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  • How to decrypt a string in C# that was encrypted in Delphi

    - by Simon Linder
    Hi all, we have a project written in Delphi that we want to convert to C#. Problem is that we have some passwords and settings that are encrypted and written into the registry. When we need a specified password we get it from the registry and decrypt it so we can use it. For the conversion into C# we have to do it the same way so that the application can also be used by users that have the old version and want to upgrade it. Here is the code we use to encrypt/decrypt strings in Delphi: unit uCrypt; interface function EncryptString(strPlaintext, strPassword : String) : String; function DecryptString(strEncryptedText, strPassword : String) : String; implementation uses DCPcrypt2, DCPblockciphers, DCPdes, DCPmd5; const CRYPT_KEY = '1q2w3e4r5t6z7u8'; function EncryptString(strPlaintext) : String; var cipher : TDCP_3des; strEncryptedText : String; begin if strPlaintext <> '' then begin try cipher := TDCP_3des.Create(nil); try cipher.InitStr(CRYPT_KEY, TDCP_md5); strEncryptedText := cipher.EncryptString(strPlaintext); finally cipher.Free; end; except strEncryptedText := ''; end; end; Result := strEncryptedText; end; function DecryptString(strEncryptedText) : String; var cipher : TDCP_3des; strDecryptedText : String; begin if strEncryptedText <> '' then begin try cipher := TDCP_3des.Create(nil); try cipher.InitStr(CRYPT_KEY, TDCP_md5); strDecryptedText := cipher.DecryptString(strEncryptedText); finally cipher.Free; end; except strDecryptedText := ''; end; end; Result := strDecryptedText; end; end. So for example when we want to encrypt the string asdf1234 we get the result WcOb/iKo4g8=. We now want to decrypt that string in C#. Here is what we tried to do: public static void Main(string[] args) { string Encrypted = "WcOb/iKo4g8="; string Password = "1q2w3e4r5t6z7u8"; string DecryptedString = DecryptString(Encrypted, Password); } public static string DecryptString(string Message, string Passphrase) { byte[] Results; System.Text.UTF8Encoding UTF8 = new System.Text.UTF8Encoding(); // Step 1. We hash the passphrase using MD5 // We use the MD5 hash generator as the result is a 128 bit byte array // which is a valid length for the TripleDES encoder we use below MD5CryptoServiceProvider HashProvider = new MD5CryptoServiceProvider(); byte[] TDESKey = HashProvider.ComputeHash(UTF8.GetBytes(Passphrase)); // Step 2. Create a new TripleDESCryptoServiceProvider object TripleDESCryptoServiceProvider TDESAlgorithm = new TripleDESCryptoServiceProvider(); // Step 3. Setup the decoder TDESAlgorithm.Key = TDESKey; TDESAlgorithm.Mode = CipherMode.ECB; TDESAlgorithm.Padding = PaddingMode.None; // Step 4. Convert the input string to a byte[] byte[] DataToDecrypt = Convert.FromBase64String(Message); // Step 5. Attempt to decrypt the string try { ICryptoTransform Decryptor = TDESAlgorithm.CreateDecryptor(); Results = Decryptor.TransformFinalBlock(DataToDecrypt, 0, DataToDecrypt.Length); } finally { // Clear the TripleDes and Hashprovider services of any sensitive information TDESAlgorithm.Clear(); HashProvider.Clear(); } // Step 6. Return the decrypted string in UTF8 format return UTF8.GetString(Results); } Well the result differs from the expected result. After we call DecryptString() we expect to get asdf1234but we get something else. Does anyone have an idea of how to decrypt that correctly? Thanks in advance Simon

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  • Migrate from MySQL to PostgreSQL on Linux (Kubuntu)

    - by Dave Jarvis
    A long time ago in a galaxy far, far away... Trying to migrate a database from MySQL to PostgreSQL. All the documentation I have read covers, in great detail, how to migrate the structure. I have found very little documentation on migrating the data. The schema has 13 tables (which have been migrated successfully) and 9 GB of data. MySQL version: 5.1.x PostgreSQL version: 8.4.x I want to use the R programming language to analyze the data using SQL select statements; PostgreSQL has PL/R, but MySQL has nothing (as far as I can tell). A New Hope Create the database location (/var has insufficient space; also dislike having the PostgreSQL version number everywhere -- upgrading would break scripts!): sudo mkdir -p /home/postgres/main sudo cp -Rp /var/lib/postgresql/8.4/main /home/postgres sudo chown -R postgres.postgres /home/postgres sudo chmod -R 700 /home/postgres sudo usermod -d /home/postgres/ postgres All good to here. Next, restart the server and configure the database using these installation instructions: sudo apt-get install postgresql pgadmin3 sudo /etc/init.d/postgresql-8.4 stop sudo vi /etc/postgresql/8.4/main/postgresql.conf Change data_directory to /home/postgres/main sudo /etc/init.d/postgresql-8.4 start sudo -u postgres psql postgres \password postgres sudo -u postgres createdb climate pgadmin3 Use pgadmin3 to configure the database and create a schema. The episode continues in a remote shell known as bash, with both databases running, and the installation of a set of tools with a rather unusual logo: SQL Fairy. perl Makefile.PL sudo make install sudo apt-get install perl-doc (strangely, it is not called perldoc) perldoc SQL::Translator::Manual Extract a PostgreSQL-friendly DDL and all the MySQL data: sqlt -f DBI --dsn dbi:mysql:climate --db-user user --db-password password -t PostgreSQL > climate-pg-ddl.sql mysqldump --skip-add-locks --complete-insert --no-create-db --no-create-info --quick --result-file="climate-my.sql" --databases climate --skip-comments -u root -p The Database Strikes Back Recreate the structure in PostgreSQL as follows: pgadmin3 (switch to it) Click the Execute arbitrary SQL queries icon Open climate-pg-ddl.sql Search for TABLE " replace with TABLE climate." (insert the schema name climate) Search for on " replace with on climate." (insert the schema name climate) Press F5 to execute This results in: Query returned successfully with no result in 122 ms. Replies of the Jedi At this point I am stumped. Where do I go from here (what are the steps) to convert climate-my.sql to climate-pg.sql so that they can be executed against PostgreSQL? How to I make sure the indexes are copied over correctly (to maintain referential integrity; I don't have constraints at the moment to ease the transition)? How do I ensure that adding new rows in PostgreSQL will start enumerating from the index of the last row inserted (and not conflict with an existing primary key from the sequence)? How do you ensure the schema name comes through when transforming the data from MySQL to PostgreSQL inserts? Resources A fair bit of information was needed to get this far: https://help.ubuntu.com/community/PostgreSQL http://articles.sitepoint.com/article/site-mysql-postgresql-1 http://wiki.postgresql.org/wiki/Converting_from_other_Databases_to_PostgreSQL#MySQL http://pgfoundry.org/frs/shownotes.php?release_id=810 http://sqlfairy.sourceforge.net/ Thank you!

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  • Reuse security code between WCF and MVC.NET

    - by mrjoltcola
    First the background: I jumped into MVC.NET from the Java MVC world, so my implementation below is possibly cheating, I don't know. I avoided fooling with a custom membership provider and I just implemented the base code needed to authenticate and load roles in my LogOn action. Typically I just need to check roles programatically, and have no use for all of the other membership features, so I didn't originally think I needed a full Membership provider. I have a successful WCF project with a custom authentication and authorization layer that I did at least write per the proper API. I implemented it with custom IPrincipal, UserNamePasswordValidator and IAuthorizationPolicy classes to load from an Oracle database. In my WCF services, I use declarative security: [PrincipalPermission(SecurityAction.Demand, Role="ADMIN")]. The question (on the ASP.NET/MCV.NET side): All my reading indicates I should implement a custom Membership/Roles provider, and use [Authorize(Roles="ADMIN")] on my controller actions. At this point, I don't have a true Membership provider, but I'm using the same User class that implements the IPrincipal interface that works with the WCF security. I plan to share common code between the WCF and ASP.NET modules. So my LogOn action is not using the FormsService (and I assume this is bad). I had commented it out, and just used my "UserService" to access the Oracle db. Note my "TODO" comment below. public ActionResult LogOn(LogOnModel model, string returnUrl) { log.Info("Login attempt by " + model.UserName); if (ModelState.IsValid) { User user = userService.findByUserName(model.UserName); // Commented original MemberShipService code, this is probably bad // if (MembershipService.ValidateUser(model.UserName, model.Password)) if (user != null && user.Authenticate(model.Password) == true) { log.Info("Login success by " + model.UserName); FormsService.SignIn(model.UserName, model.RememberMe); // TODO: Override with Custom identity / roles? user.AddRoles(userService.listRolesByUser(user)); // pull in roles from db if (!String.IsNullOrEmpty(returnUrl)) return Redirect(returnUrl); else return RedirectToAction("Index", "Home"); } else { log.Info("Login failure by " + model.UserName); ModelState.AddModelError("", "The user name or password provided is incorrect."); } } // If we got this far, something failed, redisplay form return View(model); } So can I make the above work? Can I stick the IPrincipal (User) into the CurrentContext or HttpContext? Can I integrate the custom IPrincipal I've already created without writing a full Membership/Roles Provider? I currently stick the User object into the session and access it from all MVC.NET controllers with "CurrentUser" property which grabs it from the session on demand. But this doesn't work with the [Authorize] attribute; I assume that is because it knows nothing about my custom Principal in the session, and is instead using whatever FormsService.SignIn() produces. I also found that session timeouts screw up the login redirect, the user doesn't get forwarded, instead we get a null exception accessing User from the session, and I assume it is related to my "skipping steps" to get a quick implementation. Thanks.

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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • Problem in working with async and await?

    - by Vicky
    I am trying to upload files to Azure Blob Storage and after successful upload adding the filename to a list for my further operation. When i am doing synchronous it works fine but when i am doing async the error occured. Error : Collection was modified; enumeration operation may not execute. foreach(var file in files) { // ..... await blockBlob.UploadFromStreamAsync(fs); listOfMovedLabelFiles.Add(fileName); } if (listOfMovedLabelFiles.Count > 0) // error point { // my code for further operation } Is there any way to wait till all the async operations get completed.

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