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  • Database.ExecuteNonQuery does not return

    - by dan-waterbly
    I have a very odd issue. When I execute a specific database stored procedure from C# using SqlCommand.ExecuteNonQuery, my stored procedure is never executed. Furthermore, SQL Profiler does not register the command at all. I do not receive a command timeout, and no exeception is thrown. The weirdest thing is that this code has worked fine over 1,200,000 times, but for this one particular file I am inserting into the database, it just hangs forever. When I kill the application, I receive this error in the event log of the database server: "A fatal error occurued while reading the input stream from the network. The session will be terminated (input error: 64, output error: 0). Which makes me think that the database server is receiving the command, though SQL Profiler says otherwise. I know that the appropiate permissions are set, and that the connection string is right as this piece of code and stored procedure works fine with other files. Below is the code that calls the stored procedure, it may be important to note that the file I am trying to insert is 33.5MB, but I have added more than 10,000 files larger than 500MB, so I do not think the size is the issue: using (SqlConnection sqlconn = new SqlConnection(ConfigurationManager.ConnectionStrings["TheDatabase"].ConnectionString)) using (SqlCommand command = sqlconn.CreateCommand()) { command.CommandText = "Add_File"; command.CommandType = CommandType.StoredProcedure; command.CommandTimeout = 30 //should timeout in 30 seconds, but doesn't... command.Parameters.AddWithValue("@ID", ID).SqlDbType = SqlDbType.BigInt; command.Parameters.AddWithValue("@BinaryData", byteArr).SqlDbType = SqlDbType.VarBinary; command.Parameters.AddWithValue("@FileName", fileName).SqlDbType = SqlDbType.VarChar; sqlconn.Open(); command.ExecuteNonQuery(); } There is no firewall between the server making the call and the database server, and the windows firewalls have been disabled to troubleshoot this issue.

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  • How to keep form items selected after post request?

    - by Ole Jak
    I have a simple html form. On php page. A simple list is placed on form. I submit this form (selected list items) to this page so it gives me page refresh. I want items which were POSTED to be selected after form was submited. How to do such thing? For my form I use such code: <form action="FormPage.php" method="post"> <select id="Streams" class="multiselect ui-widget-content ui-corner-all" multiple="multiple" name="Streams[]"> <?php $query = " SELECT s.streamId, s.userId, u.username FROM streams AS s JOIN user AS u ON s.userId = u.id LIMIT 0 , 30 "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); $streams_count = mysql_num_rows($streams_set); while ($row = mysql_fetch_array($streams_set)){ echo '<option value="' , $row['streamId'] , '"> ' , $row['username'] , ' (' , $row['streamId'] ,')' ,'</option> '; } ?> </select> <br/> <input type="submit" class="ui-state-default ui-corner-all" name="submitForm" id="submitForm" value="Play Stream from selected URL's!"/> </fieldset> </form>

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  • How does the receiver of a cipher text know the IV used for encryption?

    - by PatrickL
    If a random IV is used in encrypting plain text, how does the receiver of the cipher text know what the IV is in order to decrypt it? This is a follow-up question to a response to the previous stackoverflow question on IVs here. The IV allows for plaintext to be encrypted such that the encrypted text is harder to decrypt for an attacker. Each bit of IV you use will double the possibilities of encrypted text from a given plain text. The point is that the attacker doesn't know what the IV is and therefore must compute every possible IV for a given plain text to find the matching cipher text. In this way, the IV acts like a password salt. Most commonly, an IV is used with a chaining cipher (either a stream or block cipher). ... So, if you have a random IV used to encrypt the plain text, how do you decrypt it? Simple. Pass the IV (in plain text) along with your encrypted text. Wait. You just said the IV is randomly generated. Then why pass it as plain text along with the encrypted text?

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  • What hash algorithms are paralellizable? Optimizing the hashing of large files utilizing on mult-co

    - by DanO
    I'm interested in optimizing the hashing of some large files (optimizing wall clock time). The I/O has been optimized well enough already and the I/O device (local SSD) is only tapped at about 25% of capacity, while one of the CPU cores is completely maxed-out. I have more cores available, and in the future will likely have even more cores. So far I've only been able to tap into more cores if I happen to need multiple hashes of the same file, say an MD5 AND a SHA256 at the same time. I can use the same I/O stream to feed two or more hash algorithms, and I get the faster algorithms done for free (as far as wall clock time). As I understand most hash algorithms, each new bit changes the entire result, and it is inherently challenging/impossible to do in parallel. Are any of the mainstream hash algorithms parallelizable? Are there any non-mainstream hashes that are parallelizable (and that have at least a sample implementation available)? As future CPUs will trend toward more cores and a leveling off in clock speed, is there any way to improve the performance of file hashing? (other than liquid nitrogen cooled overclocking?) or is it inherently non-parallelizable?

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  • PHP Outputting File Attachments with Headers

    - by OneNerd
    After reading a few posts here I formulated this function which is sort of a mishmash of a bunch of others: function outputFile( $filePath, $fileName, $mimeType = '' ) { // Setup $mimeTypes = array( 'pdf' => 'application/pdf', 'txt' => 'text/plain', 'html' => 'text/html', 'exe' => 'application/octet-stream', 'zip' => 'application/zip', 'doc' => 'application/msword', 'xls' => 'application/vnd.ms-excel', 'ppt' => 'application/vnd.ms-powerpoint', 'gif' => 'image/gif', 'png' => 'image/png', 'jpeg' => 'image/jpg', 'jpg' => 'image/jpg', 'php' => 'text/plain' ); // Send Headers //-- next line fixed as per suggestion -- header('Content-Type: ' . $mimeTypes[$mimeType]); header('Content-Disposition: attachment; filename="' . $fileName . '"'); header('Content-Transfer-Encoding: binary'); header('Accept-Ranges: bytes'); header('Cache-Control: private'); header('Pragma: private'); header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); readfile($filePath); } I have a php page (file.php) which does something like this (lots of other code stripped out): // I run this thru a safe function not shown here $safe_filename = $_GET['filename']; outputFile ( "/the/file/path/{$safe_filename}", $safe_filename, substr($safe_filename, -3) ); Seems like it should work, and it almost does, but I am having the following issues: When its a text file, I am getting a strange symbol as the first letter in the text document When its a word doc, it is corrupt (presumably that same first bit or byte throwing things off). I presume all other file types will be corrupt - have not even tried them Any ideas on what I am doing wrong? Thanks - UPDATE: changed line of code as suggested - still same issue.

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  • Accents in uploaded file being replaced with '?'

    - by Katfish
    I am building a data import tool for the admin section of a website I am working on. The data is in both French and English, and contains many accented characters. Whenever I attempt to upload a file, parse the data, and store it in my MySQL database, the accents are replaced with '?'. I have text files containing data (charset is iso-8859-1) which I upload to my server using CodeIgniter's file upload library. I then read the file in PHP. My code is similar to this: $this->upload->do_upload() $data = array('upload_data' => $this->upload->data()); $fileHandle = fopen($data['upload_data']['full_path'], "r"); while (($line = fgets($fileHandle)) !== false) { echo $line; } This produces lines with accents replaced with '?'. Everything else is correct. If I download my uploaded file from my server over FTP, the charset is still iso-8850-1, but a diff reveals that the file has changed. However, if I open the file in TextEdit, it displays properly. I attempted to use PHP's stream_encoding method to explicitly set my file stream to iso-8859-1, but my build of PHP does not have the method. After running out of ideas, I tried wrapping my strings in both utf8_encode and utf8_decode. Neither worked. If anyone has any suggestions about things I could try, I would be extremely grateful.

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  • Data structure to build and lookup set of integer ranges

    - by actual
    I have a set of uint32 integers, there may be millions of items in the set. 50-70% of them are consecutive, but in input stream they appear in unpredictable order. I need to: Compress this set into ranges to achieve space efficient representation. Already implemented this using trivial algorithm, since ranges computed only once speed is not important here. After this transformation number of resulting ranges is typically within 5 000-10 000, many of them are single-item, of course. Test membership of some integer, information about specific range in the set is not required. This one must be very fast -- O(1). Was thinking about minimal perfect hash functions, but they do not play well with ranges. Bitsets are very space inefficient. Other structures, like binary trees, has complexity of O(log n), worst thing with them that implementation make many conditional jumps and processor can not predict them well giving poor performance. Is there any data structure or algorithm specialized in integer ranges to solve this task?

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  • create and write to a text file in vb.net

    - by woolardz
    I'm creating a small vb.net application, and I'm trying trying to write a list of results from a listview to a text file. I've looked online and found the code to open the save file dialog and write the text file. When I click save on the save file dialog, I receive an IOException with the message "The process cannot access the file 'C:\thethe.txt' because it is being used by another process." The text file is created in the correct location, but is empty. The application quits at this line "Dim fs As New FileStream(saveFileDialog1.FileName, FileMode.OpenOrCreate, FileAccess.Write)" Thanks in advance for any help. Private Sub btnSave_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnSave.Click Dim myStream As Stream Dim saveFileDialog1 As New SaveFileDialog() saveFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" saveFileDialog1.FilterIndex = 2 saveFileDialog1.RestoreDirectory = True If saveFileDialog1.ShowDialog() = DialogResult.OK Then myStream = saveFileDialog1.OpenFile() If (myStream IsNot Nothing) Then Dim fs As New FileStream(saveFileDialog1.FileName, FileMode.OpenOrCreate, FileAccess.Write) Dim m_streamWriter As New StreamWriter(fs) m_streamWriter.Flush() 'Write to the file using StreamWriter class m_streamWriter.BaseStream.Seek(0, SeekOrigin.Begin) 'write each row of the ListView out to a tab-delimited line in a file For i As Integer = 0 To Me.ListView1.Items.Count - 1 m_streamWriter.WriteLine(((ListView1.Items(i).Text & vbTab) + ListView1.Items(i).SubItems(0).ToString() & vbTab) + ListView1.Items(i).SubItems(1).ToString()) Next myStream.Close() End If End If End Sub

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  • Add "Become-a-Fan" button to my website

    - by vamsivanka
    I am trying to add "Become-a-Fan" button to my 2008 asp.net website (vb) Here is an example i have followed http://www.docstoc.com/docs/9646635/Add-a-Facebook-Fan-Button-to-your-Website/ posted this following script to my website: <script type="text/javascript" src="http://static.ak.connect.facebook.com/connect.php/en_US"></script><div id="fb-root"></div> <script> window.fbAsyncInit = function() { FB.init({ appId: "myapiid", xfbml: true }); }; (function() { var e = document.createElement('script'); e.async = true; e.src = document.location.protocol + '//connect.facebook.net/en_US/all.js'; document.getElementById('fb-root').appendChild(e); }()); </script> <fb:fan profile_id="253787057811" stream="0" connections="0" logobar="1" width="300"></fb:fan> <div style="font-size:8px; padding-left:10px"> <a href="http://www.facebook.com/pages/FanPageName/myapiid">FanPageName</a> on Facebook </div> But Not able to see the Image button "Become-a-Fan" Please Let me know.

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  • What's causing this permission's error and how can I work around it?

    - by Scott B
    Warning: move_uploaded_file(/home/site/public_html/wp-content/themes/mytheme/upgrader.zip) [function.move-uploaded-file]: failed to open stream: Permission denied in /home/site/public_html/wp-content/themes/mytheme/uploader.php on line 79 Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move '/tmp/phptempfile' to '/home/site/public_html/wp-content/themes/mytheme/upgrader.zip' in /home/site/public_html/wp-content/themes/mytheme/uploader.php on line 79 There was a problem. Sorry! Code is below for that line... // permission settings for newly created folders $chmod = 0755; // Ensures that the correct file was chosen $accepted_types = array('application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/s-compressed'); foreach($accepted_types as $mime_type) { if($mime_type == $type) { $okay = true; break; } } $okay = strtolower($name[1]) == 'zip' ? true: false; if(!$okay) { die("This upgrader requires a zip file. Please make sure your file is a valid zip file with a .zip extension"); } //mkdir($target); $saved_file_location = $target . $filename; //Next line is 79 if(move_uploaded_file($source, $saved_file_location)) { openZip($saved_file_location); } else { die("There was a problem. Sorry!"); }

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  • Resize my image on upload not working - what must my buffer be?

    - by Etienne
    This is the code i got from This Link I want the user to upload a picture and then resize it............. Public Sub ResizeFromStream(ByVal ImageSavePath As String, ByVal MaxSideSize As Integer, ByVal Buffer As System.IO.Stream) Dim intNewWidth As Integer Dim intNewHeight As Integer Dim imgInput As System.Drawing.Image = System.Drawing.Image.FromStream(Buffer) 'Determine image format Dim fmtImageFormat As ImageFormat = imgInput.RawFormat 'get image original width and height Dim intOldWidth As Integer = imgInput.Width Dim intOldHeight As Integer = imgInput.Height 'determine if landscape or portrait Dim intMaxSide As Integer If (intOldWidth >= intOldHeight) Then intMaxSide = intOldWidth Else intMaxSide = intOldHeight End If If (intMaxSide > MaxSideSize) Then 'set new width and height Dim dblCoef As Double = MaxSideSize / CDbl(intMaxSide) intNewWidth = Convert.ToInt32(dblCoef * intOldWidth) intNewHeight = Convert.ToInt32(dblCoef * intOldHeight) Else intNewWidth = intOldWidth intNewHeight = intOldHeight End If 'create new bitmap Dim bmpResized As Drawing.Bitmap = New Drawing.Bitmap(imgInput, intNewWidth, intNewHeight) 'save bitmap to disk bmpResized.Save(ImageSavePath, fmtImageFormat) 'release used resources imgInput.Dispose() bmpResized.Dispose() Buffer.Close() End Sub Now when i click on my submit button it must execute my code but i am not sure what the input must be for the Buffer field? Protected Sub btnUpload_Click() Handles btnUpload.Click ResizeFromStream("~Pics", 200, ??????????) End Sub Thanks in advanced! Edit I need to get my Image from the File Upload control!

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  • Saving ntext data from SQL Server to file directory using asp

    - by April
    A variety of files (pdf, images, etc.) are stored in a ntext field on a MS SQL Server. I am not sure what type is in this field, other than it shows question marks and undefined characters, I am assuming they are binary type. The script is supposed to iterate through the rows and extract and save these files to a temp directory. "filename" and "contenttype" are given, and "data" is whatever is in the ntext field. I have tried several solutions: 1) data.SaveToFile "/temp/"&filename, 2 Error: Object required: '????????????????????' ??? 2) File.WriteAllBytes "/temp/"&filename, data Error: Object required: 'File' I have no idea how to import this, or the Server for MapPath. (Cue: what a noob!) 3) Const adTypeBinary = 1 Const adSaveCreateOverWrite = 2 Dim BinaryStream Set BinaryStream = CreateObject("ADODB.Stream") BinaryStream.Type = adTypeBinary BinaryStream.Open BinaryStream.Write data BinaryStream.SaveToFile "C:\temp\" & filename, adSaveCreateOverWrite Error: Arguments are of the wrong type, are out of acceptable range, or are in conflict with one another. 4) Response.ContentType = contenttype Response.AddHeader "content-disposition","attachment;" & filename Response.BinaryWrite data response.end This works, but the file should be saving to the server instead of popping up save-as dialog. I am not sure if there is a way to save the response to file. Thanks for shedding light on any of these problems!

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  • How can I write an XML on my hard drive to GetRequestStream

    - by swolff1978
    I need to post raw xml to a site and read the response. With the following code I keep getting an "Unknown File Format" error and I'm not sure why. XmlDocument sampleRequest = new XmlDocument(); sampleRequest.Load(@"C:\SampleRequest.xml"); byte[] bytes = Encoding.UTF8.GetBytes(sampleRequest.ToString()); string uri = "https://www.sample-gateway.com/gw.aspx"; req = WebRequest.Create(uri); req.Method = "POST"; req.ContentLength = bytes.Length; req.ContentType = "text/xml"; using (var requestStream = req.GetRequestStream()) { requestStream.Write(bytes, 0, bytes.Length); } // Send the data to the webserver rsp = req.GetResponse(); XmlDocument responseXML = new XmlDocument(); using (var responseStream = rsp.GetResponseStream()) { responseXML.Load(responseStream); } I am fairly certain my issue is what/how I am writing to the requestStream so.. How can I modify that code so that I may write an xml located on the hard drive to the request stream?

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  • Having trouble uploading a file

    - by neo skosana
    Hi I am having trouble uploading a file. First of all I have a class: class upload { private $name; private $document; public function __construct($nme,$doc) { $this->setName($nme); $this->setDocument($doc); } public function setName($nme) { $this->name = $nme; } public function setDocument($doc) { $this->document = $doc; } public function fileNotPdf() { /* Was the file a PDF? */ if ($this->document['type'] != "application/pdf") { return true; } else { return false; } } public function fileNotUploaded() { /* Make sure that the file was POSTed. */ if (!(is_uploaded_file($this->document['tmp_name']))) { return true; } else { return false; } } public function fileNotMoved($repositry) { /* move uploaded file to final destination. */ $result = move_uploaded_file($this->document['tmp_name'], "$repositry/$this->name.pdf"); if($result) { return false; } else { return true; } } } Now for my main page: $docName = $_POST['name']; $page = $_FILES['doc']; if($_POST['submit']) { /* Set a few constants */ $filerepository = "np"; $uploadObj = new upload($docName, $page); if($uploadObj->fileNotUploaded()) { promptUser("There was a problem uploading the file.",""); } elseif($uploadObj->fileNotPdf()) { promptUser("File must be in pdf format.",""); } elseif($uploadObj->fileNotMoved($filerepository)) { promptUser("File could not be uploaded to final destination.",""); } else { promptUser("File has been successfully uploaded.",""); } } The errors that I get: Warning: move_uploaded_file(about.pdf)[function.move-uploaded-file]: failed to open stream: No such file or directory in... Warning: move_uploaded_file()[function.move-uploaded-file]: Unable to move 'c:\xampp\tmp\php13.tmp' to 'about.pdf' in... File could not be uploaded to final destination.

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  • My C# UploadFile method successfully uploads a file, but then my UI hangs...

    - by kyrathaba
    I have a simple WinForms test application in C#. Using the following method, I'm able to upload a file when I invoke the method from my button's Click event handler. The only problem is: my Windows Form "freezes". I can't close it using the Close button. I have to end execution from within the IDE (Visual C# 2010 Express edition). Here are the two methods: public void UploadFile(string FullPathFilename) { string filename = Path.GetFileName(FullPathFilename); try { FtpWebRequest request = (FtpWebRequest)WebRequest.Create(_remoteHost + filename); request.Method = WebRequestMethods.Ftp.UploadFile; request.Credentials = new NetworkCredential(_remoteUser, _remotePass); StreamReader sourceStream = new StreamReader(FullPathFilename); byte[] fileContents = Encoding.UTF8.GetBytes(sourceStream.ReadToEnd()); request.ContentLength = fileContents.Length; Stream requestStream = request.GetRequestStream(); requestStream.Write(fileContents, 0, fileContents.Length); FtpWebResponse response = (FtpWebResponse)request.GetResponse(); response.Close(); requestStream.Close(); sourceStream.Close(); } catch (Exception ex) { MessageBox.Show(ex.Message, "Upload error"); } finally { } } which gets called here: private void btnUploadTxtFile_Click(object sender, EventArgs e) { string username = "my_username"; string password = "my_password"; string host = "ftp://mywebsite.com"; try { clsFTPclient client = new clsFTPclient(host + "/httpdocs/non_church/", username, password); client.UploadFile(Path.GetDirectoryName(Application.ExecutablePath) + "\\myTextFile.txt"); } catch (Exception ex) { MessageBox.Show(ex.Message, "Upload problem"); } }

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  • Login to website and use cookie to get source for another page

    - by Stu
    I am trying to login to the TV Rage website and get the source code of the My Shows page. I am successfully logging in (I have checked the response from my post request) but then when I try to perform a get request on the My Shows page, I am re-directed to the login page. This is the code I am using to login: private string LoginToTvRage() { string loginUrl = "http://www.tvrage.com/login.php"; string formParams = string.Format("login_name={0}&login_pass={1}", "xxx", "xxxx"); string cookieHeader; WebRequest req = WebRequest.Create(loginUrl); req.ContentType = "application/x-www-form-urlencoded"; req.Method = "POST"; byte[] bytes = Encoding.ASCII.GetBytes(formParams); req.ContentLength = bytes.Length; using (Stream os = req.GetRequestStream()) { os.Write(bytes, 0, bytes.Length); } WebResponse resp = req.GetResponse(); cookieHeader = resp.Headers["Set-cookie"]; String responseStream; using (StreamReader sr = new StreamReader(resp.GetResponseStream())) { responseStream = sr.ReadToEnd(); } return cookieHeader; } I then pass the cookieHeader into this method which should be getting the source of the My Shows page: private string GetSourceForMyShowsPage(string cookieHeader) { string pageSource; string getUrl = "http://www.tvrage.com/mytvrage.php?page=myshows"; WebRequest getRequest = WebRequest.Create(getUrl); getRequest.Headers.Add("Cookie", cookieHeader); WebResponse getResponse = getRequest.GetResponse(); using (StreamReader sr = new StreamReader(getResponse.GetResponseStream())) { pageSource = sr.ReadToEnd(); } return pageSource; } I have been using this previous question as a guide but I'm at a loss as to why my code isn't working.

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  • How do I set the LRECL in C#.NET?

    - by donde
    I have been trying to ftp a dtl file from .net to, what I beleive, is an AS400. The error being reported back to me is: "One or more lines have been truncated" and the admin is saying the file is coming over with 256 lines that have variable length columns. I found this explanation online: we have to establish defaults because no specifics about the file exist. The default RECFM is V and LRECL is 256. This means that SAS will scan the input record looking for the CR & LF to tell us that we are at the EOR. If the marker isn't found within the limit of the LRECL, SAS discards the data from the LRECL value to the end of the record and adds a message to the Log that "One or more lines have been truncated". So I need to set the LRECL. How do I do this in C# .NET? FtpWebRequest ftp = (FtpWebRequest)FtpWebRequest.Create(ftpfullpath); ftp.Credentials = new NetworkCredential(user, pwd); ftp.KeepAlive = false; ftp.UseBinary = false; ftp.Method = WebRequestMethods.Ftp.UploadFile; FileStream fs = File.OpenRead(inputfilepath + ftpfileName); byte[] buffer = new byte[fs.Length]; fs.Read(buffer, 0, buffer.Length); fs.Close(); Stream ftpstream = ftp.GetRequestStream(); int i = 0; int intBlock = 1786; int intBuffLeft = buffer.Length; while (i < buffer.Length) { if (intBuffLeft >= 1786) { ftpstream.Write(buffer, i, intBlock); } else { ftpstream.Write(buffer, i, intBuffLeft); } i += intBlock; intBuffLeft -= 1786; } ftpstream.Close();

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  • how to get selected chracter position in JTextArea?

    - by Reddy
    Hi! Here is a challenging question! Let me first tell you my scenario how am i implementing a solution to a problem. I am reading a log file and displaying it on the JTextArea. Log file is cp037 character coded. I was reading each file as a byte stream or byte array from the log file & displaying it. Anyways, i managed to display the text properly in JTextArea by cp037 character coding. Now, User may select a set of characters in the JTextArea. All i want is the position of first character of the user's selected text, from a nearest special character '+'(its character code in cp037 is 4E), which is prior to the selected text. This character may occur at several places in the JTextArea. In simple sentence, i want the first character location(of user selected text) from nearset '+' which should be occuring prior to the user's selected text. PS: cp037 is a type of character encoding scheme which is created by IBM & used for IBM Mainframes. Please fell free to ask me if the question is not clear...:-

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  • How can i get a file from remote machine?

    - by programmerist
    How can i get a file from remote computer? i know remote computer ip and 51124 port is open. i need this algorith: 1) Connect 192.xxx.x.xxx ip via 51124 port 2) filename:123456 (i want to search it on remote machine) 3) Get File 4) Save C:\ 51124 port is open. can i access and can i search any file according to filename? My code is below: IPEndPoint ipEnd = new IPEndPoint(IPAddress.Any, 51124); Socket sock = new Socket(AddressFamily.InterNetwork, SocketType.Stream, ProtocolType.IP); sock.Bind(ipEnd); sock.Listen(maxConnections); Socket serverSocket = sock.Accept(); byte[] data = new byte[bufferSize]; int received = serverSocket.Receive(data); int filenameLength = BitConverter.ToInt32(data, 0); string filename = Encoding.ASCII.GetString(data, 4, filenameLength); BinaryWriter bWrite = new BinaryWriter(File.Open(outPath + filename, FileMode.Create)); bWrite.Write(data, filenameLength + 4, received - filenameLength - 4); int received2 = serverSocket.Receive(data); while (received2 0) { bWrite.Write(data, 0, received2); received2 = serverSocket.Receive(data); } bWrite.Close(); serverSocket.Close(); sock.Close();

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  • Why can't I open a JBoss vfs:/ URL?

    - by skiphoppy
    We are upgrading our application from JBoss 4 to JBoss 6. A couple of pieces of our application get delivered to the client in an unusual way: jars are looked up inside of our application and sent to the client from a servlet, where the client extracts them in order to run certain support functions. In JBoss 4 we would look these jars up with the classloader and find a jar:// URL which would be used to read the jar and send its contents to the client. In JBoss 6 when we perform the lookup we get a vfs:/ URL. I understand that this is from the org.jboss.vfs package. Unfortunately when I call openStream() on this URL and read from the stream, I immediately get an EOF (read() returns -1). What gives? Why can't I read the resource this URL refers to? I've tried trying to access the underlying VFS packages to open the file through the JBoss VFS API, but most of the API appears to be private, and I couldn't find a routine to translate from a vfs:/ URL to a VFS VirtualFile object, so I couldn't get anywhere. I can try to find the file on disk within JBoss, but that approach sounds very failure prone on upgrade. Our old approach was to use Java Web Start to distribute the jars to the client and then look them up within Java Web Start's cache to extract them. But that broke on every minor upgrade of Java because the layout of the cache changed.

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  • php lampp permissions of fopen function

    - by marmoushismail
    hi i'm programming php using: netbeans 6.8 lampp for ubuntu (xampp) apache which came with xampp $fh = fopen("testfile2.txt", 'w') or die("Failed to create file"); $text ="hello man cool good"; fwrite($fh, $text) or die("Could not write to file"); fclose($fh); echo "File 'testfile.txt' written successfully"; //i get the next error: Warning: fopen(testfile2.txt) [function.fopen]: failed to open stream: Permission denied in /home/marmoush/allprojects/phpprojects/myindex.php on line 91 Failed to create file anyway i know what this error is; it's about folder and files permissions; i looked into the folder permission tab made access available for "others" group ( to read and write) the program worked result was a file (test.txt) so i looked at the created file permission it appears to be that (php , xampp or whoever) creates file with (nobody permission) I have 2 QUESTIONS: 1- what if i need the file created by (php code and xampp ) to have the "root or user or myname" permissions ?? where to set this setting 2-also my concern (what if i send this files to actual web server will it make nobody permissions also nobody ? when they create files

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  • How do I get Firefox to launch Visio when I click on a linked .vsd file?

    - by Dean
    On our intranet site, we have various MS Office documents linked. When I click on a Word, Excel or PowerPoint file, Firefox gives me the option to Open, Save or Cancel. When I click on Open, the appropriate app is launched and the file is loaded. This is perfect. But for some reason, when I click on a linked Visio file, I only get the option to Save, which is inconvenient. I know that Firefox knows the linked file is a Visio file because it tells me so in the dialog box: "You have chosen to open example.vsd which is a: Microsoft Visio Drawing". How can I make Firefox launch Visio when I click on a linked Visio file? Update: Firefox is not launching Visio when I click on a linked Visio file because the web server does not identify the content-type correctly. It identifies the Visio file as application/octet-stream instead of application/x-visio. (Thanks Grant Wagner.) This explains why it doesn't work. And in my case, I may be able to get the Apache config file changed, but this is not certain. However, I would love to know if there is a way to configure Firefox itself to launch Visio based on some other criteria, like file name extension. This way I can open Visio files even if I don't have access to the Apache configuration.

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  • How to Avoid Server Error 414 for Very Long QueryString Values

    - by Registered User
    I had a project that required posting a 2.5 million character QueryString to a web page. The server itself only parsed URI's that were 5,400 characters or less. After trying several different sets of code for WebRequest/WebResponse, WebClient, and Sockets, I finally found the following code that solved my problem: HttpWebRequest webReq; HttpWebResponse webResp = null; string Response = ""; Stream reqStream = null; webReq = (HttpWebRequest)WebRequest.Create(strURL); Byte[] bytes = Encoding.UTF8.GetBytes("xml_doc=" + HttpUtility.UrlEncode(strQueryString)); webReq.ContentType = "application/x-www-form-urlencoded"; webReq.Method = "POST"; webReq.ContentLength = bytes.Length; reqStream = webReq.GetRequestStream(); reqStream.Write(bytes, 0, bytes.Length); reqStream.Close(); webResp = (HttpWebResponse)webReq.GetResponse(); if (webResp.StatusCode == HttpStatusCode.OK) { StreamReader loResponseStream = new StreamReader(webResp.GetResponseStream(), Encoding.UTF8); Response = loResponseStream.ReadToEnd(); } webResp.Close(); reqStream = null; webResp = null; webReq = null;

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  • Python + Expat: Error on &#0; entities

    - by clacke
    I have written a small function, which uses ElementTree and xpath to extract the text contents of certain elements in an xml file: #!/usr/bin/env python2.5 import doctest from xml.etree import ElementTree from StringIO import StringIO def parse_xml_etree(sin, xpath): """ Takes as input a stream containing XML and an XPath expression. Applies the XPath expression to the XML and returns a generator yielding the text contents of each element returned. >>> parse_xml_etree( ... StringIO('<test><elem1>one</elem1><elem2>two</elem2></test>'), ... '//elem1').next() 'one' >>> parse_xml_etree( ... StringIO('<test><elem1>one</elem1><elem2>two</elem2></test>'), ... '//elem2').next() 'two' >>> parse_xml_etree( ... StringIO('<test><null>&#0;</null><elem3>three</elem3></test>'), ... '//elem2').next() 'three' """ tree = ElementTree.parse(sin) for element in tree.findall(xpath): yield element.text if __name__ == '__main__': doctest.testmod(verbose=True) The third test fails with the following exception: ExpatError: reference to invalid character number: line 1, column 13 Is the � entity illegal XML? Regardless whether it is or not, the files I want to parse contain it, and I need some way to parse them. Any suggestions for another parser than Expat, or settings for Expat, that would allow me to do that?

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  • extra new lines with several outputStream.write

    - by Sam
    Hi All, I am writing jsp to export data in excel format to user. An excel could be recieved on the cient side. However, since there's large amount of data, and I don't want to keep it in the server memory and write them at the end. I try to divide them and write serveral times. However, each extra write(..) will cause an extra new lines at the top of the excel worksheet and then the extra data is placed after these new lines. Does anyone know the reasons? The code is something like this: response.setHeader("Content-disposition","attachment;filename=DocuShareSearch.xls"); response.setHeader("Content-Type", "application/octet-stream"); responseContent ="<table><tr><td>12131</td></tr>......."; byte[] responseByte1 = responseContent.getBytes("utf-16"); outputStream.write(responseByte1, 0, responseByte1.length ); responseContent =".....<tr><td>12131</td></tr></table>"; byte[] responseByte2 = responseContent.getBytes("utf-16"); outputStream.write(responseByte2, 0, responseByte2.length ); outputStream.close();

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