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  • Requirements for connecting to Oracle with JDBC?

    - by Lord Torgamus
    I'm a newbie to Java-related web development, and I can't seem to get a simple program with JDBC working. I'm using off-the-shelf Oracle 10g XE and the Eclipse EE IDE. From the books and web pages I've checked so far, I've narrowed the problem down to either an incorrectly written database URL or a missing JAR file. I'm getting the following error: java.sql.SQLException: No suitable driver found for jdbc:oracle://127.0.0.1:8080 with the following code: import java.sql.*; public class DatabaseTestOne { public static void main(String[] args) { String url = "jdbc:oracle://127.0.0.1:8080"; String username = "HR"; String password = "samplepass"; String sql = "SELECT EMPLOYEE_ID FROM EMPLOYEES WHERE LAST_NAME='King'"; Connection connection; try { connection = DriverManager.getConnection(url, username, password); Statement statement = connection.createStatement(); System.out.println(statement.execute(sql)); connection.close(); } catch (SQLException e) { System.err.println(e); } } } What is the proper format for a database URL, anyways? They're mentioned a lot but I haven't been able to find a description. Thanks! EDIT (the answer): Based on duffymo's answer, I got ojdbc14.jar from http://www.oracle.com/technology/software/tech/java/sqlj_jdbc/htdocs/jdbc_10201.html and dropped it in the Eclipse project's Referenced Libraries. Then I changed the start of the code to ... try { Class.forName("oracle.jdbc.driver.OracleDriver"); } catch (ClassNotFoundException e) { System.err.println(e); } // jdbc:oracle:thin:@<hostname>:<port>:<sid> String url = "jdbc:oracle:thin:@GalacticAC:1521:xe"; ... and it worked.

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  • Configurating JOOMLA's e-mail notification for new account

    - by Dion
    I'm using Joomla 1.5 to create a local site for my office. The site will be accessed locally via intranet, and my PC will be the localhost for the site. I'm using a Login pluggin, so that anyone who wanted to enter the site should create an account. In JOOMLA, all user who created their account for the first time will receive a notification e-mail like : "Hello pras, You have been added as a User to Information Center by an Administrator. This e-mail contains your username and password to log in to http://localhost/yaddayadda/ Username: hadisuryo.prasetio Password: xxxx Please do not respond to this message as it is automatically generated and is for information purposes only." but if the user click the URL in the mail, which is, "localhost/yaddayadda/" they will not be directed to my site, but to their own PC's localhost.... My question is : How can I Modified the e-mail or the site configuration so that the URL will not be "localhost/yaddayadda/" anymore, but will be "(My-IP adress)/yaddayadda" I'm not going to host my site to a web hosting service, just using my PC as a host. I've been trying to trace on each config and .ini files...it seems that i have to do something with the "JURI" function or the "$mosConfig_live_site" on the backlink.php file $mosConfig_absolute_path = JPATH_SITE; $mosConfig_live_site = JURI :: base(); $url_array = explode('/', $_SERVER['REQUEST_URI']); Can anyone give me assistance ? Thank You

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  • What is wrong with accessing DBI directly?

    - by canavanin
    Hi everyone! I'm currently reading Effective Perl Programming (2nd edition). I have come across a piece of code which was described as being poorly written, but I don't yet understand what's so bad about it, or how it should be improved. It would be great if someone could explain the matter to me. Here's the code in question: sub sum_values_per_key { my ( $class, $dsn, $user, $password, $parameters ) = @_; my %results; my $dbh = DBI->connect( $dsn, $user, $password, $parameters ); my $sth = $dbh->prepare( 'select key, calculate(value) from my_table'); $sth->execute(); # ... fill %results ... $sth->finish(); $dbh->disconnect(); return \%results; } The example comes from the chapter on testing your code (p. 324/325). The sentence that has left me wondering about how to improve the code is the following: Since the code was poorly written and accesses DBI directly, you'll have to create a fake DBI object to stand in for the real thing. I have probably not understood a lot of what the book has so far been trying to teach me, or I have skipped the section relevant for understanding what's bad practice about the above code... Well, thanks in advance for your help!

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  • Flex : providing data with a PHP Class

    - by Tristan
    Hello, i'm a very new user to flex (never use flex, nor flashbuilder, nor action script before), but i want to learn this langage because of the beautiful RIA and chart it can do. I watched the video on adobe : 1 hour to build your first program but i'm stuck : On the video it says that we have to provide a PHP class for accessing data and i used the example that flash builder gave (with zend framework and mysqli). I never used those ones and it makes a lot to learn if i count zen + mysqli. My question is : can i use a PHP class like this one ? What does flash builder except in return ? i hear that was automatic. example it may be wrong, i'm not very familiar with classes when acessing to database : <?php class DBConnection { protected $server = "localhost"; protected $username = "root"; protected $password = "root"; protected $dbname = "something"; protected $connection; function __construct() { $this->connection = mysql_connect($this->server, $this->username, $this->password); mysql_select_db($this->dbname,$this->connection); mysql_query("SET NAMES 'utf8'", $this->connection); } function query($query) { $result = mysql_query($query, $this->connection); if (!$result) { echo 'request error ' . mysql_error($this->connection); exit; } return $result; } function getAll() { $req = "select * from servers"; $result = query($req) return $result } function num_rows() { return mysql_num_rows($result); } function end() { mysql_close($this->connection); } } ?> Thank you,

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  • i dont understand error while connecting php and mysql? user denied ? plz help me out to solve. ?

    - by user309381
    class MySQLDatabase { public $connection; function _construct() { $this->open_connection();} public function open_connection() {$this->connection = mysql_connect(DB_SERVER,DB_USER,DB_PASS); if(!$this->connection){die("Database Connection Failed" . mysql_error());} else{$db_select = mysql_select_db(DB_NAME,$this->connection); if(!$db_select){die("Database Selection Failed" . mysql_error()); } }} public function close_connection({ if(isset($this->connection)){ mysql_close($this->connection); unset($this->connection);}} public function query(/*$sql*/){ $sql = "SELECT*FROM users where id = 1"; $result = mysql_query($sql); $this->confirm_query($result); //return $result;while( $found_user = mysql_fetch_assoc($result)) { echo $found_user ['username']; } } private function confirm_query($result) { if(!$result) { die("The Query has problem" . mysql_error()); } } } $database = new MySQLDatabase(); $database->open_connection(); $database->query(); $database->close_connection(); I am getting error like denied for user system@locahost(using password no).i have also other database but it runs fine and i dont also i have set the password after encountered the error what else can do to solve plz help ?

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  • Lack of security in many PHP applications?

    - by John
    Over the past year of freelancing, I inherited two web projects, both of them built in PHP, both of them with sensitive information like credit card info, bank info, etc... In one application, when I typed http://thecompany.com/admin/, and without being asked for a username and password, I saw every user's sensitive information, including credit card numbers, bank account numbers etc... In another application, I was able to bypass the login screen by simply typing http://the2ndcompany.com/customer.php?user_id=777, and again, without any prompts for username and password, i was able to see user 777's credit card info. I cycled through a few more user_ids (any integer) and saw each person's credit card info. Is something wrong here? Or is this the quality of work that the "average" programmer produces? Because if this is what the average programmer produces, does that means I'm an...gasp...elite programmer?? No..that can't be right....something doesn't make sense. So my question is, is it just coincidence that I inherited two applications both of which are dangerously lacking in security? Or are there are a lot of bad PHP programmers out there?

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  • Java Mvc And Hibernate

    - by GigaPr
    Hi i am trying to learn Java, Hibernate and the MVC pattern. Following various tutorial online i managed to map my database, i have created few Main methods to test it and it works. Furthermore i have created few pages using the MVC patter and i am able to display some mock data as well in a view. the problem is i can not connect the two. this is what i have My view Looks like this <%@ include file="/WEB-INF/jsp/include.jsp" %> <html> <head> <title>Users</title> <%@ include file="/WEB-INF/jsp/head.jsp" %> </head> <body> <%@ include file="/WEB-INF/jsp/header.jsp" %> <img src="images/rss.png" alt="Rss Feed"/> <%@ include file="/WEB-INF/jsp/menu.jsp" %> <div class="ContainerIntroText"> <img src="images/usersList.png" class="marginL150px" alt="Add New User"/> <br/> <br/> <div class="usersList"> <div class="listHeaders"> <div class="headerBox"> <strong>FirstName</strong> </div> <div class="headerBox"> <strong>LastName</strong> </div> <div class="headerBox"> <strong>Username</strong> </div> <div class="headerAction"> <strong>Edit</strong> </div> <div class="headerAction"> <strong>Delete</strong> </div> </div> <br><br> <c:forEach items="${users}" var="user"> <div class="listElement"> <c:out value="${user.firstName}"/> </div> <div class="listElement"> <c:out value="${user.lastName}"/> </div> <div class="listElement"> <c:out value="${user.username}"/> </div> <div class="listElementAction"> <input type="button" name="Edit" title="Edit" value="Edit"/> </div> <div class="listElementAction"> <input type="image" src="images/delete.png" name="image" alt="Delete" > </div> <br /> </c:forEach> </div> </div> <a id="addUser" href="addUser.htm" title="Click to add a new user">&nbsp;</a> </body> </html> My controller public class UsersController implements Controller { private UserServiceImplementation userServiceImplementation; public ModelAndView handleRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { ModelAndView modelAndView = new ModelAndView("users"); List<User> users = this.userServiceImplementation.get(); modelAndView.addObject("users", users); return modelAndView; } public UserServiceImplementation getUserServiceImplementation() { return userServiceImplementation; } public void setUserServiceImplementation(UserServiceImplementation userServiceImplementation) { this.userServiceImplementation = userServiceImplementation; } } My servelet definitions <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd"> <!-- the application context definition for the springapp DispatcherServlet --> <bean name="/home.htm" class="com.rssFeed.mvc.HomeController"/> <bean name="/rssFeeds.htm" class="com.rssFeed.mvc.RssFeedsController"/> <bean name="/addUser.htm" class="com.rssFeed.mvc.AddUserController"/> <bean name="/users.htm" class="com.rssFeed.mvc.UsersController"> <property name="userServiceImplementation" ref="userServiceImplementation"/> </bean> <bean id="userServiceImplementation" class="com.rssFeed.ServiceImplementation.UserServiceImplementation"> <property name="users"> <list> <ref bean="user1"/> <ref bean="user2"/> </list> </property> </bean> <bean id="user1" class="com.rssFeed.domain.User"> <property name="firstName" value="firstName1"/> <property name="lastName" value="lastName1"/> <property name="username" value="username1"/> <property name="password" value="password1"/> </bean> <bean id="user2" class="com.rssFeed.domain.User"> <property name="firstName" value="firstName2"/> <property name="lastName" value="lastName2"/> <property name="username" value="username2"/> <property name="password" value="password2"/> </bean> <bean id="viewResolver" class="org.springframework.web.servlet.view.InternalResourceViewResolver"> <property name="viewClass" value="org.springframework.web.servlet.view.JstlView"></property> <property name="prefix" value="/WEB-INF/jsp/"></property> <property name="suffix" value=".jsp"></property> </bean> </beans> and finally this class to access the database public class HibernateUserDao extends HibernateDaoSupport implements UserDao { public void addUser(User user) { getHibernateTemplate().saveOrUpdate(user); } public List<User> get() { User user1 = new User(); user1.setFirstName("FirstName"); user1.setLastName("LastName"); user1.setUsername("Username"); user1.setPassword("Password"); List<User> users = new LinkedList<User>(); users.add(user1); return users; } public User get(int id) { throw new UnsupportedOperationException("Not supported yet."); } public User get(String username) { return null; } } the database connection occurs in this file <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd"> <bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"> <property name="driverClassName" value="org.hsqldb.jdbcDriver"/> <property name="url" value="jdbc:hsqldb:hsql://localhost/rss"/> <property name="username" value="sa"/> <property name="password" value=""/> </bean> <bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean" > <property name="dataSource" ref="dataSource" /> <property name="mappingResources"> <list> <value>com/rssFeed/domain/User.hbm.xml</value> </list> </property> <property name="hibernateProperties" > <props> <prop key="hibernate.dialect">org.hibernate.dialect.HSQLDialect</prop> </props> </property> </bean> <bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager"> <property name="sessionFactory" ref="sessionFactory"/> </bean> <bean id="userDao" class="com.rssFeed.dao.hibernate.HibernateUserDao"> <property name="sessionFactory" ref="sessionFactory"/> </bean> </beans> Could you help me to solve this problem i spent the last 4 days and nights on this issue without any success Thanks

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  • How can I log any login operation in case of "Remember Me" option ?

    - by Space Cracker
    I have an asp.net login web form that have ( username textBox - password textBox ) plus Remember Me CheckBox option When user login i do the below code if (provider.ValidateUser(username, password)) { int timeOut = 0x13; DateTime expireDate = DateTime.Now.AddMinutes(19.0); if (rememberMeCheckBox.Checked) { timeOut = 0x80520; expireDate = DateTime.Now.AddYears(1); } FormsAuthenticationTicket ticket = new FormsAuthenticationTicket(username, true, timeOut); string cookieValue = FormsAuthentication.Encrypt(ticket); HttpCookie cookie = new HttpCookie(FormsAuthentication.FormsCookieName, cookieValue); cookie.Expires = expireDate; HttpContext.Current.Response.Cookies.Add(cookie); AddForLogin(username); Response.Redirect("..."); } as in code after user is authenticated i log that he login in db by calling method AddForLogin(username); But if user choose remember me in login and then he try to go to site any time this login method isn't executed as it use cookies ... so i have many questions: 1- Is this the best way to log login operation or is there any other better ? 2- In my case how to log login operation in case of remember me chosen by user ?

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  • changing WCF endpoint does not persist data.

    - by Vinay Pandey
    Hi All, I have an application that has reference of a WCF service on machine A, now on certain situation I want tu use similar service hosted on machine B. When I changed the endpoint using following:- EndpointAddress endpoint = new EndpointAddress(new Uri(ConfigurationManager.AppSettings["ServiceURLForMachineB"])); BasicHttpBinding binding = new BasicHttpBinding(); binding.SendTimeout = TimeSpan.FromMinutes(1); binding.OpenTimeout = TimeSpan.FromMinutes(1); binding.CloseTimeout = TimeSpan.FromMinutes(1); binding.ReceiveTimeout = TimeSpan.FromMinutes(10); binding.AllowCookies = false; binding.BypassProxyOnLocal = false; binding.HostNameComparisonMode = HostNameComparisonMode.StrongWildcard; binding.MessageEncoding = WSMessageEncoding.Mtom; binding.TextEncoding = System.Text.Encoding.UTF8; binding.TransferMode = TransferMode.Buffered; binding.UseDefaultWebProxy = true; repositoryService = new WorkflowRepositoryServiceClient(binding, endpoint); When I call login method although method is called from machine B, but username and password in Login(string username,string password) are coming null on machine B. Any Idea what I am doing wrong here?

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  • how i can retrive files from folder on hard-disk and how to display uplaoded file data into a textar

    - by Deepak Narwal
    I have made a application form in which i am asking for username,password,email id and user's resume.Now after uploading resume i am storing it into hard disk into htdocs/uploadedfiles/..in a format something like this username_filename.In database i am storing file name,file size,file type.Some coading for this i am showing here $filesize=$_FILES['file']['size']; $filename=$_FILES['file']['name']; $filetype=$_FILES['file']['type']; $temp_name=$_FILES['file']['tmp_name']; //temporary name of uploaded file $pwd_hash = hash('sha1',$_POST['password']); $target_path = "uploadedfiles/"; $target_path = $target_path.$_POST['username']."_".basename( $_FILES['file']['name']); move_uploaded_file($_FILES['file']['tmp_name'], $target_path) ; $sql="insert into employee values ('NULL','{$_POST[username]}','{$pwd_hash}','{$filename}','{$filetype}','$filesize',NOW())"; Now i have two questions 1.NOw how i can display this file data into a textarea(something like naukri.com resume section) 2.How one can retrive that resume file from folder on hard-disk.What query should i write to fetch this file from that folder.I know how to retrive data from database but i dont know how to retrive data from a folder in hard-disk like in the case if user want to delete this file or he wnat to download this file.How i can do this

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  • Spring Security 3.0- Customise basic http Authentication Dialog

    - by gav
    Rather than reading; A user name and password are being requested by http://localhost:8080. The site says: "Spring Security Application" I want to change the prompt, or at least change what the "site says". Does anyone know how to do this via resources.xml? In my Grails App Spring configuration, my current version is as follows; <?xml version="1.0" encoding="UTF-8"?> <beans:beans xmlns="http://www.springframework.org/schema/security" xmlns:beans="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.0.xsd"> <http auto-config="true" use-expressions="true"> <http-basic/> <intercept-url pattern="/**" access="isAuthenticated()" /> </http> <authentication-manager alias="authenticationManager"> <authentication-provider> <user-service> <user name="admin" password="admin" authorities="ROLE_ADMIN"/> </user-service> </authentication-provider> </authentication-manager> </beans:beans>

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  • Fix a 404: missing parameters error from a GET request to CherryPy

    - by norabora
    I'm making a webpage using CherryPy for the server-side, HTML, CSS and jQuery on the client-side. I'm also using a mySQL database. I have a working form for users to sign up to the site - create a username and password. I use jQuery to send an AJAX POST request to the CherryPy which queries the database to see if that username exists. If the username exists, alert the user, if it doesn't, add it to the database and alert success. $.post('submit', postdata, function(data) { alert(data); }); Successful jQuery POST. I want to change the form so that instead of checking that the username exists on submit, a GET request is made as on the blur event from the username input. The function gets called, and it goes to the CherryPy, but then I get an error that says: HTTPError: (404, 'Missing parameters: username'). $.get('checkUsername', getdata, function(data) { alert(data); }); Unsuccessful jQuery GET. The CherryPy: @cherrypy.expose def submit(self, **params): cherrypy.response.headers['Content-Type'] = 'application/json' e = sqlalchemy.create_engine('mysql://mysql:pw@localhost/6470') c = e.connect() com1 = "SELECT * FROM `users` WHERE `username` = '" + params["username"] + "'" b = c.execute(com1).fetchall() if not len(b) > 0: com2 = "INSERT INTO `6470`.`users` (`username` ,`password` ,`website` ,`key`) VALUES ('" com2 += params["username"] + "', MD5( '" + params["password"] + "'), '', NULL);" a = c.execute(com2) c.close() return simplejson.dumps("Success!") #login user and send them to home page c.close() return simplejson.dumps("This username is not available.") @cherrypy.expose def checkUsername(self, username): cherrypy.response.headers['Content-Type'] = 'application/json' e = sqlalchemy.create_engine('mysql://mysql:pw@localhost/6470') c = e.connect() command = "SELECT * FROM `users` WHERE `username` = '" + username + "'" a = c.execute(command).fetchall(); c.close() sys.stdout.write(str(a)) return simplejson.dumps("") I can't see any differences between the two so I don't know why the GET request is giving me a problem. Any insight into what I might be doing wrong would be helpful. If you have ideas about the jQuery, CherryPy, config files, anything, I'd really appreciate it.

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  • How do I use Perl's LWP to log in to a web application?

    - by maxjackie
    I would like to write a script to login to a web application and then move to other parts of the application: use HTTP::Request::Common qw(POST); use LWP::UserAgent; use Data::Dumper; $ua = LWP::UserAgent->new(keep_alive=>1); my $req = POST "http://example.com:5002/index.php", [ user_name => 'username', user_password => "password", module => 'Users', action => 'Authenticate', return_module => 'Users', return_action => 'Login', ]; my $res = $ua->request($req); print Dumper(\$res); if ( $res->is_success ) { print $res->as_string; } When I try this code I am not able to login to the application. The HTTP status code returned is 302 that is found, but with no data. If I post username/password with all required things then it should return the home page of the application and keep the connection live to move other parts of the application.

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  • Sharing runtime variables between files

    - by nightcracker
    I have a project with a few files that all include the header global.hpp. Those files want to share and update information that is relevant for the whole program during runtime (that data is gathered progressively during the program runs but the fields of data are known at compile-time). Now my idea was to use a struct like this: global.hpp #include <string> #ifndef _GLOBAL_SESSION_STRUCT #define _GLOBAL_SESSION_STRUCT struct session_struct { std::string username; std::string password; std::string hostname; unsigned short port; // more data fields as needed }; #endif extern struct session_struct session; main.cpp #include "global.hpp" struct session_struct session; int main(int argc, char* argv[]) { session.username = "user"; session.password = "secret"; session.hostname = "example.com"; session.port = 80; // other stuff, etc return 0; } Now every file that includes global.hpp can just read & write the fields of the session struct and easily share information. Is this the correct way to do this? NOTE: For this specific project no threading is used. But please (for future projects and other people reading) clarify in your answer how this (or your proposed) solution works when threaded. Also, for this example/project session variables are shared. But this should also apply to any other form of shared variables.

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  • Creating a RESTful service in CakePHP

    - by NathanGaskin
    I'm attempting to create a RESTful service in CakePHP but I've hit a bit of a brick wall. I've enabled the default RESTful routing using Router::mapResources('users') and Router::parseExtensions(). This works well if I make a GET request, and returns some nicely formatted XML. So far so good. The problem is if I want to make a POST or PUT request. CakePHP doesn't seem to be able to read the data from the request. At the moment my add(), edit() and delete() actions don't contain any logic, they're simply setting $this-data to the view. I'm testing with the following cURL command: curl -v -d "<user><username>blahblah</username><password>blahblah</password>" http://localhost/users.xml --header 'content-type: text/xml' Which only returns a 404 header. If I remove the --header parameter then it returns the view but no data is set. It feels like I'm missing something obvious here. Any ideas?

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  • Is there a FAST way to export and install an app on my phone, while signing it with my own keystore?

    - by Alexei Andreev
    So, I've downloaded my own application from the market and installed it on my phone. Now, I am trying to install a temporary new version from Eclipse, but here is the message I get: Re-installation failed due to different application signatures. You must perform a full uninstall of the application. WARNING: This will remove the application data! Please execute 'adb uninstall com.applicationName' in a shell. Launch canceled! Now, I really really don't want to uninstall the application, because I will lose all my data. One solution I found is to Export my application, creating new .apk, and then install it via HTC Sync (probably a different program based on what phone you have). The problem is this takes a long time to do, since I need to enter the password for the keystore each time and then wait for HTC Sync. It's a pain in the ass! So the question is: Is there a way to make Eclipse automatically use my keystore to sign the application (quickly and automatically)? Or perhaps to replace debug keystore with my own? Or perhaps just tell it to remember the password, so I don't have to enter it every time...? Or some other way to solve this problem?

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  • How to avoid hard-coded credentials in Sharepoint webpart?

    - by Bryan
    I am building a Sharepoint web part that will be used by all users, but can only be modified by admins. The web part connects to a web service which needs credentials. I hard coded credentials in the web part's code. query.Credentials = new System.Net.NetworkCredential("username", "password", "domain"); query is an instance of the web service class This may not be a good approach. In regard with security, the source code of the web apart is available to people who are not allowed to see the credentials. In normal ASP.net applications, credentials can be written into web.config and encrypted. A web part doesn't have a .config file associated. There is a application-level .config file for the whole sharepoint site, but I don't want to modify it for a single webpart. I wonder if there is a webpart-specific way to solve the credential problem? Say we provide a WebBrowsable property of that web part so that privileged users can modify credentials. If this is desirable, how should I make the property displayed in a password ("*") rather than in plain text? Thanks.

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  • Problem with impersonating a specific user in WCF service

    - by aJ
    I am having a WCF service hosted in IIS on WindowsServer 2008. This service needs to write to a shared folder present on another machine(Windows XP). The shared folder has write permissions for a particular user say "X" which is present on both the machines .i.e on the server where the service is running as well as the machine where the shared folder is present. The service runs under the NETWORK SERVICE account. For the service to access the shared folder I have added code to impersonate the user "X"in the service so that it gets the permission to write to the shared folder. Since I want to impersonate user "X" only when I run a particular section of code I have used the sample code. Even after the impersonation the service fails to write to the shared folder sometimes. It works sporadically. Whereas if I add tag in the Web.config file it works perfectly fine. <identity impersonate="true" userName="accountname" password="password" /> But the above is not desirable since it impersonates a specific user for all the requests. What I need is to impersonate a specific user only when I run a particular section of code. Also, the impersonation code works absolutely fine when the shared folder is present on another WindowsServer 2008. Could anyone give me ideas on what's going wrong here.

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  • PhpBB3: adding background to specific php generated text input without affecting the other text inputs

    - by user1780055
    I have created a custom PhpBB3 style and desperately since a few hours tried to add a background image to a specific comment text area. With firebug I checked if the comment text area had a class and it does, so I tried some css variations and finally tried: sn-inputComment { background: url("{T_THEME_PATH}/images/pencil.png") repeat-x left top #FFFFFF;} { I also tried to find and manipulate the php generated text area but no success. Non of my methods worked. I will provide you all with a tinylink url to my forum with a test user and password access. User: test Password: 123456 url: http://tinyurl.com/9yqpxdb Now when you are logged in you should be redirected to the correct url and you will see a a few text boxes with "Write a comment...". I would be very happy if you could tell me what I did wrong, why im not able to add a background to the text input without having my search boxes and "what is on your mind box" affected. I appreciate your time and hope that this can be somehow solved. Sincerely, Daniel

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  • problem configure JBoss to work with JNDI(2)

    - by Spiderman
    in continuation to the question from last week: http://stackoverflow.com/questions/2828237/problem-configure-jboss-to-work-with-jndi I'm trying to bind datasource in JBoss and use it in my application. In my struggling, I already managed to avoid the javax.naming.NameNotFoundException by: 1. using in java new InitialContext().lookup(connection); instead of new JndiObjectFactoryBean().setJndiName(connection); 2. changing the connection name from: 'jndi-name' to 'java:jndi-name' Now the problem is that the datasouce that I get from the lookup is null. I created the datsource file: <datasources> <local-tx-datasource> <jndi-name>bilby</jndi-name> <connection-url>jdbc:oracle:myURL</connection-url> <driver-class>oracle.jdbc.OracleDriver </driver-class> <user-name>myUsername</user-name> <password>myPassword</password> <exception-sorter-class- name>org.jboss.resource.adapter.jdbc.vendor.OracleExceptionSorter</exception-sorter-class-name> <metadata> <type-mapping>Oracle9i</type-mapping> </metadata> </local-tx-datasource> </datasources> and put it under \server\default\deploy\oracle-ds.xml I get during runtime the line: 18:37:56,560 INFO [ConnectionFactoryBindingService] Bound ConnectionManager 'jb oss.jca:service=DataSourceBinding,name=bilby' to JNDI name 'java:bilby' So my question is - why do I get null as my datasource???

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  • Servlet Security question about j_security_check, j_username and j_password

    - by Nitesh Panchal
    Hello, I used jdbcRealm in my web application and it's working fine. I defined all constraints also in my web.xml. Like all pages of url pattern /Admin/* should be accessed by only admin. I have a login form with uses standard j_security_check, j_username and j_password. Now, when i type Admin/home.jsf it rightly redirects me login.jsf and there when i type the password i am redirected to home.jsf. This works alright but problem comes i directly go to login.jsf and then type password and username. This time it again redirects me to login.jsf. Is there any way through which i can specify which page to go when successful login is there? I need to specify different different pages for different roles. For Admin, it is /Admin/home.jsf for general users it is /General/home.jsf because login form is shared between different type of users. Where do i specify all these things? Secondly, i want to have a remember me checkbox at the end of login form. How do i do this? By default, it is submitted to j_security_check servlet and i have no control over its execution. Please help. This doesn't seem so hard but looks like i am missing something.

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  • Using $this when not in object context--i am using the latest version of php and mysql

    - by user309381
    This is user.php: include("databse.php");//retrieving successfully first name and lastname from databse file into user.php class user { public $first_name; public $last_name; public static function full_name() { if(isset($this->first_name) && isset($this->last_name)) { return $this->first_name . " " . $this->last_name; } else { return ""; } } } ?> Other php file, index.php: <?php include(databse.php); include(user.php); $record = user::find_by_id(1); $object = new user(); $object->id = $record['id']; $object->username = $record['username']; $object->password = $record['password']; $object->first_name = $record['first_name']; $object->last_name = $record['last_name']; // echo $object->full_name(); echo $object->id;// successfully print the id echo $object->username;//success fully print the username echo->$object->full_name();//**ERROR:Using $this when not in object context** ?>

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  • INSERT INTO table doesn't work???

    - by Joann
    I found a tutorial from nettuts and it has a source code in it so tried implementing it in my site.. It is working now. However, it doesn't have a Registration system so I am making one. The thing is, as I have expected, my code is not working... It doesn't seem to know how to INSERT into the database. Here's the function that inserts data into the db. function register_User($un, $email, $pwd) { $query = "INSERT INTO users( username, password, email ) VALUES(:uname, :pwd, :email) LIMIT 1"; if($stmt = $this->conn->prepare($query)) { $stmt->bind_param(':uname', $un); $stmt->bind_param(':pwd', $pwd); $stmt->bind_param(':email', $email); $stmt->execute(); if($stmt->fetch()) { $stmt->close(); return true; } else return "The username or email you entered is already in use..."; } } I have debugged the connection to the database from within the class, it says it's connected. I tried using this method instead: function register($un, $email, $pwd) { $registerquery = $this->conn->query( "INSERT INTO users(uername, password, email) VALUES('".$un."', '".$pwd."', '".$email."')"); if($registerquery) { echo "<h4>Success</h4>"; } else { echo "<h4>Error</h4>"; } } And it echos "Error"... Can you please help me pen point the error in this??? :(

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  • c# Sending emails with authentication. standard approach not working

    - by Ready Cent
    I am trying to send an email using the following very standard code. However, I get the error that follow... MailMessage message = new MailMessage(); message.Sender = new MailAddress("[email protected]"); message.To.Add("[email protected]"); message.Subject = "test subject"; message.Body = "test body"; SmtpClient client = new SmtpClient(); client.Host = "mail.myhost.com"; //client.Port = 587; NetworkCredential cred = new NetworkCredential(); cred.UserName = "[email protected]"; cred.Password = "correct password"; cred.Domain = "mail.myhost.com"; client.Credentials = cred; client.UseDefaultCredentials = false; client.Send(message); Mailbox unavailable. The server response was: No such user here. This recipient email address definitely works. To make this account work I had to do some special steps in outlook. Specifically, I had to do change account settings - more settings - outgoing server - my outgoing server requires authentication & use same settings. I am wondering if there is some other strategy. I think the key here is that my host is Server Intellect and I know that some people on here use them so hopefully someone else has been able to get through this. I did talk to support but they said with coding issues I am on my own :o

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  • odd nullreference error at foreach when rendering view

    - by giddy
    This error is so weird I Just can't really figure out what is really wrong! In UserController I have public virtual ActionResult Index() { var usersmdl = from u in RepositoryFactory.GetUserRepo().GetAll() select new UserViewModel { ID = u.ID, UserName = u.Username, UserGroupName = u.UserGroupMain.GroupName, BranchName = u.Branch.BranchName, Password = u.Password, Ace = u.ACE, CIF = u.CIF, PF = u.PF }; if (usersmdl != null) { return View(usersmdl.AsEnumerable()); } return View(); } My view is of type @model IEnumerable<UserViewModel> on the top. This is what happens: Where and what exactly IS null!? I create the users from a fake repository with moq. I also wrote unit tests, which pass, to ensure the right amount of mocked users are returned. Maybe someone can point me in the right direction here? Top of the stack trace is : at lambda_method(Closure , User ) at System.Linq.Enumerable.WhereSelectArrayIterator`2.MoveNext() at ASP.Index_cshtml.Execute() Is it something to do with linq here? Tell me If I should include the full stack trace.

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