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• Missing on-screen keyboard for Flash/Flex web application on Android?

  - by HDave

  I have an enterprise Flex web application, served up over https, that runs fine on Flash player 10.0 and beyond from a desktop computer/browser. However, when I run it from my HTC Incredible with Android 2.2 the app loads fine, but there is no on-screen keyboard and so I cannot log in. I can see the blinking cursor inside the username and password text fields. I can switch between them. I can even hit the login button and see an authentication error! According to Adobe this should Just Work. Any ideas? I wanna show off our spanking new app to strangers at the bus stop!

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• Creating a Variable of Present Date Minus a Past Timestamp

  - by John

  Hello, In the code below, "created" is a field in a MySQL table. This field is of the type "timestamp" and the default is set to "CURRENT_TIMESTAMP" of whenever a given row is created. In the query below, I would like to create a new variable that equals the present date minus the timestamp of "created", rounded off to units of days. I would like the present date to be whenever the query is run. How could I do this? Thanks in advance, John $sqlStr = "SELECT l.loginid, l.username, l.created, ...

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• Allowing New Users to Invite Their Gmail Contacts

  - by John

  Hello, For my site, I would like to give new users the option to invite all of their Gmail contacts to join. What is the basic step-by-step process to set this up? (Also, is it necessary to buy an SSL for this?) Thanks in advance, John EDIT: My site has a basic login where users set up a username and password. I would like to give users the option to invite their Gmail contacts right after they create their new profile. I would also like to give them the option to invite their Gmail contacts anytime they want.

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• Station ID - more than IP

  - by bensiu

  Hello I am working on internal PHP application where users login only from our network (we are on dedicated IP and application is checking $_SERVER['REMOTE_ADDR'] and if is match our IP go thru if not - Bye However how I can identify from which station user is login ? Using cookie not gonna work - those are windows station working under control MS StedyState and all cookies are deleted every login I don't need nothing fancy like full MAC address just any unique ID (can not be application username - because users use different stations) Any ideas ? Maybe Javascript is able to grab some Unique detail from station's browser (IE 8) and pass this as hidden input? bensiu

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• phpMyAdmin - can't connect - invalid setings - ever since I added a root password - locked out

  - by OrangeRind

  I run XAMPP, a few days back i had set up a password for the root password through phpmyadmin I am not able to access phpMyAdmin ever since that moment I followed help on this link but everything seems fine there (in config.inc.php). I even tried unistalling xampp fully, restarting windows and then reinstalling xampp, but still pointing to localhost/phpmyadmin I get the following error MySQL said: Cannot connect: invalid settings. phpMyAdmin tried to connect to the MySQL server, and the server rejected the connection. You should check the host, username and password in your configuration and make sure that they correspond to the information given by the administrator of the MySQL server. I Also tried to reset root password through mysqld.bat as given on mysql's website help but to no avail Please Help!

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• SharePoint form-based authentication with custom database

  - by Clodin

  Hi, I have SharePoint site and I want to use form-based authentication, not Windows how it is by default. For this I read that I have to modify the web.config from Central Administration and web.config from my site with the membership and roleManager tags configured properly. But if I use this: <membership> <providers> <add name="MyProvider" type="System.Web.Security.SqlMembershipProvider, System.Web, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b03f5f7f11d50a3a" .../> </providers> </membership> System.Web.Security.SqlMembershipProvider requires a database generated with ASP.NET SQL Server Setup Wizard (aspnet_regsql.exe), and this is my problem! I want to use another database with cunstom table 'Users' from where to take the username and password for authentication. How can I do this? Thank you in advance

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• Programmatically login to a website and redirect the user to the logged in page?

  - by Santhosh

  Hi, Right now, I have all the employees of my company login to an external website using the company id, username and a password. We are trying to integrate it into an intranet portal which should provide seamless access to this website without requiring the user to enter these credentials. Is there any way of doing this programmatically (.NET C#)? Very similar to screenscraping, Can I simulate the appropriate POST action and then redirect the user to the logged in page? Any help is appreciated. Thanks.

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• PHP registration script: verifying and telling the user what went wrong

  - by Maxime

  Hi, I'm building a registration script and I want to insert the user's input (username, mail) ONLY if it's not already in the database. What I usually do in such cases is a request to see if something's already there, something like: "SELECT * FROM things WHERE thing_name = '$treated_user_input'". I have two fields that need to be unique this time though. Is there a way to do only one SELECT request and still be able to tell the user exactly what field went wrong? Or do I have to do one request per unique field? Thanks for your answers.

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• SQL Server Reporting Services 2008: How to set the credentials property properly?

  - by wgpubs

  No matter how I configure the Credentials property I get a 401 exception when I try to Render the report. Here is my (latest) code: var rs = new ReportExecutionService(); rs.Url = "https://myserver/reportserver/reportexecution2005.asmx"; var myCache = new System.Net.CredentialCache(); myCache.Add(new Uri(rs.Url), "kerberos" , new System.Net.NetworkCredential("username", "password", "Domain")); rs.Credentials = myCache; The URL and credentials are all correct. But still getting a 401 when I cal rs.Render(...). The Reporting Services install is sitting on a Windows Server 2008 box and requires integrated authentication. Thanks

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• What is wrong with my Basic Authentication in FireFox?

  - by Pure.Krome

  Hi folks, i'm trying to goto the following url :- http://user1:pass1@localhost:1234/api/users?format=xml nothing to complex. Notice how i've got the username/password in the url? this, i believe, is for basic authentication. When i do that, the Request Headers are MISSING the 'Authorize' header. Er... that's not right :( I have anonymous authentication only setup on the site. I don't want to have anon off and basic turned on .. because not all of the site requires basic.. only a few action methods. So .. why is this not working? Is this something to do with the fact my code is not sending a 401 challenge or some crap? For What It's Worth, my site is ASP.NET MVC1 running on IIS7 (and the same thing happens when i run it on cassini).

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• Multiple ports listed in SQL Server connection string

  - by BBlake

  I have a legacy VB6 app where the servername, databasename, username, etc are defined in an INI file, but the port number for the connection string (the default 1433) is hard coded in the app. It's being moved to a new sql server back end that runs off a different port number. I'm trying to avoid having to alter and recompile the application which entails signifigant retesting, documentation, etc. I tried altering the INI file so that for the new server I have put in: SERVERNAME\INSTANCE,NEWPORTNUMBER This effectively builds the connection with Data Source = SERVERNAME\INSTANCE,NEWPORTNUMBER,1433; This appears to work correctly as it connects to the database when I run the app. It appears to me that the ,1433 portion is being ignored. Is this a valid assumption or will this cause me some problem I'm not seeing here?

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• Using sessions & session variables in a PHP Login Script

  - by user1325291

  I have just finished creating an entire login and register systsem in PHP, but my problem is I haven't used any sessions yet. (Like, there's not the word "session" in the entire script). I'm kind of a newbie in PHP and I've never used sessions before. What I want to do it, after the user registers and fills out the login form, they will still stay on the same page. So, like one half of the page will be if the session is logged_in and the other part will be else (the user is not logged in so display the login form). I also want to create session variables which will hold the username and password given at login. Thanks in advance! BTW this is my website - CP Cheats

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• Error in executing

  - by Saranya.R

  Hi........ I am interested in sending tweets using Java program.I wrote the following prgram.It doen't show any error in compiling.But while executing it shows "Exception in thread "main" java.lang.NoClassDefFoundError: hi Caused by: java.lang.ClassNotFoundException: hi at java.net.URLClassLoader$1.run(URLClassLoader.java:200) at java.security.AccessController.doPrivileged(Native Method) at java.net.URLClassLoader.findClass(URLClassLoader.java:188) at java.lang.ClassLoader.loadClass(ClassLoader.java:307) at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301) at java.lang.ClassLoader.loadClass(ClassLoader.java:252) at java.lang.ClassLoader.loadClassInternal(ClassLoader.java:320) Could not find the main class: hi. Program will exit." This is my code.. package twitter; //import java.lang.object; import net.unto.twitter.*; import net.unto.twitter.TwitterProtos.Status; public class hi { public static void main(String args[]) { Api api = Api.builder().username("usename").password("password").build(); api.updateStatus("This is a test message.").build().post(); } } Can anybody help me ..Pls

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• http authenitcation in xcode

  - by user313100

  I am trying to make twitter work in my app and everything works fine except the code does not seem to recognize an error from twitter. If the username/password are not valid, I get an error message through this function: - (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data { NSString* strData = [[[NSString alloc] initWithBytes:[data bytes] length:[data length] encoding:NSASCIIStringEncoding] autorelease]; NSLog(@"Received data: %@", strData ) ; return ; } It prints: Received data: Could not authenticate you. . However, the app continues to the post a tweet view I have and ignores the error. Obviously, I do not have something setup right to detect such an error from twitter so my question is how do I get xcode to recognize an error like this? This uses basic http auth btw and don't mention anything about OAuth...just trying to get this to work for now.

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• Error creating a table : "There is already an object named ... in the database", but not object with

  - by DavRob60

  Hi, I'm trying to create a table on a Microsoft SQL Server 2005 (Express). When i run this query USE [QSWeb] GO /****** Object: Table [dbo].[QSW_RFQ_Log] Script Date: 03/26/2010 08:30:29 ******/ SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ON GO SET ANSI_PADDING ON GO CREATE TABLE [dbo].[QSW_RFQ_Log]( [RFQ_ID] [int] NOT NULL, [Action_Time] [datetime] NOT NULL, [Quote_ID] [int] NULL, [UserName] [nvarchar](256) NOT NULL, [Action] [int] NOT NULL, [Parameter] [int] NULL, [Note] [varchar](255) NULL, CONSTRAINT [QSW_RFQ_Log] PRIMARY KEY CLUSTERED ( [RFQ_ID] ASC, [Action_Time] ASC )WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY] ) ON [PRIMARY] GO I got this error message Msg 2714, Level 16, State 4, Line 2 There is already an object named 'QSW_RFQ_Log' in the database. Msg 1750, Level 16, State 0, Line 2 Could not create constraint. See previous errors. but if i try to find the object in question using this query: SELECT * FROM QSWEB.sys.all_objects WHERE upper(name) like upper('QSW_RFQ_%') I got this (0 row(s) affected) What is going on????

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• Making a Link Appear if a Condition is Met

  - by John

  Hello, The PHP code below echoes a link if the "loginid" of the logged-in user is on a list determined by getEditorList();. It works fairly well, but I think it might work better if I were to do it with Javascript instead. How could I accomplish the same thing with Javascript? Thanks in advance, John $editors = getEditorsList(); foreach($editors as $editor) { $editorids[] = $editor['loginid']; } if(in_array($_SESSION['loginid'], $editorids)) { echo "<div class='footervote'><a href='http://www...com/.../footervote.php'>Vote</a></div>"; } Login function: <?php if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { show_userbox(); } ?>

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• Connect to a MySQL database and count the number of rows.

  - by Hugo

  Hi there! I need to connect to a MySQL database and then show the number of rows. This is what I've got so far; <?php include "connect.php"; db_connect(); $result = mysql_query("SELECT * FROM hacker"); $num_rows = mysql_num_rows($result); echo $num_rows; ?> When I use that code I end up with this error; Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\username\Desktop\xammp\htdocs\news2\results.php on line 10 Thanks in advance :D

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• A column ID occurred more than once in the specification

  - by Puzzle84

  Recently i've picked up my EF 4.1 / MVC 3 project again and started building in actual frontend capabilities. Now i'm developing a "simple" message system but upon going to that page i get the error as stated in the title EDIT It creates the database just not the models. Stack trace: [NullReferenceException: Object reference not set to an instance of an object.] ASP._Page_Views_Inbox_Index_cshtml.Execute() in c:\Development\MVC\DOCCL\Views\Inbox\Index.cshtml:18 System.Web.WebPages.WebPageBase.ExecutePageHierarchy() +197 System.Web.Mvc.WebViewPage.ExecutePageHierarchy() +81 System.Web.WebPages.StartPage.RunPage() +17 System.Web.WebPages.StartPage.ExecutePageHierarchy() +62 System.Web.WebPages.WebPageBase.ExecutePageHierarchy(WebPageContext pageContext, TextWriter writer, WebPageRenderingBase startPage) +76 System.Web.Mvc.RazorView.RenderView(ViewContext viewContext, TextWriter writer, Object instance) +222 System.Web.Mvc.BuildManagerCompiledView.Render(ViewContext viewContext, TextWriter writer) +115 System.Web.Mvc.ViewResultBase.ExecuteResult(ControllerContext context) +295 System.Web.Mvc.ControllerActionInvoker.InvokeActionResult(ControllerContext controllerContext, ActionResult actionResult) +13 System.Web.Mvc.<c_DisplayClass1c.b_19() +23 System.Web.Mvc.ControllerActionInvoker.InvokeActionResultFilter(IResultFilter filter, ResultExecutingContext preContext, Func1 continuation) +242 System.Web.Mvc.<>c__DisplayClass1e.<InvokeActionResultWithFilters>b__1b() +21 System.Web.Mvc.ControllerActionInvoker.InvokeActionResultWithFilters(ControllerContext controllerContext, IList1 filters, ActionResult actionResult) +177 System.Web.Mvc.ControllerActionInvoker.InvokeAction(ControllerContext controllerContext, String actionName) +324 System.Web.Mvc.Controller.ExecuteCore() +106 System.Web.Mvc.ControllerBase.Execute(RequestContext requestContext) +91 System.Web.Mvc.ControllerBase.System.Web.Mvc.IController.Execute(RequestContext requestContext) +10 System.Web.Mvc.<c_DisplayClassb.b_5() +34 System.Web.Mvc.Async.<c_DisplayClass1.b_0() +19 System.Web.Mvc.Async.<c_DisplayClass81.<BeginSynchronous>b__7(IAsyncResult _) +10 System.Web.Mvc.Async.WrappedAsyncResult1.End() +62 System.Web.Mvc.<c_DisplayClasse.b_d() +48 System.Web.Mvc.SecurityUtil.b_0(Action f) +7 System.Web.Mvc.SecurityUtil.ProcessInApplicationTrust(Action action) +22 System.Web.Mvc.MvcHandler.EndProcessRequest(IAsyncResult asyncResult) +60 System.Web.Mvc.MvcHandler.System.Web.IHttpAsyncHandler.EndProcessRequest(IAsyncResult result) +9 System.Web.CallHandlerExecutionStep.System.Web.HttpApplication.IExecutionStep.Execute() +9478661 System.Web.HttpApplication.ExecuteStep(IExecutionStep step, Boolean& completedSynchronously) +178 InnerException : {"A column ID occurred more than once in the specification."} The recently added code is. Controller: // // GET: /Inbox/Index/5/1 public ActionResult Index(int? Id, int Page = 1) { try { const int pageSize = 10; var messages = from m in horseTracker.Messages where m.ReceiverId.Equals(Id) select m; var paginatedMessages = new PaginatedList<Message>(messages, Page, pageSize); return View(paginatedMessages); } catch (Exception ex) { } return View(); } Models public class Message { [Key] public int Id { get; set; } [Required(ErrorMessage = "Subject is required")] [Display(Name = "Subject")] public string Subject { get; set; } [Required(ErrorMessage = "Message is required")] [Display(Name = "Message")] public string Content { get; set; } [Required] [Display(Name = "Date")] public DateTime Created { get; set; } public Boolean Read { get; set; } [Required(ErrorMessage = "Can't create a message without a user")] public int SenderId { get; set; } public virtual User Sender { get; set; } [Required(ErrorMessage = "Please pick a recipient")] public int ReceiverId { get; set; } public virtual User Receiver { get; set; } } public class User { [Key] public int Id { get; set; } [Required] [Display(Name = "Username")] public string UserName { get; set; } [Required] [Display(Name = "First Name")] public string FirstName { get; set; } [Required] [Display(Name = "Last Name")] public string LastName { get; set; } [Required] [Display(Name = "E-Mail")] public string Email { get; set; } [Required] [Display(Name = "Password")] public string Password { get; set; } [Required] [Display(Name = "Country")] public string Country { get; set; } public string EMail { get; set; } //Races public virtual ICollection<Message> Messages { get; set; } } modelBuilder.Entity<User>() .HasMany(u => u.Messages) .WithRequired(m => m.Receiver) .HasForeignKey(m => m.ReceiverId) .WillCascadeOnDelete(false); Anyone have a clue on why i might be getting that error? Before i added these classes it was working fine.

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• I want to keep the values on textbox after onchange function

  - by user1908045

  hello i have problem with this code..I want to keep the values of the textboxes when the pageload and didnt write again the values. I have two drop down list. the First one is the country when the country selected then the page load and appear the city to select but afte the pageload the values on textbox is empty. I want to keep the values of textbox when the page load. This is the code <head> <script type="text/javascript"> function Load_id() { var Count = document.getElementById("Count").value; var Count_txt = "?Count=" location = Count_txt + Count } </script> <meta charset="UTF-8"> </head> <body> <div class="main"> <div class="headers"> <table> <tr><td rowspan="2"><img alt="unipi" src="/Images/logo.jpeg" height="75" width="52"></td> <td>University</td></tr> <tr><td>Data</td></tr> </table> </div> <div class="form"> <h3>Personal</h3><br/><br/><br/> <form id="Page1" name="Page1" action="Form1Sub.php" method="Post"> <table style="width:520px;text-align:left;"> <tr><td><label>Number:</label></td> <td><input type="text" required="required" id="AM" name="AM" value=""/></td> </tr> <tr><td><label>Name:</label></td> <td><input type="text" required="required" name="Name"/></td> </tr> <?php $host="localhost"; $username=""; $password=""; $dbName="Database"; $connection = mysql_connect($host, $username, $password) or die("Couldn't Connect to the Server"); $db = mysql_select_db($dbName, $connection) or die("cannot select DataBase"); $Count = $_GET['Count']; echo "<tr><td><label>Country</label></td>\n"; $country = mysql_query("select DISTINCT Country FROM lut_country_city "); echo " <td><select id=\"Count\" name=\"cat\" onChange=\"Load_id(this)\">\n"; echo " `<option>Select Country</option>\n"; while($nt=mysql_fetch_array($country)){ $selected = ($nt["Country"] == $Count)? "SELECTED":""; echo"<option value=\"".$nt['Country']."\"". $selected." >".$nt['Country']."</option>"; } echo " </select></td></tr>\n"; echo"<tr><td><label>City:</label></td>\n"; $q2 = mysql_query("Select id,City,Country FROM lut_country_city WHERE Country = '$Count'"); echo"<td><select name=\"SelectCity\">\n"; while($row = mysql_fetch_array($q2)) { echo"<option value=\"".$row['id']."\">".$row['City']."</option>"; } echo " </select></td></tr>\n"; ?> </table> <p> <button type="submit" id="Next">Next</button> </form> <form id="form1" action="index.php"> <button id="Back" type="submit">Back</button> </form> </p> </div> </div> </body> </html>

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• MySQL with Java: Open connection only if possible

  - by emempe

  I'm running a database-heavy Java application on a cluster, using Connector/J 5.1.14. Therefore, I have up to 150 concurrent tasks accessing the same MySQL database. I get the following error: Exception in thread "main" com.mysql.jdbc.exceptions.jdbc4.MySQLNonTransientConnectionException: Too many connections This happens because the server can't handle so many connections. I can't change anything on the database server. So my question is: Can I check if a connection is possible BEFORE I actually connect to the database? Something like this (pseudo code): check database for open connection slots if (slot is free) { Connection cn = DriverManager.getConnection(url, username, password); } else { wait ... } Cheers

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• git global config issue

  - by Andrew Bullock

  Somehow, my global git (msysgit) settings for user.name and user.email (and god knows what else) are set to a recent ex-colleague's details. When I try and change them i get could not commit to u://.gitconfig If I try and create u://.gitconfig through git bash then i get permission denied. C:\Users\<My Username>\ contains no references to git. I've tried uninstalling, searching the registry and my file system for all references to git and I can't find any (windows file search is crap though). What the hell is going on? Why even after reinstalling are this guys details still the global settings??? Thanks

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• Inheritance in tables - structure problem

  - by Naor

  I have 3 types of users in my system. each type has different information I created the following tables: BaseUser(base_user_id, username, password, additional common data) base_user_id is PK and Identity UserType1(user_id, data related to type1 only) user_id is PK and FK to base_user_id UserType2(user_id, data related to type2 only) user_id is PK and FK to base_user_id UserType3(user_id, data related to type3 only) user_id is PK and FK to base_user_id Now I have relation from each type of user to warehouses table. Users from type1 and type2 should have only warehouse_id and users from type3 should have warehouse_id and customer_id. I thought about this structure: WarehouseOfUser(base_user_id,warehouse_id) base_user_id is FK to base_user_id in BaseUser WarehouseOfTyp3User(base_user_id,warehouse_id, customer_id) base_user_id is FK to base_user_id in BaseUser The problem is that such structure allows 2 things I want to prevent: 1. add to WarehouseOfTyp3User data of user from type2 or type1. 2. add to WarehouseOfUser data of user from type3. what is the best structure for such case?

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• my jComboBox does not react to my keyListener and actionPerform perfroms weired stuff

  - by aladdin

  hi I am trying to search for UserName and return values onto jComboBox, here is the code public void actionPerformed(java.awt.event.ActionEvent e) { sr = new Search(((String) jComboBoxReceiver.getSelectedItem())); usrList = sr.searchUser(); String[] userList = new String[usrList.size()] ; for(int i=0;i<usrList.size();i++){ userList[i]= usrList.get(i).getUserName(); } model = new DefaultComboBoxModel(userList); jComboBoxReceiver.setModel(model); } However, if i do that, it does perform correctly, however, it will go search for the first item again, which is very confusing... then i tried using key Pressed if(e.getKeyCode()==13){ sr = new Search(((String) jComboBoxReceiver.getSelectedItem())); usrList = sr.searchUser(); String[] userList = new String[usrList.size()] ; for(int i=0;i<usrList.size();i++){ userList[i]= usrList.get(i).getUserName(); } model = new DefaultComboBoxModel(userList); jComboBoxReceiver.setModel(model); } And this one does not react at all ...

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• Change password in Task Scheduler in script

  - by titanium

  I'm changing password every month for all scheduled tasks I created in Task Scheduler. This is because our security policy expires our password every month. Due to increasing number of scheduled tasks I'm creating, it eats up a lot of time in just changing password within Task Scheduler. My question is: Is there a way in script to change password in one run specifying the tasks, DOMAIN\username, and password? I know there's a security risk in putting the password in script. The password in script will be removed after the running it.

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• How to use avg function?

  - by Marcelo

  I'm new at php and mysql stuff and i'm trying to use an avg function but i don't know how to. I'm trying to do something like this: mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die ("Did not connect to $database"); mysql_query("AVG(column1) FROM table1 ") or die(mysql_error()); mysql_close(); echo AVG(column1); (Q1)I'd like to see the value printed in the screen, but i'm getting nothing but an error message. How could I print this average on the screen ? (Q2)If I had a column month in my table1, how could I print the averages by the months ? Sorry for any bad English, and thanks for the attention.

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