Search Results

Search found 12055 results on 483 pages for 'password complexity'.

Page 232/483 | < Previous Page | 228 229 230 231 232 233 234 235 236 237 238 239  | Next Page >

  • 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

    Read the article

  • How to write custom (odd) authentication plugins for Wordpress, Joomla and MediaWiki?

    - by Bart van Heukelom
    On our network (a group of related websites - not a LAN) we have a common authentication system which works like this: On a network site ("consumer") the user clicks on a login link This redirects the user to a login page on our auth system ("RAS"). Upon successful login the user is directed back to the consumer site. Extra data is passed in the query string. This extra data does not include any information about the user yet. The consumer site's backend contacts RAS, with this extra data, to get the information about the logged in user (id, name, email, preferences, etc.). So as you can see, the consumer site knows nothing about the authentication method. It doesn't know if it's by username/password, fingerprint, smartcard, or winning a game of poker. This is the main problem I'm encountering when trying to find out how I could write custom authentication plugins for these packages, acting as consumer sites: Wordpress Joomla MediaWiki For example Joomla offers a pretty simple auth plugin system, but it depends on a username/password entered on the Joomla site. Any hints on where to start?

    Read the article

  • best practice - loging events (general) and changes (database)

    - by b0x0rz
    need help with logging all activities on a site as well as database changes. requirements: * should be in database * should be easily searchable by initiator (user name / session id), event (activity type) and event parameters i can think of a database design but either it involves a lot of tables (one per event) so i can log each of the parameters of an event in a separate field OR it involves one table with generic fields (7 int numeric and 7 text types) and log everything in one table with event type field determining what parameter got written where (and hoping that i don't need more than 7 fields of a certain type, or 8 or 9 or whatever number i choose)... example of entries (the usual things): [username] login failed @datetime [username] login successful @datetime [username] changed password @datetime, estimated security of password [low/ok/high/perfect] @datetime [username] clicked result [result number] [result id] after searching for [search string] and got [number of results] @datetime [username] clicked result [result number] [result id] after searching for [search string] and got [number of results] @datetime [username] changed profile name from [old name] to [new name] @datetime [username] verified name with [credit card type] credit card @datetime datbase table [table name] purged of old entries @datetime etc... so anyone dealt with this before? any best practices / links you can share? i've seen it done with the generic solution mentioned above, but somehow that goes against what i learned from database design, but as you can see the sheer number of events that need to be trackable (each user will be able to see this info) is giving me headaches, BUT i do LOVE the one event per table solution more than the generic one. any thoughts? edit: also, is there maybe an authoritative list of such (likely) events somewhere? thnx stack overflow says: the question you're asking appears subjective and is likely to be closed. my answer: probably is subjective, but it is directly related to my issue i have with designing a database / writing my code, so i'd welcome any help. also i tried narrowing down the ideas to 2 so hopefully one of these will prevail, unless there already is an established solution for these kinds of things.

    Read the article

  • KeyStore, HttpClient, and HTTPS: Can someone explain this code to me?

    - by stormin986
    I'm trying to understand what's going on in this code. KeyStore trustStore = KeyStore.getInstance(KeyStore.getDefaultType()); FileInputStream instream = new FileInputStream(new File("my.keystore")); try { trustStore.load(instream, "nopassword".toCharArray()); } finally { instream.close(); } SSLSocketFactory socketFactory = new SSLSocketFactory(trustStore); Scheme sch = new Scheme("https", socketFactory, 443); httpclient.getConnectionManager().getSchemeRegistry().register(sch); My Questions: trustStore.load(instream, "nopassword".toCharArray()); is doing what exactly? From reading the documentation load() will load KeyStore data from an input stream (which is just an empty file we just created), using some arbitrary "nopassword". Why not just load it with null as the InputStream parameter and an empty string as the password field? And then what is happening when this empty KeyStore is being passed to the SSLSocketFactory constructor? What's the result of such an operation? Or -- is this simply an example where in a real application you would have to actually put a reference to an existing keystore file / password?

    Read the article

  • 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.

    Read the article

  • Best practice - logging events (general) and changes (database)

    - by b0x0rz
    need help with logging all activities on a site as well as database changes. requirements: * should be in database * should be easily searchable by initiator (user name / session id), event (activity type) and event parameters i can think of a database design but either it involves a lot of tables (one per event) so i can log each of the parameters of an event in a separate field OR it involves one table with generic fields (7 int numeric and 7 text types) and log everything in one table with event type field determining what parameter got written where (and hoping that i don't need more than 7 fields of a certain type, or 8 or 9 or whatever number i choose)... example of entries (the usual things): [username] login failed @datetime [username] login successful @datetime [username] changed password @datetime, estimated security of password [low/ok/high/perfect] @datetime [username] clicked result [result number] [result id] after searching for [search string] and got [number of results] @datetime [username] clicked result [result number] [result id] after searching for [search string] and got [number of results] @datetime [username] changed profile name from [old name] to [new name] @datetime [username] verified name with [credit card type] credit card @datetime datbase table [table name] purged of old entries @datetime via automated process etc... so anyone dealt with this before? any best practices / links you can share? i've seen it done with the generic solution mentioned above, but somehow that goes against what i learned from database design, but as you can see the sheer number of events that need to be trackable (each user will be able to see this info) is giving me headaches, BUT i do LOVE the one event per table solution more than the generic one. any thoughts? edit: also, is there maybe an authoritative list of such (likely) events somewhere? thnx stack overflow says: the question you're asking appears subjective and is likely to be closed. my answer: probably is subjective, but it is directly related to my issue i have with designing a database / writing my code, so i'd welcome any help. also i tried narrowing down the ideas to 2 so hopefully one of these will prevail, unless there already is an established solution for these kinds of things.

    Read the article

  • 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.

    Read the article

  • 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>

    Read the article

  • How to ensure DB security for a Windows Forms application?

    - by Vilx-
    The basic setup is classic - you're creating a Windows Forms application that connects to a DB and does all kinds of enterprise-y stuff. Naturally, such an application will have many users with different access rights in the DB, and each with their own login name and password. So how do you implement this? One way is to create a DB login for every application user, but that's a pretty serious thing to do, which even requires admin rights on the DB server, etc. If the DB server hosts several applications, the admins are quite likely not to be happy with this. In the web world typically one creates his own "Users" table which contains all the necessary info, and uses one fixed DB login for all interaction. That is all nice for a web app, but a windows forms can't hide this master login information, negating security altogether. (It can try to hide, but all such attempts are easily broken with a bit of effort). So... is there some middle way? Perhaps logging in with a fixed login, and then elevating priviledges from a special stored procedure which checks the username and password?

    Read the article

  • Image appearing in the wrong place.

    - by Luke
    I have a list that I want to precede on each line with a small image. The CSS: .left div.arrow{background-image: url(../images/design/box.png); float:left; height:15px; width:15px; margin-right:2px;} .left a.list:link, a:visited, a:active {color:black; font-size:0.8em; text-decoration:none; display:block; float:left;} The HTML: <div class="panel">My quick links</div> <div class="arrow"></div> <a href="/messages.php?p=new" class="list">Send a new message</a> <br /> <div class="arrow"></div> <a href="/settings.php#password" class="list">Change my password</a> <br /> <div class="arrow"></div> <a href="/settings.php#picture" class="list">Upload a new site image</a> <br /> As you can see, each image should go before the writing. On the first line, the text "Send a new message" has the preceeding image. However, each line afterwards has the image at the end. So "Send a new message" has an image at the start and finish. It is like the images are staying at the end of the previous line. Any ideas?

    Read the article

  • 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 ?

    Read the article

  • ASP.NET Login Control rejects users who exist

    - by rsteckly
    Hi, I'm having some trouble with the ASP.NET 2.0 Login Control. I've setup a database with the aspI.net regsql tool. I've checked the application name. It is set to "/". The application can access the SQL Server. In fact, when I go to retrieve the password, it will even send me the password. Despite this, the login control continues to reject logins. I added this to the web.config: <membership defaultProvider="AspNetSqlProvider"> <providers> <clear/> <add name="AspNetSqlProvider" connectionStringName="LocalSqlServer" applicationName="/" type="System.Web.Security.SqlMembershipProvider, System.Web, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a"/> </providers> And I added the following to my connection strings: <remove name="LocalSqlServer" /> <add name="LocalSqlServer" connectionString="Data Source=IDC-4\EXCALIBUR;Initial Catalog=allied_nr;Integrated Security=True;Asynchronous Processing=True"/> (Note the "remove name" is to get rid of the default connection string in the App_Data directory.) Why won't the login control authenticate users?

    Read the article

  • 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 ?

    Read the article

  • OleDB connection only working when debugging

    - by Francesc
    I have a C# application that connects to a named SQL Express instance on the local machine using OleDBConnection: _connection = new OleDbConnection(_strConn); _connection.Open(); _strConn is something like this: "Provider=sqloledb;Data Source=.\NAMEDINSTANCE;Initial Catalog=dbname;User Id=sa;Password=password;" If I debug the application, the connection works fine. If I run the application from Windows Explorer (the same debug compilation), I get an "OleDBException: Login timeout expired" in the Open() line after 30 seconds. The strange thing is the exception happens even if I attach the debugger to the exe. I can see that the connection string is correct and everything seems fine. I can't fine any extra information in the SQL Express error log or SQL Activity Monitor either. If it helps, here is the exception: System.Data.OleDb.OleDbException: Login timeout expired at System.Data.OleDb.OleDbConnectionInternal..ctor(OleDbConnectionString constr, OleDbConnection connection) at System.Data.OleDb.OleDbConnectionFactory.CreateConnection(DbConnectionOptions options, Object poolGroupProviderInfo, DbConnectionPool pool, DbConnection owningObject) at System.Data.ProviderBase.DbConnectionFactory.CreateNonPooledConnection(DbConnection owningConnection, DbConnectionPoolGroup poolGroup) at System.Data.ProviderBase.DbConnectionFactory.GetConnection(DbConnection owningConnection) at System.Data.ProviderBase.DbConnectionClosed.OpenConnection(DbConnection outerConnection, DbConnectionFactory connectionFactory) at System.Data.OleDb.OleDbConnection.Open() I imagine that find the issue with the information I give here might be difficult, but I don't know where else to look or what other tests to do, so any ideas on what it could be or what test I could do to find out will be really appreciated.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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,

    Read the article

  • PHP Transferring Photos From One Oracle Database Table to Another

    - by Jonathan Swift
    I am attempting to transfer a set of photos (blobs) from one table to another across databases. I'm nearly there, except for binding the photo parameter. I have the following code: $conn_db1 = oci_pconnect('username', 'password', 'db1'); $conn_db2 = oci_pconnect('username', 'password', 'db2'); $parse_db1_select = oci_parse($conn_db1, "SELECT REF PID, BINARY_OBJECT PHOTOGRAPH FROM BLOBS"); $parse_db2_insert = oci_parse($conn_db2, "INSERT INTO PHOTOGRAPHS (PID, PHOTOGRAPH) VALUES (:pid, :photo)"); oci_execute($parse_db1_select); while ($row = oci_fetch_assoc($parse_db1_select)) { $pid = $row['PID']; $photo = $row['PHOTOGRAPH']; oci_bind_by_name($parse_db2_insert, ':pid', $pid, -1, OCI_B_INT); // This line causes an error oci_bind_by_name($parse_db_insert, ':photo', $photo, -1, OCI_B_BLOB); oci_execute($parse_db2_insert); } oci_close($db1); oci_close($db2); But I get the following error, on the error line commented above: Warning: oci_execute() [function.oci-execute]: ORA-03113: end-of-file on communication channel Process ID: 0 Session ID: 790 Serial number: 118 Does anyone know the right way to do this?

    Read the article

  • 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

    Read the article

  • Remove this URL string when login fails and simply show div error

    - by Anagio
    My developer built our registration page to display a div when logins failed based on a string in the URL. When logins fail this is added to the URL /login?msg=invalid The PHP in my login.phtml which displays the error messages based on the msg= parameter is <?php $msg = ""; $msg = $_GET['msg']; if($msg==""){ $showMsg = ""; } elseif($msg=="invalid"){ $showMsg = ' <div class="alert alert-error"> <a class="close" data-dismiss="alert">×</a> <strong>Error!</strong> Login or password is incorrect! </div>'; } elseif($msg=="disabled"){ $showMsg = "Your account has been disabled."; } elseif($msg==2){ $showMsg = "Your account is not activated. Please check your email."; } ?> In the controller the redirect to that URL is else //email id does not exist in our database { //redirecting back with invalid email(invalid) msg=invalid. $this->_redirect($url."?msg=invalid"); } I know there are a few other validation types for disabled accounts etc. I'm in the process of redesigning the entire interface and would like to get rid of this kind of validation so that the div tags display when logins fail but not show the URL strings. If it matters the new div I want to display is <div class="alert alert-error alert-login"> Email or password incorrect </div> I'd like to replace the php my self in my login.phtml and controller but not a good programmer. What can I replace $this->_redirect($url."?msg=invalid"); with so that no strings are added to the URL and display the appropriate div tags? Thanks

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • 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.

    Read the article

  • 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

    Read the article

  • 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?

    Read the article

< Previous Page | 228 229 230 231 232 233 234 235 236 237 238 239  | Next Page >