entities. And the entities are similar to rows or records in the tradtional database. An entity should had a partition key, a row key, a timestamp and set of properties. You can treat the partition key as a group name, the row key as a primary key and the timestamp as the identifer for solving the concurrency problem. Different with a table in a database, the table service does not enforce the schema for tables, which means you can have 2 entities in the same table with different property sets. The partition key is being used for the load balance of the Azure OS and the group entity transaction. As you know in the cloud you will never know which machine is hosting your application and your data. It could be moving based on the transaction weight and the number of the requests. If the Azure OS found that there are many requests connect to your Book entities with the partition key equals “Novel” it will move them to another idle machine to increase the performance. So when choosing the partition key for your entities you need to make sure they indecate the category or gourp information so that the Azure OS can perform the load balance as you wish.   Consuming the Table Although the table service looks like a database, you cannot access it through the way you are using now, neither ADO.NET nor ODBC. The table service exposed itself by ADO.NET Data Service protocol, which allows you can consume it through the RESTful style by Http requests. The Azure SDK provides a sets of classes for us to connect it. There are 2 classes we might need: TableServiceContext and TableServiceEntity. The TableServiceContext inherited from the DataServiceContext, which represents the runtime context of the ADO.NET data service. It provides 4 methods mainly used by us: CreateQuery: It will create a IQueryable instance from a given type of entity. AddObject: Add the specified entity into Table Service. UpdateObject: Update an existing entity in the Table Service. DeleteObject: Delete an entity from the Table Service. Beofre you operate the table service you need to provide the valid account information. It’s something like the connect string of the database but with your account name and the account key when you created the storage service on the Windows Azure Development Portal. After getting the CloudStorageAccount you can create the CloudTableClient instance which provides a set of methods for using the table service. A very useful method would be CreateTableIfNotExist. It will create the table container for you if it’s not exsited. And then you can operate the eneities to that table through the methods I mentioned above. Let me explain a bit more through an exmaple. We always like code rather than sentence.   Straightforward Accessing to the Table Here I would like to build a WCF service on the Windows Azure platform, and for now just one requirement: it would allow the client to create an account entity on the table service. The WCF service would have a method named Register and accept an instance of the account which the client wants to create. After perform some validation it will add the entity into the table service. So the first thing I should do is to create a Cloud Application on my VIstial Studio 2010 RC. (The Azure SDK 1.1 only supports VS2008 and VS2010 RC.) The solution should be like this below. Then I added a configuration items for the storage account through the Settings section under the cloud project. (Double click the Services file under Roles folder and navigate to the Setting section.) This setting will be used when to retrieve my storage account information. Since for now I just in the development phase I will select “UseDevelopmentStorage=true”. And then I navigated to the WebRole.cs file under my WCF project. If you have read my previous posts you would know that this file defines the process when the application start, and terminate on the cloud. What I need to do is to when the application start, set the configuration publisher to load my config file with the config name I specified. So the code would be like below. I removed the original service and contract created by the VS template and add my IAccountService contract and its implementation class - AccountService. And I add the service method Register with the parameters: email, password and it will return a boolean value to indicates the result which is very simple. At this moment if I press F5 the application will be established on my local development fabric and I can see my service runs well through the browser. Let’s implement the service method Rigister, add a new entity to the table service. As I said before the entities you want to store in the table service must have 3 properties: partition key, row key and timespan. You can create a class with these 3 properties. The Azure SDK provides us a base class for that named TableServiceEntity in Microsoft.WindowsAzure.StorageClient namespace. So what we need to do is more simply, create a class named Account and let it derived from the TableServiceEntity. And I need to add my own properties: Email, Password, DateCreated and DateDeleted. The DateDeleted is a nullable date time value to indecate whether this entity had been deleted and when. Do you notice that I missed something here? Yes it’s the partition key and row key I didn’t assigned. The TableServiceEntity base class defined 2 constructors one was a parameter-less constructor which will be used to fill values into the properties from the table service when retrieving data. The other was one with 2 parameters: partition key and row key. As I said below the partition key may affect the load balance and the row key must be unique so here I would like to use the email as the parition key and the email plus a Guid as the row key. OK now we finished the entity class we need to store onto the table service. The next step is to create a data access class for us to add it. Azure SDK gives us a base class for it named TableServiceContext as I mentioned below. So let’s create a class for operate the Account entities. The TableServiceContext need the storage account information for its constructor. It’s the combination of the storage service URI that we will create on Windows Azure platform, and the relevant account name and key. The TableServiceContext will use this information to find the related address and verify the account to operate the storage entities. Hence in my AccountDataContext class I need to override this constructor and pass the storage account into it. All entities will be saved in the table storage with one or many tables which we call them “table containers”. Before we operate an entity we need to make sure that the table container had been created on the storage. There’s a method we can use for that: CloudTableClient.CreateTableIfNotExist. So in the constructor I will perform it firstly to make sure all method will be invoked after the table had been created. Notice that I passed the storage account enpoint URI and the credentials to specify where my storage is located and who am I. Another advise is that, make your entity class name as the same as the table name when create the table. It will increase the performance when you operate it over the cloud especially querying. Since the Register WCF method will add a new account into the table service, here I will create a relevant method to add the account entity. Before implement, I should add a reference - System.Data.Services.Client to the project. This reference provides some common method within the ADO.NET Data Service which can be used in the Windows Azure Table Service. I will use its AddObject method to create my account entity. Since the table service are not fully implemented the ADO.NET Data Service, there are some methods in the System.Data.Services.Client that TableServiceContext doesn’t support, such as AddLinks, etc. Then I implemented the serivce method to add the account entity through the AccountDataContext. You can see in the service implmentation I load the storage account information through my configuration file and created the account table entity from the parameters. Then I created the AccountDataContext. If it’s my first time to invoke this method the constructor of the AccountDataContext will create a table container for me. Then I use Add method to add the account entity into the table. Next, let’s create a farely simple client application to test this service. I created a windows console application and added a service reference to my WCF service. The metadata information of the WCF service cannot be retrieved if it’s deployed on the Windows Azure even though the <serviceMetadata httpGetEnabled="true"/> had been set. If we need to get its metadata we can deploy it on the local development service and then changed the endpoint to the address which is on the cloud. In the client side app.config file I specified the endpoint to the local development fabric address. And the just implement the client to let me input an email and a password then invoke the WCF service to add my acocunt. Let’s run my application and see the result. Of course it should return TRUE to me. And in the local SQL Express I can see the data had been saved in the table.   Summary In this post I explained more about the Windows Azure Table Storage Service. I also created a small application for demostration of how to connect and consume it through the ADO.NET Data Service Managed Library provided within the Azure SDK. I only show how to create an eneity in the storage service. In the next post I would like to explain about how to query the entities with conditions thruogh LINQ. I also would like to refactor my AccountDataContext class to make it dyamic for any kinds of entities.   Hope this helps, Shaun   All documents and related graphics, codes are provided "AS IS" without warranty of any kind. Copyright © Shaun Ziyan Xu. This work is licensed under the Creative Commons License.

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