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  • Retrieving information with Python's urllib from a page that is done via __doPostBack()?

    - by Omar
    I'm trying to parse a page that has different sections that are loaded with a Javascript __doPostBack() function. An example of a link is: javascript:__doPostBack('ctl00$cphMain$ucOemSchPicker$dlSch$ctl03$btnSch','') As soon as this is clicked, the browser doesn't fetch a new URL but a section of webpage is updated to reflect new information. What would I pass into a urllib function to complete the operation?

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  • Embedding Flash & Quicktime Via JavaScript

    - by thesunneversets
    I have a JavaScript function that loads a flash movie into a webpage div using swfobject.embedSWF(). I want to be able to, alternatively, load a .mov file into the same div, in the event that this is the file found instead of the .swf. Is there a close equivalent to swfobject.embedSWF for the purposes of embedding a .mov file? If not, what is an efficient route to doing this using JavaScript?

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  • Problem in hiding scrollbar of webbrowser control

    - by Royson
    I have web browser control in panel and panel is placed in table layout. i have tried to hide scroll bar by webBrowser.ScrollBarsEnabled = false; but still there is scrollbar are visible. I want to hide it irrespective of page width. User should see page area which are best fitted on panel. How to do this.

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  • How to update the column of datagridview from the text contents of textbox in c# Windows form

    - by user286546
    I have a datagridview with contents from a table. In that I have a column for Remarks which will be 1-2 lines. When I click on the remarks column, I want to open another form that contains the text box. I have linked the text box with the table using the table adapter. Now when I close the form with the text box, I want to show that in the datagridview column. Please help me

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  • Kubuntu - Can't move/max/min windows [closed]

    - by GregH
    All of a sudden it seems when ever I open a window on my Kubuntu (9.10) system, the windows dock in the upper left corner and can't be moved. There is nor border on the windows, no min/max/close buttons in the upper right corner of the windows. I tried opening a term window but it seems I can't type in the window. Any ideas what might be causing this?

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  • How to download file into string with progress callback?

    - by Kaminari
    I would like to use the WebClient (or there is another better option?) but there is a problem. I understand that opening up the stream takes some time and this can not be avoided. However, reading it takes a strangely much more amount of time compared to read it entirely immediately. Is there a best way to do this? I mean two ways, to string and to file. Progress is my own delegate and it's working good. FIFTH UPDATE: Finally, I managed to do it. In the meantime I checked out some solutions what made me realize that the problem lies elsewhere. I've tested custom WebResponse and WebRequest objects, library libCURL.NET and even Sockets. The difference in time was gzip compression. Compressed stream lenght was simply half the normal stream lenght and thus download time was less than 3 seconds with the browser. I put some code if someone will want to know how i solved this: (some headers are not needed) public static string DownloadString(string URL) { WebClient client = new WebClient(); client.Headers["User-Agent"] = "Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US) AppleWebKit/532.5 (KHTML, like Gecko) Chrome/4.1.249.1045 Safari/532.5"; client.Headers["Accept"] = "application/xml,application/xhtml+xml,text/html;q=0.9,text/plain;q=0.8,image/png,*/*;q=0.5"; client.Headers["Accept-Encoding"] = "gzip,deflate,sdch"; client.Headers["Accept-Charset"] = "ISO-8859-2,utf-8;q=0.7,*;q=0.3"; Stream inputStream = client.OpenRead(new Uri(URL)); MemoryStream memoryStream = new MemoryStream(); const int size = 32 * 4096; byte[] buffer = new byte[size]; if (client.ResponseHeaders["Content-Encoding"] == "gzip") { inputStream = new GZipStream(inputStream, CompressionMode.Decompress); } int count = 0; do { count = inputStream.Read(buffer, 0, size); if (count > 0) { memoryStream.Write(buffer, 0, count); } } while (count > 0); string result = Encoding.Default.GetString(memoryStream.ToArray()); memoryStream.Close(); inputStream.Close(); return result; } I think that asyncro functions will be almost the same. But i will simply use another thread to fire this function. I dont need percise progress indication.

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  • Why sockets does not die when server dies? Why socket dies when server is alive?

    - by Roman
    I try to play with sockets a bit. For that I wrote very simple "client" and "server" applications. Client: import java.net.*; public class client { public static void main(String[] args) throws Exception { InetAddress localhost = InetAddress.getLocalHost(); System.out.println("before"); Socket clientSideSocket = null; try { clientSideSocket = new Socket(localhost,12345,localhost,54321); } catch (ConnectException e) { System.out.println("Connection Refused"); } System.out.println("after"); if (clientSideSocket != null) { clientSideSocket.close(); } } } Server: import java.net.*; public class server { public static void main(String[] args) throws Exception { ServerSocket listener = new ServerSocket(12345); while (true) { Socket serverSideSocket = listener.accept(); System.out.println("A client-request is accepted."); } } } And I found a behavior that I cannot explain: I start a server, than I start a client. Connection is successfully established (client stops running and server is running). Then I close the server and start it again in a second. After that I start a client and it writes "Connection Refused". It seems to me that the server "remember" the old connection and does not want to open the second connection twice. But I do not understand how it is possible. Because I killed the previous server and started a new one! I do not start the server immediately after the previous one was killed (I wait like 20 seconds). In this case the server "forget" the socket from the previous server and accepts the request from the client. I start the server and then I start the client. Connection is established (server writes: "A client-request is accepted"). Then I wait a minute and start the client again. And server (which was running the whole time) accept the request again! Why? The server should not accept the request from the same client-IP and client-port but it does!

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  • Why do sockets not die when server dies? Why does a socket die when server is alive?

    - by Roman
    I try to play with sockets a bit. For that I wrote very simple "client" and "server" applications. Client: import java.net.*; public class client { public static void main(String[] args) throws Exception { InetAddress localhost = InetAddress.getLocalHost(); System.out.println("before"); Socket clientSideSocket = null; try { clientSideSocket = new Socket(localhost,12345,localhost,54321); } catch (ConnectException e) { System.out.println("Connection Refused"); } System.out.println("after"); if (clientSideSocket != null) { clientSideSocket.close(); } } } Server: import java.net.*; public class server { public static void main(String[] args) throws Exception { ServerSocket listener = new ServerSocket(12345); while (true) { Socket serverSideSocket = listener.accept(); System.out.println("A client-request is accepted."); } } } And I found a behavior that I cannot explain: I start a server, than I start a client. Connection is successfully established (client stops running and server is running). Then I close the server and start it again in a second. After that I start a client and it writes "Connection Refused". It seems to me that the server "remember" the old connection and does not want to open the second connection twice. But I do not understand how it is possible. Because I killed the previous server and started a new one! I do not start the server immediately after the previous one was killed (I wait like 20 seconds). In this case the server "forget" the socket from the previous server and accepts the request from the client. I start the server and then I start the client. Connection is established (server writes: "A client-request is accepted"). Then I wait a minute and start the client again. And server (which was running the whole time) accept the request again! Why? The server should not accept the request from the same client-IP and client-port but it does!

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  • Why is this <p> expanding the whole page?

    - by George Edison
    If you visit this page and shrink your browser window, you will see my problem. [If you want to open the page in a new window, just hold down shift when you click the link.] The answers to the question extend beyond the page margin instead of wrapping. I have spent the last half hour working with Chrome's Inspector and Firefox's DOM inspector - all to no avail. I just cannot figure out why it's doing this.

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  • C#/.NET Project - Am I setting things up correctly?

    - by JustLooking
    1st solution located: \Common\Controls\Controls.sln and its project: \Common\Controls\Common.Controls\Common.Controls.csproj Description: This is a library that contains this class: public abstract class OurUserControl : UserControl { // Variables and other getters/setters common to our UserControls } 2nd solution located: \AControl\AControl.sln and its project: \AControl\AControl\AControl.csproj Description: Of the many forms/classes, it will contain this class: using Common.Controls; namespace AControl { public partial class AControl : OurUserControl { // The implementation } } A note about adding references (not sure if this is relevant): When I add references (for projects I create), using the names above: 1. I add Common.Controls.csproj to AControl.sln 2. In AControl.sln I turn off the build of Common.Controls.csproj 3. I add the reference to Common.Controls (by project) to AControl.csproj. This is the (easiest) way I know how to get Debug versions to match Debug References, and Release versions to match Release References. Now, here is where the issue lies (the 3rd solution/project that actually utilizes the UserControl): 3rd solution located: \MainProj\MainProj.sln and its project: \MainProj\MainProj\MainProj.csproj Description: Here's a sample function in one of the classes: private void TestMethod<T>() where T : Common.Controls.OurUserControl, new() { T TheObject = new T(); TheObject.OneOfTheSetters = something; TheObject.AnotherOfTheSetters = something_else; // Do stuff with the object } We might call this function like so: private void AnotherMethod() { TestMethod<AControl.AControl>(); } This builds, runs, and works. No problem. The odd thing is after I close the project/solution and re-open it, I have red squigglies everywhere. I bring up my error list and I see tons of errors (anything that deals with AControl will be noted as an error). I'll see errors such as: The type 'AControl.AControl' cannot be used as type parameter 'T' in the generic type or method 'MainProj.MainClass.TestMethod()'. There is no implicit reference conversion from 'AControl.AControl' to 'Common.Controls.OurUserControl'. or inside the actual method (the properties located in the abstract class): 'AControl.AControl' does not contain a definition for 'OneOfTheSetters' and no extension method 'OneOfTheSetters' accepting a first argument of type 'AControl.AControl' could be found (are you missing a using directive or an assembly reference?) Meanwhile, I can still build and run the project (then the red squigglies go away until I re-open the project, or close/re-open the file). It seems to me that I might be setting up the projects incorrectly. Thoughts?

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  • I Get the error message while write data in file --Cannot access the disposable object .Object name

    - by kunal rai
    This was the code: public static void SaveFile(Stream stream, string fileName = "") { using (IsolatedStorageFile file = IsolatedStorageFile.GetUserStoreForApplication()) { IsolatedStorageFileStream fs = file.CreateFile(fileName); var filesize = stream.Length; var getContent = new byte[(int)filesize]; stream.Read(getContent, 0, (int)filesize); fs.Write(getContent, 0, (int)filesize); fs.Close(); } }

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  • Question about InputMismatchException while using Scanner

    - by aser
    The question : Input file: customer’s account number, account balance at beginning of month, transaction type (withdrawal, deposit, interest), transaction amount Output: account number, beginning balance, ending balance, total interest paid, total amount deposited, number of deposits, total amount withdrawn, number of withdrawals package sentinel; import java.io.*; import java.util.*; public class Ex7 { /** * @param args the command line arguments */ public static void main(String[] args) throws FileNotFoundException { int AccountNum; double BeginningBalance; double TransactionAmount; int TransactionType; double AmountDeposited=0; int NumberOfDeposits=0; double InterestPaid=0.0; double AmountWithdrawn=0.0; int NumberOfWithdrawals=0; boolean found= false; Scanner inFile = new Scanner(new FileReader("Account.in")); PrintWriter outFile = new PrintWriter("Account.out"); AccountNum = inFile.nextInt(); BeginningBalance= inFile.nextDouble(); while (inFile.hasNext()) { TransactionAmount=inFile.nextDouble(); TransactionType=inFile.nextInt(); outFile.printf("Account Number: %d%n", AccountNum); outFile.printf("Beginning Balance: $%.2f %n",BeginningBalance); outFile.printf("Ending Balance: $%.2f %n",BeginningBalance); outFile.println(); switch (TransactionType) { case '1': // case 1 if we have a Deposite BeginningBalance = BeginningBalance + TransactionAmount; AmountDeposited = AmountDeposited + TransactionAmount; NumberOfDeposits++; outFile.printf("Amount Deposited: $%.2f %n",AmountDeposited); outFile.printf("Number of Deposits: %d%n",NumberOfDeposits); outFile.println(); break; case '2':// case 2 if we have an Interest BeginningBalance = BeginningBalance + TransactionAmount; InterestPaid = InterestPaid + TransactionAmount; outFile.printf("Interest Paid: $%.2f %n",InterestPaid); outFile.println(); break; case '3':// case 3 if we have a Withdraw BeginningBalance = BeginningBalance - TransactionAmount; AmountWithdrawn = AmountWithdrawn + TransactionAmount; NumberOfWithdrawals++; outFile.printf("Amount Withdrawn: $%.2f %n",AmountWithdrawn); outFile.printf("Number of Withdrawals: %d%n",NumberOfWithdrawals); outFile.println(); break; default: System.out.println("Invalid transaction Tybe: " + TransactionType + TransactionAmount); } } inFile.close(); outFile.close(); } } But is gives me this : Exception in thread "main" java.util.InputMismatchException at java.util.Scanner.throwFor(Scanner.java:840) at java.util.Scanner.next(Scanner.java:1461) at java.util.Scanner.nextInt(Scanner.java:2091) at java.util.Scanner.nextInt(Scanner.java:2050) at sentinel.Ex7.main(Ex7.java:36) Java Result: 1

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  • What is href=javascript:;

    - by vetri
    in a code which i am goin through there is a link has href=javascript:; in code.when it is clicked it opens a lightbox to show some msg with close button.how is it done.I think this uses dojo

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  • Can I use rest-client to POST a binary file to HTTP without multipart?

    - by Angela
    I have tried to do the following, but the web-service is NOT REST and does not take multi-part. What do I do in order to POST the image? @response = RestClient.post('http://www.postful.com/service/upload', {:upload => { :file => File.new("#{@postalcard.postalimage.path}",'rb') } }, {"Content-Type" => @postalcard.postalimage.content_type, "Content-Length" => @postalcard.postalimage.size, "Authorization" => 'Basic xxxxxx' } # end headers ) #close arguments to Restclient.post

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  • Javascript - Canvas image never appears on first function run

    - by Matt
    I'm getting a bit of a weird issue, the image never shows the first time you run the game in your browser, after that you see it every time. If you close your browser and re open it and run the game again, the same issue occurs - you don't see the image the first time you run it. Here's the issue in action, just hit a wall and there's no image the first time on the end game screen. Any help would be appreciated. Regards, Matt function showGameOver() { ctx.clearRect(0, 0, canvas.width, canvas.height); ctx.fillStyle = "black"; ctx.font = "16px sans-serif"; ctx.fillText("Game Over!", ((canvas.width / 2) - (ctx.measureText("Game Over!").width / 2)), 50); ctx.font = "12px sans-serif"; ctx.fillText("Your Score Was: " + score, ((canvas.width / 2) - (ctx.measureText("Your Score Was: " + score).width / 2)), 70); myimage = new Image(); myimage.src = "xcLDp.gif"; var size = [119, 26], //set up size coord = [443, 200]; ctx.font = "12px sans-serif"; ctx.fillText("Restart", ((canvas.width / 2) - (ctx.measureText("Restart").width / 2)), 197); ctx.drawImage( //draw it on canvas myimage, coord[0], coord[1], size[0], size[1] ); $("canvas").click(function(e) { //when click.. if ( testIfOver(this, e, size, coord) ) { startGame(); //reload } }); $("canvas").mousemove(function(e) { //when mouse moving if ( testIfOver(this, e, size, coord) ) { $(this).css("cursor", "pointer"); //change the cursor } else { $(this).css("cursor", "default"); //change it back } }); function testIfOver(ele,ev,size,coord){ if ( ev.pageX > coord[0] + ele.offsetLeft && ev.pageX < coord[0] + size[0] + ele.offsetLeft && ev.pageY > coord[1] + ele.offsetTop && ev.pageY < coord[1] + size[1] + ele.offsetTop ) { return true; } return false; } }

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  • How to properly encode "[" and "]" in queries using Apache HttpClient?

    - by Jason Nichols
    I've got a GET method that looks like the following: GetMethod method = new GetMethod("http://host/path/?key=[\"item\",\"item\"]"); Such a path works just fine when typed directly into a browser, but the above line when run causes an IllegalArgumentException : Invalid URI. I've looked at using the URIUtils class, but without success. Is there a way to automatically encode this (or to add a query string onto the URL without causing HttpClient to barf?).

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  • .htaccess, two consecutive rewrites?

    - by Matthew Haworth
    I need to take a url, "/ServiceSearch/r.php?n=blahblah", and have it go to "/search/blahblah/" so that it appears in the browser as "/search/blahblah", but I actually want it to REALLY be going to "r.php?n=ServiceSearch&n=blahblah".. So I was thinking I'll need to rewrite the first URL to "/ServiceSearch/r.php?n=blahblah" and then the second url, "/search/blahblah/", to the third, "r.php?n=ServiceSearch&n=blahblah". Well, I know this is wrong, but it's my best guess. I'm really struggling with it.

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  • finding and returning a string with a specified prefix

    - by tipu
    I am close but I am not sure what to do with the restuling match object. If I do p = re.search('[/@.* /]', str) I'll get any words that start with @ and end up with a space. This is what I want. However this returns a Match object that I dont' know what to do with. What's the most computationally efficient way of finding and returning a string which is prefixed with a @? For example, "Hi there @guy" After doing the proper calculations, I would be returned guy

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  • file output in python giving me garbage

    - by Richard
    When I write the following code I get garbage for an output. It is just a simple program to find prime numbers. It works when the first for loops range only goes up to 1000 but once the range becomes large the program fail's to output meaningful data output = open("output.dat", 'w') for i in range(2, 10000): prime = 1 for j in range(2, i-1): if i%j == 0: prime = 0 j = i-1 if prime == 1: output.write(str(i) + " " ) output.close() print "writing finished"

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  • my layout breaks in IE7 and javascript page reloads make the screen blink

    - by chibineku
    My layout breaks if I change the window size in IE7/AOL, so I added a simple javascript function that fires on window.onresize, but no matter how I change the location I get problems. It was suggested I post a link and here it is: link text I already use PHP to detect browser and include an IE7-only inline stylesheet (and for mobile browsers), and my page looks nearly identical to the way it does in FF, Opera, Chrome, Safari and IE8, but when I change the window size, some things go wonky, and come back into line if you refresh. Any advice is welcome :)

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