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  • To bring the IE window in front of the Screen

    - by Parameswari
    I am creating new IE browser instance dynamically, and opening a aspx page from there. Everything works fine , but the browser is not popping in the Front of the screen .Able to see the Aspx page in the task bar when I click it from there it comes to the Front . How to bring that page in front of all the Screen as soon as IE is created. I have pasted the code I used to create new IE instance. public class IEInstance { public SHDocVw.InternetExplorer IE1; public void IEInstanceCls(string check) { IE1 = new SHDocVw.InternetExplorer(); object Empty = 0; string urlpath = " "; urlpath = "http://localhost/TestPage.aspx?"; object URL = urlpath; IE1.Top = 260; IE1.Left = 900; IE1.Width = 390; IE1.Height = 460; IE1.StatusBar = false; IE1.ToolBar = 0; IE1.MenuBar = false; IE1.Visible = true; IE1.Navigate2(ref URL, ref Empty, ref Empty, ref Empty, ref Empty); } } Help me to solve this problem. Thank You

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  • My cookies won't stay (PHP).

    - by RemiX
    I'm building an autologin system using cookies, but one fundamental part of the functionality of the cookies fails: they are non-persistent over different sessions - or even pages! In my login script, I set the cookies like this: setcookie('userID', $userID, time()+86400); // (edited after replies) $userID has a value. Then I print the $_COOKIE variable and it says array(['base_usid'] = 1); So that's good, but when I click the home page and print the $_COOKIE variable there, it says NULL. Does anyone see the problem?

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  • rails separate login for an api

    - by Squadrons
    I have a very simple api that is part of a rails app that requires logging in. I just need a way to make the api part accessible with a simple form that allows the user to enter parameters like a key (just a simple one stored in the DB, no OAuth or anything), a userId to find and return a user via json, and maybe some other parameters like asking for their schedule. How can I keep this seperate from the rest of the app, making it a public facing form that will grant access only to the api? Thanks.

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  • Keeping track of the action before a login?

    - by soybie
    I'm trying to do the following: User can vote for an item (controller: item, action: vote) 2a. If the user is logged in, then vote action goes through. 2b. If user is not logged in, then user needs to log in/creates an account (handled by user controller), then vote action goes through. How do I do 2b such that once the user logs in/creates account, the vote action automatically goes through without having the user vote for the item again?

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  • How can I login linux using C or C++

    - by jnblue
    I need to programmely switch the current user to another,then the followed code should be executed in the environment(such as path,authority..) of another user. I've find the 'chroot()','setuid()' may be associated with my case, but these functions need the root authority, I don't have root authority, but I have the password of the second user. what should I do? I have tried shell "su - " can switch current user, can this command help me in my C++ code? Don't laugh at me if my question is very stupid, I'm a true freshman on linux. :) Thanks!

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  • The best way to store username and password without a database

    - by Mokuchan
    Hello everyone, I want to build a simple single user login "library" in PHP, but I'm facing the following dilemma: how should I store username and password if I'm not using a database? A simple plain text file could be easily read and then the password could be easily decripted so that's not an option. If I create a php file with just <?php $Username= "username"; $Password= "password"; ?> then no one should be able to read it and I could simply include this file where I need it, but I'm not really sure you can't find a better way to do this! So what's, in your opinion, the best solution to this problem (and why)? Thanks

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  • How to use Twitter4j With JSF2 for Login? [on hold]

    - by subodh
    I am trying to do login with Twitter and using Twitter4j for that and wrote this code In JSF <h:commandButton id="twitterbutton" value="Sign up with Twitter" action="#{twitterLoginBean.redirectTwitterLogin}" immediate="true" styleClass="twitterbutton"/> In ManagedBean public String redirectTwitterLogin() throws ServletException, IOException, TwitterException { HttpServletRequest request = (HttpServletRequest) FacesContext .getCurrentInstance().getExternalContext().getRequest(); HttpServletResponse response = (HttpServletResponse) FacesContext .getCurrentInstance().getExternalContext().getResponse(); Twitter twitter = TwitterFactory.getSingleton(); twitter.setOAuthConsumer(apiKey, apiSecret); RequestToken requestToken = twitter.getOAuthRequestToken(); if (requestToken != null) { AccessToken accessToken = null; BufferedReader br = new BufferedReader(new InputStreamReader( System.in)); while (null == accessToken) { try { String pin = br.readLine(); accessToken = twitter .getOAuthAccessToken(requestToken, pin); } catch (TwitterException te) { System.out .println("Failed to get access token, caused by: " + te.getMessage()); System.out.println("Retry input PIN"); } } request.setAttribute(IS_AUTHENTICATED, true); if (accessToken != null) { LOGGER.debug("We have a valid oauth token! Make the facebook request"); doApiCall(twitter, request, response); return null; } } else { LOGGER.debug("We don't have auth code yet, fetching the Authorization URL..."); String authorizationUrl = requestToken.getAuthorizationURL(); LOGGER.debug("Redirecting to the Authorization URL: {}", authorizationUrl); request.setAttribute(IS_AUTHENTICATED, false); redirect(authorizationUrl, response); return null; } return null; } In above code i want first Login window of twitter will show and then again same method will call and after user will login i can show user information userId,Handel,location etc. Redirect private void redirect(String url, HttpServletResponse response) throws IOException { String urlWithSessionID = response.encodeRedirectURL(url); response.sendRedirect(urlWithSessionID); } But this code is not working Can anyone tell better Solution for this ?

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  • Son of Suckerfish ie6 problem - right-most dropdown menu also appearing on left side of screen

    - by Kevin Burke
    I'm interning for an NGO in India and trying to fix their website, including updating their menu so it's not the last item on the page to load, and it's centered on the screen. Everything works well enough but when I try out my new menu in IE6, I get this weird error where the content below the menu is padded an extra 30px or so and the material in the right-most drop down appears on the far left of the screen, always visible. When I drop down the rightmost link ("Publications") the content appears both in the correct location and in the same spot on the far left of the screen, and changes color when I hover as well. It's tough to describe, so it would probably be best if you took a look: visit http://sevamandir.org/a30/index.htm in your Internet Explorer 6 browser to see for yourself. I really appreciate your help. Also I'm using a 1000px wide monitor, if there's more hijinks going on outside that space I'd like to know about that too. Here's the relevant code: in the html head: <script> sfHover = function() { var sfEls = document.getElementById("nav").getElementsByTagName("LI"); for (var i=0; i<sfEls.length; i++) { sfEls[i].onmouseover=function() { this.className+=" sfhover"; } sfEls[i].onmouseout=function() { this.className=this.className.replace(new RegExp(" sfhover\\b"), ""); } } } if (window.attachEvent) window.attachEvent("onload", sfHover); </script> text surrounding the menu - the menu is simply <ul id="nav"><li></li></ul> etc. <!--begin catchphrase--> <div style="float:left; height:27px; width:520px; margin:0px; font:16px Arial, Helvetica, sans-serif; font-weight:bold; color:#769841;"> Transforming lives through democratic &amp; participatory development </div> <?php include("menu.php"); ?> </div><!-- end header --> <!--begin main text div--> <div id="maincontent"> Relevant menu CSS: #nav, #nav ul { font:bold 11px Verdana, sans-serif; float: left; width: 980px; list-style: none; line-height: 1; background: white; font-weight: bold; padding: 0; border: solid #769841; border-width: 0; margin: 0 0 1em 0; } #nav a { display: block; width: 140px; /*this is the total width of the upper menu*/ w\idth: 120px; /*this is the width less horizontal padding */ padding: 5px 10px 5px 10px; /*horiz padding is the 2nd & 4th items here - goes Top Right Bottom Left */ color: #ffffff; background:#b6791e; text-decoration: none; } #nav a.daddy { background: url(rightarrow2.gif) center right no-repeat; } #nav li { float: left; padding: 0; width: 140px; /*this needs to be updated to match top #nav a */ background:#b6791e; } #nav li:hover, #nav li a:hover, #nav li:hover a { background:#769841; } #nav li:hover li a { background:#ffffff; color:#769841; } #nav li ul { position: absolute; left: -999em; height: auto; width: 14.4em; w\idth: 13.9em; font-weight: bold; border-width: 0.25em; /*green border around dropdown menu*/ margin: 0; } #nav li ul a { background:#ffffff; color:#769841; } #nav li li { padding-right: 1em; width: 13em; background:#ffffff; } #nav li ul a { width: 13em; w\idth: 9em; } #nav li ul ul { margin: -1.75em 0 0 14em; } #nav li:hover ul ul, #nav li:hover ul ul ul, #nav li.sfhover ul ul, #nav li.sfhover ul ul ul { left: -999em; } #nav li:hover ul, #nav li li:hover ul, #nav li li li:hover ul, #nav li.sfhover ul, #nav li li.sfhover ul, #nav li li li.sfhover ul { left: auto; } #nav li:hover, #nav li.sfhover, { background: #769841; color:#ffe400; } #nav li a:hover, #nav li li a:hover, #nav li:hover li:hover, #nav li.sfhover a:hover { background: #769841; color:#ffe400; }

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  • Trying to get around this Webservice call from Android using AsycTask

    - by Kevin Rave
    I am a fairly beginner in Android Development. I am developing an application that extensively relays on Webservice calls. First screen takes username and password and validates the user by calling the Webservice. If U/P is valid, then I need to fire up the 2nd activity. In that 2nd activity, I need to do 3 calls. But I haven't gotten to the 2nd part yet. In fact, I haven't completed the full coding yet. But I wanted to test if the app is working as far as I've come through. When calling webserivce, I am showing alert dialog. But the app is crashing somewhere. The LoginActivity shows up. When I enter U/P and press Login Button, it crashes. My classes: TaskHandler.java public class TaskHandler { private String URL; private User userObj; private String results; private JSONDownloaderTask task; ; public TaskHandler( String url, User user) { this.URL = url; this.userObj = user; } public String handleTask() { Log.d("Two", "In the function"); task = new JSONDownloaderTask(); Log.d("Three", "In the function"); task.execute(URL); return results; } private class JSONDownloaderTask extends AsyncTask<String, Void, String> { private String username;// = userObj.getUsername(); private String password; //= userObj.getPassword(); public HttpStatus status_code; public JSONDownloaderTask() { Log.d("con", "Success"); this.username = userObj.getUsername(); this.password = userObj.getPassword(); Log.d("User" + this.username , " Pass" + this.password); } private AsyncProgressActivity progressbar = new AsyncProgressActivity(); @Override protected void onPreExecute() { progressbar.showLoadingProgressDialog(); } @Override protected String doInBackground(String... params) { final String url = params[0]; //getString(R.string.api_staging_uri) + "Authenticate/"; // Populate the HTTP Basic Authentitcation header with the username and password HttpAuthentication authHeader = new HttpBasicAuthentication(username, password); HttpHeaders requestHeaders = new HttpHeaders(); requestHeaders.setAuthorization(authHeader); requestHeaders.setAccept(Collections.singletonList(MediaType.APPLICATION_JSON)); // Create a new RestTemplate instance RestTemplate restTemplate = new RestTemplate(); restTemplate.getMessageConverters().add(new MappingJacksonHttpMessageConverter()); try { // Make the network request Log.d(this.getClass().getName(), url); ResponseEntity<Message> response = restTemplate.exchange(url, HttpMethod.GET, new HttpEntity<Object>(requestHeaders), Message.class); status_code = response.getStatusCode(); return response.getBody().toString(); } catch (HttpClientErrorException e) { status_code = e.getStatusCode(); return new Message(0, e.getStatusText(), e.getLocalizedMessage(), "error").toString(); } catch ( Exception e ) { Log.d(this.getClass().getName() ,e.getLocalizedMessage()); return "Unknown Exception"; } } @Override protected void onPostExecute(String result) { progressbar.dismissProgressDialog(); switch ( status_code ) { case UNAUTHORIZED: result = "Invalid username or passowrd"; break; case ACCEPTED: result = "Invalid username or passowrd" + status_code; break; case OK: result = "Successful!"; break; } } } } AsycProgressActivity.java public class AsyncProgressActivity extends Activity { protected static final String TAG = AsyncProgressActivity.class.getSimpleName(); private ProgressDialog progressDialog; private boolean destroyed = false; @Override protected void onDestroy() { super.onDestroy(); destroyed = true; } public void showLoadingProgressDialog() { Log.d("Here", "Progress"); this.showProgressDialog("Authenticating..."); Log.d("Here", "afer p"); } public void showProgressDialog(CharSequence message) { Log.d("Here", "Message"); if (progressDialog == null) { progressDialog = new ProgressDialog(this); progressDialog.setIndeterminate(true); } Log.d("Here", "Message 2"); progressDialog.setMessage(message); progressDialog.show(); } public void dismissProgressDialog() { if (progressDialog != null && !destroyed) { progressDialog.dismiss(); } } } LoginActivity.java public class LoginActivity extends AsyncProgressActivity implements OnClickListener { Button login_button; HttpStatus status_code; /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); //this.requestWindowFeature(Window.FEATURE_NO_TITLE); setContentView(R.layout.main); login_button = (Button) findViewById(R.id.btnLogin); login_button.setOnClickListener(this); ViewServer.get(this).addWindow(this); } public void onDestroy() { super.onDestroy(); ViewServer.get(this).removeWindow(this); } public void onResume() { super.onResume(); ViewServer.get(this).setFocusedWindow(this); } public void onClick(View v) { if ( v.getId() == R.id.btnLogin ) { User userobj = new User(); String result; userobj.setUsername( ((EditText) findViewById(R.id.username)).getText().toString()); userobj.setPassword(((EditText) findViewById(R.id.password)).getText().toString() ); TaskHandler handler = new TaskHandler(getString(R.string.api_staging_uri) + "Authenticate/", userobj); Log.d(this.getClass().getName(), "One"); result = handler.handleTask(); Log.d(this.getClass().getName(), "After two"); Utilities.showAlert(result, LoginActivity.this); } } Utilities.java public class Utilities { public static void showAlert(String message, Context context) { AlertDialog.Builder alertDialogBuilder = new AlertDialog.Builder(context); alertDialogBuilder.setTitle("Login"); alertDialogBuilder.setMessage(message) .setCancelable(false) .setPositiveButton("OK",new DialogInterface.OnClickListener() { public void onClick(DialogInterface dialog,int id) { dialog.dismiss(); //dialog.cancel(); } }); alertDialogBuilder.setIcon(drawable.ic_dialog_alert); // create alert dialog AlertDialog alertDialog = alertDialogBuilder.create(); // show it alertDialog.show(); } }

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  • How to know when a user console is locked or has logged "back into" windows

    - by Paul Kohler
    This is in regards to applications that run in the taskbar but should be applicable to standard apps, Winforms, WPF, etc. Question: I am after some method (preferably via managed code) to be notified when a user either has their screen "locked" while my app is running and/or know when they log back in. GMail Notifier does this sort of thing for example, if my PC is locked for a while when I log in again it shows a list of emails that arrived since locking the PC. I'm looking to replicate that kind of functionality. Does anyone have any ideas on how to accomplish this?

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  • Java: Handling cookies when logging in with POST

    - by Cris Carter
    I'm having quite some trouble logging in to any site in Java. I'm using the default URLconnection POST request, but I'm unsure how to handle the cookies properly. I tried this guide: http://www.hccp.org/java-net-cookie-how-to.html But couldn't get it working. I've been trying basically for days now, and I really need help if anyone wants to help me. I'll probably be told that it's messy and that I should use a custom library meant for this stuff. I tried downloading one, but wasn't sure how to get it set up and working. I've been trying various things for hours now, and it just won't work. I'd rather do this with a standard URLconnection, but if anyone can help me get another library working that's better for this, that would be great, too. I would really appreciate if someone could post a working source that I could study. What I need is: POST login data to site - Get and store the cookie from the site - use cookie with next URLconnection requests to get logged-in version of the site. Can anyone help me with this? Would be EXTREMELY appreciated. It really does mean a lot. If anyone wants to actually help me out live, please leave an instant-messenger address. Thank you a lot for your time.

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  • Error 1053: the service did not respond to the start or control request in a timely fashion

    - by deejjaayy
    i know this is very much a "how long is a piece of string" type of question, however i have recently inherited a couple of applications that run as windows services, and i am having problems providing a gui (accessible from a context menu in system tray) with both of them. before you ask, the reason why we need a gui for a windows service is in order to be able to re-configure the behaviour of the windows service(s) without resorting to stopping/re-starting. my code works fine in debug mode, and i get the context menu come up, and everything behaves correctly etc. when i install the service via "installutil" using a named account (i.e., not Local System Account), the service runs fine, but doesn't display the icon in the system tray (i know this is normal behaviour because i don't have the "interact with desktop" option). here is the problem though - when i choose the "LocalSystemAccount" option, and check the "interact with desktop" option, the service takes AGES to start up for no obvious reason, and i just keep getting "Could not start the ... service on Local Computer. Error 1053: the service did not respond to the start or control request in a timely fashion". incidentally, i increased the windows service timeout from the default 30 seconds to 2 minutes via a registry hack (see http://support.microsoft.com/kb/824344, search for TimeoutPeriod in section 3), however the service start up still times out. my first question is - why might the "Local System Account" login takes SOOOOO MUCH LONGER than when the service logs in with the non-LocalSystemAccount, causing the windows service time-out? what's could the difference be between these two to cause such different behaviour at start up? secondly - taking a step back, all i'm trying to achieve, is simply a windows service that provides a gui for configuration - I'd be quite happy to run using the non-Local System Account (with named user/pwd), if I could get the service to interact with the desktop (that is, have a context menu available from the system tray). is this possible, and if so how? any pointers to the above questions would be very much appreciated! thanks in advance for your help.

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  • Uploading an xml direct to ftp

    - by Joshua Maerten
    i put this direct below a button: XmlDocument doc = new XmlDocument(); XmlElement root = doc.CreateElement("Login"); XmlElement id = doc.CreateElement("id"); id.SetAttribute("userName", usernameTxb.Text); id.SetAttribute("passWord", passwordTxb.Text); XmlElement name = doc.CreateElement("Name"); name.InnerText = nameTxb.Text; XmlElement age = doc.CreateElement("Age"); age.InnerText = ageTxb.Text; XmlElement Country = doc.CreateElement("Country"); Country.InnerText = countryTxb.Text; id.AppendChild(name); id.AppendChild(age); id.AppendChild(Country); root.AppendChild(id); doc.AppendChild(root); // Get the object used to communicate with the server. FtpWebRequest request = (FtpWebRequest)WebRequest.Create("ftp://users.skynet.be"); request.Method = WebRequestMethods.Ftp.UploadFile; request.UsePassive = false; // This example assumes the FTP site uses anonymous logon. request.Credentials = new NetworkCredential("fa490002", "password"); // Copy the contents of the file to the request stream. StreamReader sourceStream = new StreamReader(); byte[] fileContents = Encoding.UTF8.GetBytes(sourceStream.ReadToEnd()); sourceStream.Close(); request.ContentLength = fileContents.Length; Stream requestStream = request.GetRequestStream(); requestStream.Write(fileContents, 0, fileContents.Length); requestStream.Close(); FtpWebResponse response = (FtpWebResponse)request.GetResponse(); response.Close(); MessageBox.Show("Created SuccesFully!"); this.Close(); but i always get an error of the streamreader path, what do i need to place there ? the meening is, creating an account and when i press the button, an xml file is saved to, ftp://users.skynet.be/testxml/ the filename is from usernameTxb.Text + ".xml".

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  • PHP SDK Requires two logins...doesn't recognize first

    - by Jay Konieczny
    I am using the following code to authenticate whether users are logged in or not. While users can log in, they have to click the login button twice. Additionally, sometimes even after they click the log-in button twice, my "user info" part of the page (earlier in the page than the content) shows them as logged out while the actual page shows them as logged in. Here is the code. Could someone suggest a better way of handling log-ins? function isLoggedIn($facebook) { if (isset($facebook) and $facebook->getUser() != 0) { // UserID exists, but user may still not be logged in. Let's check: try { $facebook->api('/me', 'GET'); // If this succeeds, then they are logged in. return true; } catch(FacebookApiException $e) { // Some kind of error, so not logged in. if(session_id() === '') session_destroy(); return false; } } else { if(session_id() === '') session_destroy(); return false; } } Thanks!

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  • BlackBerry - waiting screen

    - by Pankaj Pareek
    Hello , i am developing one application in blackberry java development. I am requesting to http means i am connecting to web service .response of web service taking some time .That time i want to display some waiting screen. Could you tell me how can i do that.... Regards Pankaj Pareek

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  • Enable iPhone accelerometer while screen is locked..

    - by kamziro
    So apparently it is possible to keep the processor going processing stuff while the screen is locked, as indicated here: http://stackoverflow.com/questions/1551712/running-iphone-apps-while-in-sleep-mode However, after testing with the example code, UIAccelerometer will just stop giving value as soon as the device is locked pronto. Is there a way to force otherwise?

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  • White screen between Canvases

    - by user315647
    Hello everyone, I am having, what i believe is a minor issue. I am developing a J2ME application which predominantly uses canvases for display. The problem is I have set all these canvases to fullscreen and when i navigate from one class to another i am first given a white screen and then taken to the canvas i intend to go. I am not understanding what i am doing wrong I am using the following statement for navigation javax.microedition.lcdui.Display.getDisplay(MIDlet).setCurrent(Canvas); Please help!!

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  • black screen is displaying in android

    - by Aswan
    i opened my application after i am not doing any kind of operation i leave the mobile 5 min after 5 min when i touch the my application it is showing black screen.is there any way to avoid this idle state Thanks in advance Aswan

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