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  • ASP.NET / Active Directory - Supporting auto login for domain users

    - by Krisc
    I am developing a simple ASP.NET website that will run on the intranet on a WS2008(IIS7) box and respond to users running XP/IE8. Everything is domain connected and I am trying to automatically login the users much like SharePoint does. On my dev machine (XP), when running the site through VS, everything works. I can pickup on the user perfectly. I am using the following settings: <authentication mode="Windows"/> <identity impersonate="true"/> <anonymousIdentification enabled="false"/> <authorization> <allow users="*"/> <deny users="?"/> </authorization> However, when I publish to the WS2008 box, it doesn't work. Clearly I am missing a setting in IIS7 to support this. I have the following set for Authentication on the site: Anon Auth - Enabled ASP.NET Impersonation - Enabled Basic Auth - Disabled Forms Auth - Disabled Windows Auth - Disabled What am I missing? Thanks

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  • Paypal (sandbox) buy now link redirects to paypal (sandbox) login page instead of order summary

    - by Nicolas
    Hi, Before going live I try to test the paypal process against the paypal sandbox mode, but after the summary of what the user is going to pay on my website(buy now button), the link does not redirect to a paypal summary of the prder but ask the user to connect to paypal. Even after logging into the buyer sandbox account there's no summary of the order. It just disappears. Here's is the code I use on the checkout page: <form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_s-xclick"> <input type="hidden" name="hosted_button_id" value="XXXXXXXXXXXXXXXX"> <input type="hidden" name="notify_url" value="http://www.website.com/paypal/"> <div class="suggestion"> <input type="image" src="https://www.sandbox.paypal.com/en_GB/i/btn/btn_paynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.sandbox.paypal.com/en_GB/i/scr/pixel.gif" width="1" height="1"> </div> </form> Any idea why it redirects to the payapl login page instead of the order one? Btw I'm using the Website Basic Payment (not PRO then). Cheers, Nicolas.

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  • Another Security Exception on GoDaddy after Login attempt

    - by Brian Boatright
    Host: GoDaddy Shared Hosting Trust Level: Medium The following happens after I submit a valid user/pass. The database has read/write permissions and when I remove the login requirement on an admin page that updates the database work as expected. Has anyone else had this issue or know what the problem is? Anyone? Server Error in '/' Application. Security Exception Description: The application attempted to perform an operation not allowed by the security policy. To grant this application the required permission please contact your system administrator or change the application's trust level in the configuration file. Exception Details: System.Security.SecurityException: Request for the permission of type 'System.Security.Permissions.FileIOPermission, mscorlib, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089' failed. Source Error: An unhandled exception was generated during the execution of the current web request. Information regarding the origin and location of the exception can be identified using the exception stack trace below. Stack Trace: [SecurityException: Request for the permission of type 'System.Security.Permissions.FileIOPermission, mscorlib, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089' failed.] System.Security.CodeAccessSecurityEngine.Check(Object demand, StackCrawlMark& stackMark, Boolean isPermSet) +0 System.Security.CodeAccessPermission.Demand() +59 System.IO.FileStream.Init(String path, FileMode mode, FileAccess access, Int32 rights, Boolean useRights, FileShare share, Int32 bufferSize, FileOptions options, SECURITY_ATTRIBUTES secAttrs, String msgPath, Boolean bFromProxy) +684 System.IO.FileStream..ctor(String path, FileMode mode, FileAccess access, FileShare share) +114 System.Configuration.Internal.InternalConfigHost.StaticOpenStreamForRead(String streamName) +80 System.Configuration.Internal.InternalConfigHost.System.Configuration.Internal.IInternalConfigHost.OpenStreamForRead(String streamName, Boolean assertPermissions) +115 System.Configuration.Internal.InternalConfigHost.System.Configuration.Internal.IInternalConfigHost.OpenStreamForRead(String streamName) +7 System.Configuration.Internal.DelegatingConfigHost.OpenStreamForRead(String streamName) +10 System.Configuration.UpdateConfigHost.OpenStreamForRead(String streamName) +42 System.Configuration.BaseConfigurationRecord.InitConfigFromFile() +437 Version Information: Microsoft .NET Framework Version:2.0.50727.1433; ASP.NET Version:2.0.50727.1433

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  • Unable to login using OpenID for google apps using vanity URL

    - by GeekTantra
    Unable to login using OpenID for google apps using vanity URL I keep getting the following error whenever I use ajatus.co.in/openid as the openid url: The Allow Access screen appears but followed by this error Unable to log in with your OpenID provider: The OpenID Provider issued an assertion for an Identifier whose discovery information did not match. Assertion endpoint info: ClaimedIdentifier: http://ajatus.co.in/openid?id=1134xxxxxxxxxxxxxxx39 ProviderLocalIdentifier: http://ajatus.co.in/openid?id=1134xxxxxxxxxxxxxxx39 ProviderEndpoint: https://www.google.com/a/ajatus.co.in/o8/ud?be=o8 OpenID version: 2.0 Service Type URIs: Discovered endpoint info: [{ ClaimedIdentifier: http://specs.openid.net/auth/2.0/identifier_select ProviderLocalIdentifier: http://specs.openid.net/auth/2.0/identifier_select ProviderEndpoint: https://www.google.com/a/ajatus.co.in/o8/ud?be=o8 OpenID version: 2.0 Service Type URIs: http://specs.openid.net/auth/2.0/server },] Contents of ajatus.co.in/openid <?xml version="1.0" encoding="UTF-8"?> <xrds:XRDS xmlns:xrds="xri://$xrds" xmlns="xri://$xrd*($v*2.0)"> <XRD> <Service priority="0"> <Type>http://specs.openid.net/auth/2.0/signon</Type> <URI>https://www.google.com/a/ajatus.co.in/o8/ud?be=o8</URI> </Service> <Service priority="0"> <Type>http://specs.openid.net/auth/2.0/server</Type> <URI>https://www.google.com/a/ajatus.co.in/o8/ud?be=o8</URI> </Service> </XRD> </xrds:XRDS> contents of ajatus.co.in/.well-known/host-meta is Link: <https://www.google.com/accounts/o8/site-xrds?hd=ajatus.co.in>; rel="describedby http://reltype.google.com/openid/xrd-op"; type="application/xrds+xml"

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  • Crystal Reports Failed Database Login

    - by Marlon
    Hello, After spending a good 3 to 4 hours on google trying to find any solution to my problem I haven't had much luck. Basically, we use crystal reports for our .NET applications with a sql server back end, we have many clients each with their own server and so our reports need to have their connections dynamically set. Up until a week ago this worked fine. However a few days ago a client reported they were getting a database login prompt for a report (only one report, the rest worked fine). We were quite stumped but we managed to reproduce it on a netbook which didn't have visual studio or sql server installed. In the end the dev decided to reproduce the report in the hope it was just an oddity in that particular report. Unfortunately a new client today also experienced the same problem, but this time for every crystal report they had - and also they worked on the netbook, so we are really quite lost here. Below is a screenshot of what our clients get presented with - and here is the code that I use to set the connection information in the report cI.ServerName = (string)builder["Data Source"]; cI.DatabaseName = (string)builder["Initial Catalog"]; cI.UserID = (string)builder["User ID"]; cI.Password = (string)builder["Password"]; foreach (IConnectionInfo info in cryRpt.DataSourceConnections) { info.SetConnection(cI.ServerName, cI.DatabaseName, cI.UserID, cI.Password); } foreach (ReportDocument sub in cryRpt.Subreports) { foreach (IConnectionInfo info in sub.DataSourceConnections) { info.SetConnection(cI.ServerName, cI.DatabaseName, cI.UserID, cI.Password); } } As always, any help much appreciated.

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  • ajax-based login fails: IE6 and IE7 ajax calls delete session data (session_id however is kept)

    - by mateipavel
    This question comes after two days of testing and debugging, right after the shock I had seeing that none of the websites i build using ajax-based login work in IE<8 The most simplified scenario si this: 1. mypage.php : session_start(); $_SESSION['mytest'] = 'x'; <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"> </script> <script type="text/javascript"> function loadit() { $.post('http://www.mysite.com/myajax.php', {action: 'test'}, function(result){alert(result);}, 'html'); } </script> <a href="javascript:void(0);" onclick="loadit(); return false;">test link</a> 2. myajax.php session_start(); print_r($_SESSION); print session_id(); When I click the "test link", the ajax call is made and the result is alert()-ed: IE6: weird bullet-character (&bull;) IE7: Array( ) <session_id> IE8/FF (Expected behaviour): Array( [mytest] => 'x' ) <session_id> I would really appreciate some pointers regarding: 1. why this happens 2. how to fix it Thank you.

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  • Multiuser login into winforms application

    - by schoetbi
    Hi there, i have a winforms app in C# that needs access control for certain forms. That means, the application is running under the same (default) user at system startup, but certain forms need to be secured, so that only certain windows users could have access to the additional functions after identifying themself with username and password. For that step windows authentication should be used. Now the tricky part. Although the application was started under a "normal" user I would like the superusers to "login" into the special form without restarting the entiere application. My question now is. Is this possible (i.e. create one thread with the credentials of an administrator). Or do I need to setup another appdomain for that? Please give me a hint wather the user of a running application could be changed somehow. Thank you. EDIT I replaced administrators by "certain users" since the privileged user could be just another ordinary user that is granted access to the special functionality by the configuration of the installation.

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  • php curl login problem

    - by user331329
    <?php // create a new CURL resource $file_path = '/mail'; define("COOKIE_FILE", "c:\cookie.txt"); $ch = curl_init(); // set URL and other appropriate options curl_setopt($ch, CURLOPT_URL, "https://mail.gov.in/iwc/signin"); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt ($ch, CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt ($ch, CURLOPT_USERAGENT, "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6"); curl_setopt ($ch, CURLOPT_TIMEOUT, 60); curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1); curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt ($ch, CURLOPT_COOKIESESSION, TRUE); session_write_close(); $strCookie = 'PHPSESSID=d095af0e30afc021dd3652734009' . $_COOKIE['PHPSESSID'] . '; path=/mail'; curl_setopt( $ch, CURLOPT_COOKIE, $strCookie ); curl_setopt ($ch, CURLOPT_COOKIEJAR, COOKIE_FILE); curl_setopt($ch, CURLOPT_COOKIEFILE, COOKIE_FILE); curl_setopt ($ch, CURLOPT_POST, 1); curl_setopt ($ch, CURLOPT_POSTFIELDS,'fromLogin=true&domainName=nic.in&username=&password=&button=Sign%20In'); $url = curl_getinfo($ch); // grab URL and pass it to the browser $data = curl_exec($ch); echo $data."<pre>"; echo "<pre>"; print_r($url); // close CURL resource, and free up system resources curl_close($ch); ?> whats wrong with my code why i am not able to login directly in their mail

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  • broken SQL 2008 SP1 Web Edition (can not login with SSMS)

    - by gerryLowry
    Scenario: My installation of SQL Server 2008 Web Edition SP1 was working properly. Since I've recently joined Microsoft's Website Spark*, I removed SQL2008 and installed SQL 2008 again using my Website Spark edition and license from the MSDN download site. Next, I updated SQL 2008 to SP1 (this is required because I'm running Windows 2008 Server R2 Web edition). When I launch SSMS (SQL Server Management Studio), "User name" is "myhost\Administrator" and is greyed out so it can not be changed. When I installed my Website Spark version, I did not include "myhost\Administrator" when I was configuring SQL 2008 service accounts. Instead I created an administrator account "myhost\mySQLaccount". ERROR MESSAGE: Connect to Server (X) Cannot connect to (local) Additional information: Login failed for user 'myhost'Admistrator' (Microsoft SQL Server, Error: 18456) I tried to use the SQL Server Configuration Manager to correct this problem but could not find any useful way to fix this issue. How to I fix this problem? Connect to Server ... Server type: Database Engine Server name: (local) Authentication: Windows Authentication Please advise. Thank you. Gerry * http://www.microsoft.com/web/websitespark/default.aspx

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  • how to login to criaglist through c#

    - by kosikiza
    i m using the following code to login to criaglist, but hav't successsed yet. string formParams = string.Format("inputEmailHandle={0}&inputPassword={1}", "[email protected]", "pakistan01"); //string postData = "[email protected]&inputPassword=pakistan01"; string uri = "https://accounts.craigslist.org/"; HttpWebRequest request = (HttpWebRequest)WebRequest.Create(uri); request.KeepAlive = true; request.ProtocolVersion = HttpVersion.Version10; request.Method = "POST"; byte[] postBytes = Encoding.ASCII.GetBytes(formParams); request.ContentType = "application/x-www-form-urlencoded"; request.ContentLength = postBytes.Length; Stream requestStream = request.GetRequestStream(); requestStream.Write(postBytes, 0, postBytes.Length); requestStream.Close(); HttpWebResponse response = (HttpWebResponse)request.GetResponse(); cookyHeader = response.Headers["Set-cookie"]; string pageSource; string getUrl = "https://post.craigslist.org/del"; WebRequest getRequest = WebRequest.Create(getUrl); getRequest.Headers.Add("Cookie", cookyHeader); WebResponse getResponse = getRequest.GetResponse(); using (StreamReader sr = new StreamReader(getResponse.GetResponseStream())) { pageSource = sr.ReadToEnd(); }

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  • Asp.Net Login Control very slow initial connection to Non-Trusted AD Domain

    - by Eric Brown - Cal
    ASP.NET Login control is very slow making the initial connection to AD when authenticating to a different domain than the domain the web server is a member of. Problem occurs for the IIS server and when using with the Visual Studio's built in web server. It takes about 30 seconds the first time when attempting to use the control to connect against another domain. There is no trust relationship bewteen the web server's domain and the other domains (attempted connecting to several different domains). Subsequent connections execute quickly until the connection times out. Using Systernals Process Monitor to troubleshoot, there are two OpenQuery operations right before the delay to "C:\WINDOWS\asembly\GAC_MSIL\System.DirectoryServices\2.0.0.0_b03f5f7f11d50a3a\Netapi32.dll with a result NAME NOT FOUND" and right after the 30 second delay the TCP Send and TCP Recieves indicate communication begins with the AD server. Things we have tried: Impersonating an administrator on the web server in the web.config; Granting permissions to the CryptoKeys to the NetworkService and ASPNET; Specifying by IP instead of DNS name; Multiple variations of specifying the name and ldap server with domains and OU's; Local host entries; Looked for ports being blocked (SYN_SENT) with netstat -an. Nslookup resolves all the domains and systems involved correectly. TraceRt shows the Correct routes Any Idea or hints are greately appreicated.

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  • how can I know current viewcontroller name in iphone

    - by Shikhar
    I have BaseView which implement UIViewController. Every view in project must implement this BaseView. In BaseView, I have method: -(void) checkLoginStatus { defaults = [[NSUserDefaults alloc] init]; if(![[defaults objectForKey:@"USERID"] length] > 0 ) { Login *login=[[Login alloc] initWithNibName:@"Login" bundle:nil]; [self.navigationController pushViewController:login animated:TRUE]; [login release]; } [defaults release]; } The problem is my Login view also implement BaseView, checks for login, and again open LoginView i.e. stuck in to recursive calling. Can I check in checkLoginStatus method if request is from LoginView then take no action else check login. Ex: (void) checkLoginStatus { if(SubView is NOT Login){ defaults = [[NSUserDefaults alloc] init]; if(![[defaults objectForKey:@"USERID"] length] 0 ) { Login *login=[[Login alloc] initWithNibName:@"Login" bundle:nil]; [self.navigationController pushViewController:login animated:TRUE]; [login release]; } [defaults release]; } } Please help..

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  • POSTmethod and PHP- login verification

    - by Neethusha
    I wrote a code for login verification..I got output with GET. But i need output with POST since it is more secure.pls let me know if there is any error in my code. javascript code: var xml; function verifyusernamepasswd(pass) { //pass is password that will be passed as parameter xml=new XMLHttpRequest(); var url="http://localhost/loginvalidate.php"; var para="q="+username+"&p="+pass;//username is global xml.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); xml.setRequestHeader("Content-length", para.length); xml.setRequestHeader("Connection", "close"); xml.open("POST",url,true); xml.onreadystatechange=statechanged1; xml.send(para); } function statechanged1() { if(xml.readyState==4) alert(xml.responseText); } php code: <?php $username=$_POST["q"]; $password=$_POST["p"]; $con=mysql_connect("localhost","root","blaze"); if(!$con) { die('Could not connect: '.mysql.error()); } mysql_select_db("BLAZE",$con) or die("No such Db"); $result=mysql_query("SELECT Passwword FROM USERTABLE WHERE Userhandle='$username'"); if($result==null) echo "false"; else if($result!=null) { $row=mysql_fetch_array($result); if((strcmp($row['Passwword'],$password)==0)) echo "true"; else echo "false"; } ?> the verification does not return anything, cos my alert is not displayed at all...pls tell me whats wrong....

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  • redirecting _top page from asp:login control in iframe onloggedin

    - by jumpdart
    So yeah, Im building a little authenticated content(.NET app) to a large html site managed by another group. They are only comfortable with managing html so all my app content will be contained in iframes. Everything is working fine as far as navigation and calling services and whatnot but i cant bust out of the frame with my asp:login control. Im trying to register some JS on logged in but with no success. Thanks protected void login_LoggedIn(object sender, EventArgs e) { StringBuilder strScript = new StringBuilder(); strScript.Append("<script language='javascript'>"); string sHome = ConfigurationManager.AppSettings["AppHomePageURL"].ToString(); //strScript.AppendFormat("window.navigate('{0}');", sHome); //strScript.AppendFormat("parent.location.href='{0}';", sHome); //strScript.AppendFormat("window.open('{0}', '_top', '', false);", sHome); strScript.AppendFormat("top.location.href='{0}';", sHome); strScript.Append("WTF_let_me_outa_here();"); strScript.Append("</script>"); ClientScript.RegisterClientScriptBlock(typeof(Page), "LoginGO", strScript.ToString()); }

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  • login using UITableView

    - by EquinoX
    How can I create a login for username/password just like in the skype application? I know it's a grouped table view.. but how can I do that? I searched the site and found the following code: - (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { static NSString *CellIdentifier = @"Cell"; UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier]; if (cell == nil) { cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier] autorelease]; UILabel *startDtLbl = [[UILabel alloc] initWithFrame:CGRectMake(10, 10, 80, 25)]; if (indexPath.row == 0) startDtLbl.text = @"Username"; else { startDtLbl.text = @"Password"; } startDtLbl.backgroundColor = [UIColor clearColor]; [cell.contentView addSubview:startDtLbl]; UITextField *passwordTF = [[UITextField alloc] initWithFrame:CGRectMake(100, 5, 200, 35)]; passwordTF.delegate = self; if (indexPath.row == 0) passwordTF.tag = 0; else { passwordTF.tag = 1; } [cell.contentView addSubview:passwordTF]; } return cell; } When I do: NSString * username = [((UITextField*)[self.view viewWithTag:0]) text]; NSString * password = [((UITextField*)[self.view viewWithTag:1]) text]; it gives me this error: Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[UIView text]: unrecognized selector sent to instance 0x6c3c600 Why is this?

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  • Spring.NET & Immediacy CMS (or how to inject to server side controls without using PageHandlerFactor

    - by Simon Rice
    Is there any way to inject dependencies into an Immediacy CMS control using Spring.NET, ideally without having to use to ContextRegistry when initialising the control? Update, with my own answer The issue here is that Immediacy already has a handler defined in web.config that deals with all aspx pages, & so it's not possible add an entry for Spring.NET's PageHandlerFactory in web.config as per a normal webforms app. That rules out making the control implement ISupportsWebDependencyInjection. Furthermore, most of Immediacy's generated pages are aspx pages that don't physically exist on the drive. I have changed the title of the question to reflect this. What I have done to get Dependency Injection working is: Add the usual entries to web.config for Spring.NET as outlined in the documentation, except for the adding the entry to the <httpHandlers> section. In this case I've got my object definitions in Spring.config. Create the following abstract base class that will deal with all of the Dependency Injection work: DIControl.cs public abstract class DIControl : ImmediacyControl { protected virtual string DIName { get { return this.GetType().Name; } } protected override void OnInit(EventArgs e) { if (ContextRegistry.GetContext().GetObject(DIName, this.GetType()) != null) ContextRegistry.GetContext().ConfigureObject(this, DIName); base.OnInit(e); } } For non-immediacy controls, you can make this server side control inherit from Control or whatever subclass of that you like. For any control with which you wish to use with Spring.NET's Inversion of Control container, define it to inherit from DIControl & add the relelvant entry to Spring.config, for example: SampleControl.cs public class SampleControl : DIControl, INamingContainer { public string Text { get; set; } protected string InjectedText { get; set; } public SampleControl() : base() { Text = "Hello world"; } protected override void RenderContents(HtmlTextWriter output) { output.Write(string.Format("{0} {1}", Text, InjectedText)); } } Spring.config <objects xmlns="http://www.springframework.net"> <object id="SampleControl" type="MyProject.SampleControl, MyAssembly"> <property name="InjectedText" value="from Spring.NET" /> </object> </objects> You can optionally override DIName if you wish to name your entry in Spring.config differently from the name of your class. Provided everything's done correctly, you will have the control writing out "Hello world from Spring.NET!" when used in a page. This solution uses Spring.NET's ContextRegistry from within the control, but I would be surprised if there's no way around that for Immediacy at least since the page objects themselves aren't accessible. However, can this be improved at all from a Spring.NET perspective? Is there maybe an Immediacy plugin that already does this that I'm completely unaware of? Or is there an approach that does this in a more elegant way? I'm open to suggestions.

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  • Microsoft Word Document Controls not accepting carriage returns

    - by Scott
    So, I have a Microsoft Word 2007 Document with several Plain Text Format (I have tried Rich Text Format as well) controls which accept input via XML. For carriage returns, I had the string being passed through XML containing "\r\n" when I wanted a carriage return, but the word document ignored that and just kept wrapping things on the same line. I also tried replacing the \r\n with System.Environment.NewLine in my C# mapper, but that just put in \r\n anyway, which still didn't work. Note also that on the control itself I have set it to "Allow Carriage Returns (Multiple Paragrpahs)" in the control properties. This is the XML for the listMapper <Field id="32" name="32" fieldType="SimpleText"> <DataSelector path="/Data/DB/DebtProduct"> <InputField fieldType="" path="/Data/DB/Client/strClientFirm" link="" type=""/> <InputField fieldType="" path="strClientRefDebt" link="" type=""/> </DataSelector> <DataMapper formatString="{0} Account Number: {1}" name="SimpleListMapper" type=""> <MapperData> </MapperData> </DataMapper> </Field> Note that this is the listMapper C# where I actually map the list (notice where I try and append the system.environment.newline) namespace DocEngine.Core.DataMappers { public class CSimpleListMapper:CBaseDataMapper { public override void Fill(DocEngine.Core.Interfaces.Document.IControl control, CDataSelector dataSelector) { if (control != null && dataSelector != null) { ISimpleTextControl textControl = (ISimpleTextControl)control; IContent content = textControl.CreateContent(); CInputFieldCollection fileds = dataSelector.Read(Context); StringBuilder builder = new StringBuilder(); if (fileds != null) { foreach (List<string> lst in fileds) { if (CanMap(lst) == false) continue; if (builder.Length > 0 && lst[0].Length > 0) builder.Append(Environment.NewLine); if (string.IsNullOrEmpty(FormatString)) builder.Append(lst[0]); else builder.Append(string.Format(FormatString, lst.ToArray())); } content.Value = builder.ToString(); textControl.Content = content; applyRules(control, null); } } } } } Does anybody have any clue at all how I can get MS Word 2007 (docx) to quit ignoring my newline characters??

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  • Login function runs different between local and server

    - by quangnd
    Here is my check login function: protected bool checkLoginStatus(String email, String password) { bool loginStatus = false; bool status = false; try { Connector.openConn(); String str = "SELECT * FROM [User]"; SqlCommand cmd = new SqlCommand(str, Connector.conn); SqlDataAdapter da = new SqlDataAdapter(cmd); DataSet ds = new DataSet(); da.Fill(ds, "tblUser"); //check valid foreach (DataRow dr in ds.Tables[0].Rows) { if (email == dr["Email"].ToString() && password == Connector.base64Decode(dr["Password"].ToString())) { Session["login_status"] = true; Session["username"] = dr["Name"].ToString(); Session["userId"] = dr["UserId"].ToString(); status = true; break; } } } catch (Exception ex) { } finally { Connector.closeConn(); } return status; } And call it at my aspx page: String email = Login1.UserName.Trim(); String password = Login1.Password.Trim(); if (checkLoginStatus(email, password)) Response.Redirect(homeSite); else lblFailure.Text = "Invalid!"; I ran this page at localhost successful! When I published it to server, this function only can run if email and password correct! Other, error occured: A network-related or instance-specific error occurred while establishing a connection to SQL Server. The server was not found or was not accessible. Verify that the instance name is correct and that SQL Server is configured to allow remote connections. (provider: SQL Network Interfaces, error: 26 - Error Locating Server/Instance Specified) I tried open SQL Server 2008 Configuration Manager and enable SQL Server Browser service (Logon as:NT Authority/Local Service) but it stills error. (note: here is connection string of openConn() at Localhost (run on SQLEXpress 2005) connectionString="Data Source=MYLAPTOP\SQLEXPRESS;Initial Catalog=Spider_Vcms;Integrated Security=True" /> ) At server (run on SQL Server Enterprise 2008) connectionString="Data Source=SVR;Initial Catalog=Spider_Vcms;User Id=abc;password=123456;" /> anyone have an answer for my problem :( thanks a lot!

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  • ASP.Net MVC - Models and User Controls

    - by cdotlister
    Hi guys, I have a View with a Master Page. The user control makes use of a Model: <%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<BudgieMoneySite.Models.SiteUserLoginModel>" %> This user control is shown on all screens (Part of the Master Page). If the user is logged in, it shows a certain text, and if the user isn't logged in, it offers a login box. That is working OK. Now, I am adding my first functional screen. So I created a new view... and, well, i generated the basic view code for me when I selected the controller method, and said 'Create View'. My Controller has this code: public ActionResult Transactions() { List<AccountTransactionDetails> trans = GetTransactions(); return View(trans); } private List<AccountTransactionDetails> GetTransactions() { List<AccountTransactionDto> trans = Services.TransactionServices.GetTransactions(); List<AccountTransactionDetails> reply = new List<AccountTransactionDetails>(); foreach(var t in trans) { AccountTransactionDetails a = new AccountTransactionDetails(); foreach (var line in a.Transactions) { AccountTransactionLine l = new AccountTransactionLine(); l.Amount = line.Amount; l.SubCategory = line.SubCategory; l.SubCategoryId = line.SubCategoryId; a.Transactions.Add(l); } reply.Add(a); } return reply; } So, my view was generated with this: <%@ Page Language="C#" MasterPageFile="~/Views/Shared/Site.Master" Inherits="System.Web.Mvc.ViewPage<System.Collections.Generic.List<BudgieMoneySite.Models.AccountTransactionDetails>>" %> Found <%=Model.Count() % Transactions. All I want to show for now is the number of records I will be displaying. When I run it, I get an error: "The model item passed into the dictionary is of type 'System.Collections.Generic.List`1[BudgieMoneySite.Models.AccountTransactionDetails]', but this dictionary requires a model item of type 'BudgieMoneySite.Models.SiteUserLoginModel'." It looks like the user control is being rendered first, and as the Model from the controller is my List<, it's breaking! What am I doing wrong?

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  • storing user info/pass in web.config authentication

    - by Tomaszewski
    Hello, I am trying to write a simple internal app with some simple authentication. I'm also trying to make this quick and learn about the forms authentication via web.config. So i have my authentication working if I hard code my 'user name' and 'password' into C# code and do a simple conditional. However, I'm having a tough time storing the a user/pass to be checked against in the web.config file. The MSDN manual says to put this into the web.config: <authentication mode="Forms"> <forms loginUrl="login.aspx"> <credentials passwordFormat="SHA1"> <user name="user1" password="27CE4CA7FBF00685AF2F617E3F5BBCAFF7B7403C" /> <user name="user2" password="D108F80936F78DFDD333141EBC985B0233A30C7A" /> <user name="user3" password="7BDB09781A3F23885CD43177C0508B375CB1B7E9"/> </credentials> </forms> </authentication> However, the minute I add 'credentials' into the 'authentication' section, I get this error: Server Error in '/' Application. Configuration Error Description: An error occurred during the processing of a configuration file required to service this request. Please review the specific error details below and modify your configuration file appropriately. Parser Error Message: Unrecognized element 'credentials'. Source Error: Line 44: <authentication mode="Forms"> Line 45: <forms loginUrl="login.aspx" /> Line 46: <credentials> Line 47: Line 48: </credentials> Source File: C:\inetpub\wwwroot\asp\projects\passwordCatalog\passwordCatalog\web.config Line: 46 So my question is, how and where would I add the following in the web.config file? <credentials passwordFormat="SHA1"> <user name="johndoe" password="mypass123" /> </credentials>

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  • SHA function issues

    - by Damian James
    I have this php code from my login.php if (isset($_POST['logIn'])) { $errmsg = ""; $logname = mysqli_real_escape_string($dbc, trim($_POST['usernameIn'])); $logpassword = mysqli_real_escape_string($dbc, trim($_POST['passwordIn'])); $query = "SELECT user_id, username FROM members WHERE username = '$logname' AND password = SHA('$logpassword')"; $data = mysqli_query($dbc, $query); if (mysqli_num_rows($data) == 1) { $row = mysqli_fetch_array($data); setcookie('user_id', $row['user_id'], time() + (60 * 60 * 24 * 30)); //expires after 30 days setcookie('username', $row['username'], time() + (60 * 60 * 24 * 30)); $home = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['PHP_SELF']) . '/index.php'; header('Location: ' . $home); } else { $errmsg = '<p class="errormsg">Username or password is incorrect.</p>'; } } And for some reason, it always ends up setting $errmsg in the else statement. I am sure that I'm entering information (username,password) that is correct and exists in the database. I insert my values (from a signup script) using this query: $query = "INSERT INTO members (username, password, email) VALUES ('$username', SHA('$password'), '$email')"; Anyone see the problem with this script? Thanks!

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  • How does browser know when to prompt user to save password?

    - by Eric
    This is related to the question I asked here: http://stackoverflow.com/questions/2382329/how-can-i-get-browser-to-prompt-to-save-password This is the problem: I CAN'T get my browser to prompt me to save the password for the site I'm developing. (I'm talking about the bar that appears sometimes when you submit a form on Firefox, that says "Remember the password for yoursite.com? Yes / Not now / Never") This is super frustrating because this feature of Firefox (and most other modern browsers, which I hope work in a similar fashion) seems to be a mystery. It's like a magic trick the browser does, where it looks at your code, or what you submit, or something, and if it "looks" like a login form with a username (or email address) field and a password field, it offers to save. Except in this case, where it's not offering my users that option after they use my login form, and it's making me nuts. :-) (I checked my Firefox settings-- I have NOT told the browser "never" for this site. It should be prompting.) My question: exactly what the heuristics are that Firefox (or any other modern browser) uses to know when it should prompt the user to save? This shouldn't be too difficult to answer, since it's right there in the Mozilla source (I don't know where to look or else I'd try to dig it out myself). You'd think there would be a blog post or some other similar developer note from the Mozilla developers about this but I can't find that either. (* Note that if your answer to me has anything to do with cookies, encryption or anything else that is about how I'm storing the user's passwords in the database, you've probably misread my question. :-)

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  • How to use sessions in PDO?

    - by Byakugan
    I am still redoing and getting rid of old mysql_* commands in my code. I tried to transfer my session login form old code and this is what I got so far: public function login($user, $password) { if (!empty($user) && !empty($password)) { $password = $web->doHash($user, $password); // in this function is (return sha1(strtoupper($user).':'.strtoupper($password)) $stmt = $db_login->prepare("SELECT * FROM account WHERE username=:user AND pass_hash=:password"); $stmt->bindValue(':user', $user, PDO::PARAM_STR); $stmt->bindValue(':password', $password, PDO::PARAM_STR); $stmt->execute(); $rows = $stmt->rowCount(); if ($rows > 0) { $results_login = $stmt->fetch(PDO::FETCH_ASSOC); $_SESSION['user_name'] = $results_login['username']; $_SESSION['user_id'] = $results_login['id']; return true; } else { return false; } } else { return false; } } After that I am using checks if user logged on site: public function isLogged() { return (!empty($_SESSION['user_id']) && !empty($_SESSION['user_name'])); } But it seems - this function returns always empty because $_SESSION does not exists in PDO? And of course logout is used in this form on my sites: public function logout() { unset($_SESSION['user_id']); unset($_SESSION['user_name']); } But I think PDO has different way of handling session? I did not find any so what is it can i somehow add $_SESSION in PDO withou changing code much? I am using variables $_SESSION['user_name'] and $_SESSION['user_id'] in all over my web project. Summary: 1) How to use sessions in PDO correctly? 2) What is difference between using $stmt->fetch(PDO::FETCH_ASSOC); and $stmt->fetchAll(); Thank you.

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  • How to implement an EventHandler to update controls

    - by Bill
    May I ask for help with the following? I am attempting to connect and control three pieces of household electronic equipment by computer through a GlobalCache GC-100 and iTach. As you will see in the following code, I created a class-instance of GlobalCacheAdapter that communicates with each piece of equipment. Although the code seems to work well in controlling the equipment, I am having trouble updating controls with the feedback from the equipment. The procedure "ReaderThreadProc" captures the feedback; however I don't know how to update the associated TextBox with the feedback. I believe that I need to create an EventHandler to notify the TextBox of the available update; however I am uncertain as to how an EventHandler like this would be implemented. Any help wold be greatly appreciated. using System; using System.IO; using System.Net; using System.Net.Sockets; using System.Threading; using System.Windows.Forms; namespace WindowsFormsApplication1 { public partial class Form1 : Form { // Create three new instances of GlobalCacheAdaptor and connect. // GC-100 (Elan) 192.168.1.70 4998 // GC-100 (TuneSuite) 192.168.1.70 5000 // GC iTach (Lighting) 192.168.1.71 4999 private GlobalCacheAdaptor elanGlobalCacheAdaptor; private GlobalCacheAdaptor tuneSuiteGlobalCacheAdaptor; private GlobalCacheAdaptor lutronGlobalCacheAdaptor; public Form1() { InitializeComponent(); elanGlobalCacheAdaptor = new GlobalCacheAdaptor(); elanGlobalCacheAdaptor.ConnectToDevice(IPAddress.Parse("192.168.1.70"), 4998); tuneSuiteGlobalCacheAdaptor = new GlobalCacheAdaptor(); tuneSuiteGlobalCacheAdaptor.ConnectToDevice(IPAddress.Parse("192.168.1.70"), 5000); lutronGlobalCacheAdaptor = new GlobalCacheAdaptor(); lutronGlobalCacheAdaptor.ConnectToDevice(IPAddress.Parse("192.168.1.71"), 4999); elanTextBox.Text = elanGlobalCacheAdaptor._line; tuneSuiteTextBox.Text = tuneSuiteGlobalCacheAdaptor._line; lutronTextBox.Text = lutronGlobalCacheAdaptor._line; } private void btnZoneOnOff_Click(object sender, EventArgs e) { elanGlobalCacheAdaptor.SendMessage("sendir,4:3,1,40000,4,1,21,181,21,181,21,181,21,181,21,181,21,181,21,181,21,181,21,181,21,181,21,181,21,800" + Environment.NewLine); } private void btnSourceInput1_Click(object sender, EventArgs e) { elanGlobalCacheAdaptor.SendMessage("sendir,4:3,1,40000,1,1,20,179,20,179,20,179,20,179,20,179,20,179,20,179,20,278,20,179,20,179,20,179,20,780" + Environment.NewLine); } private void btnSystemOff_Click(object sender, EventArgs e) { elanGlobalCacheAdaptor.SendMessage("sendir,4:3,1,40000,1,1,20,184,20,184,20,184,20,184,20,184,20,286,20,286,20,286,20,184,20,184,20,184,20,820" + Environment.NewLine); } private void btnLightOff_Click(object sender, EventArgs e) { lutronGlobalCacheAdaptor.SendMessage("sdl,14,0,0,S2\x0d"); } private void btnLightOn_Click(object sender, EventArgs e) { lutronGlobalCacheAdaptor.SendMessage("sdl,14,100,0,S2\x0d"); } private void btnChannel31_Click(object sender, EventArgs e) { tuneSuiteGlobalCacheAdaptor.SendMessage("\xB8\x4D\xB5\x33\x31\x00\x30\x21\xB8\x0D"); } private void btnChannel30_Click(object sender, EventArgs e) { tuneSuiteGlobalCacheAdaptor.SendMessage("\xB8\x4D\xB5\x33\x30\x00\x30\x21\xB8\x0D"); } } } public class GlobalCacheAdaptor { public Socket _multicastListener; public string _preferredDeviceID; public IPAddress _deviceAddress; public Socket _deviceSocket; public StreamWriter _deviceWriter; public bool _isConnected; public int _port; public IPAddress _address; public string _line; public GlobalCacheAdaptor() { } public static readonly GlobalCacheAdaptor Instance = new GlobalCacheAdaptor(); public bool IsListening { get { return _multicastListener != null; } } public GlobalCacheAdaptor ConnectToDevice(IPAddress address, int port) { if (_deviceSocket != null) _deviceSocket.Close(); try { _port = port; _address = address; _deviceSocket = new Socket(AddressFamily.InterNetwork, SocketType.Stream, ProtocolType.Tcp); _deviceSocket.Connect(new IPEndPoint(address, port)); ; _deviceAddress = address; var stream = new NetworkStream(_deviceSocket); var reader = new StreamReader(stream); var writer = new StreamWriter(stream) { NewLine = "\r", AutoFlush = true }; _deviceWriter = writer; writer.WriteLine("getdevices"); var readerThread = new Thread(ReaderThreadProc) { IsBackground = true }; readerThread.Start(reader); _isConnected = true; return Instance; } catch { DisconnectFromDevice(); MessageBox.Show("ConnectToDevice Error."); throw; } } public void SendMessage(string message) { try { var stream = new NetworkStream(_deviceSocket); var reader = new StreamReader(stream); var writer = new StreamWriter(stream) { NewLine = "\r", AutoFlush = true }; _deviceWriter = writer; writer.WriteLine(message); var readerThread = new Thread(ReaderThreadProc) { IsBackground = true }; readerThread.Start(reader); } catch { MessageBox.Show("SendMessage() Error."); } } public void DisconnectFromDevice() { if (_deviceSocket != null) { try { _deviceSocket.Close(); _isConnected = false; } catch { MessageBox.Show("DisconnectFromDevice Error."); } _deviceSocket = null; } _deviceWriter = null; _deviceAddress = null; } private void ReaderThreadProc(object state) { var reader = (StreamReader)state; try { while (true) { var line = reader.ReadLine(); if (line == null) break; _line = _line + line + Environment.NewLine; } // Need to create EventHandler to notify the TextBoxes to update with _line } catch { MessageBox.Show("ReaderThreadProc Error."); } } }

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  • PHP mySQL Error

    - by happyCoding25
    Hello, Im new to php so I decided to follow this tutorial for a simple login screen. I got the code setup but when I try login I get this error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in (a long file path to the script) on line 27 The code I got from the tutorial is: <?php ob_start(); $host="thehost"; // Host name $username="myusername"; // Mysql username $password="mypass"; // Mysql password $db_name="test"; // Database name $tbl_name="members"; // Table name // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // Define $myusername and $mypassword $myusername=$_POST['myusername']; $mypassword=$_POST['mypassword']; // To protect MySQL injection (more detail about MySQL injection) $myusername = stripslashes($myusername); $mypassword = stripslashes($mypassword); $myusername = mysql_real_escape_string($myusername); $mypassword = mysql_real_escape_string($mypassword); $sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($sql); // Mysql_num_row is counting table row $count=mysql_num_rows($result); // If result matched $myusername and $mypassword, table row must be 1 row if($count==1){ // Register $myusername, $mypassword and redirect to file "login_success.php" session_register("myusername"); session_register("mypassword"); header("location:login_success.php"); } else { echo "Wrong Username or Password"; } ob_end_flush(); ?> (Note: All of the mySQL database info is filled in on my version) Aslo, the author gives a php5 version and a normal php version. I have tried both and gotten the same error. If anyone knows why this is happening help is really appreciated.

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