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• Is it possible to change the file path of the video using javascript?

  - by Manish

  I have an object tag in a HTML file: <object classid="clsid:22D6F312-B0F6-11D0-94AB-0080C74C7E95"> <param name="FileName" value="../ABC/WildLife.wmv" id="mediaPlayerFile"> <param name="AutoStart" value="false" /> </object> I want to change the filename using javascript. What I have so far is this: <script type="text/javascript"> function disp_current_directory() { var val = document.getElementById('mediaPlayerFile'); val.attributes['value'].value = "D:\XYZ\WildLife.wmv"; } </script> But this doesn't work. :( Is it possible? If yes, how?

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• file_get_contents returns an empty string that is 354 bytes long.

  - by Kendall Crouch

  I'm trying to read the contents of a file and simply getting an empty string. The file exists on the server. I've tried some test with the following code and get the true to display: $filename = "includes/blah.php"; $filecontents = file_get_contents($filename, FILE_USE_INCLUDE_PATH); if ($filecontents === false) { echo(":FALSE:"); } else { echo(":TRUE:"); } var_dump($filecontents); The dump displays "string(354)" which is the correct size of the file. What am I doing wrong?

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• Problem loading RTF file into windows richTextBox

  - by Ted

  I am trying to load files into a windows (vs 2010) richTextBox but only the first line of the file is loading. I'm using: // Create an OpenFileDialog to request a file to open. OpenFileDialog openFile1 = new OpenFileDialog(); // Initialize the OpenFileDialog to look for RTF files. openFile1.DefaultExt = "*.rtf"; openFile1.Filter = "RTF Files|*.rtf"; // Determine whether the user selected a file from the OpenFileDialog. if (openFile1.ShowDialog() == System.Windows.Forms.DialogResult.OK && openFile1.FileName.Length > 0) { // Load the contents of the file into the RichTextBox. rtbTest.LoadFile(openFile1.FileName); } The test file I'm using is .cs file saved as an rtf file. Any help appreciated please.

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• Subprocess with variables & Command from different class

  - by Pastelinux

  source: http://pastebin.com/utL7Ebeq My thinking is that if i run from controller class "main" it will allow me to take the "data" from Class "model", def "filename". It doesn't seem to work. As you can see below what i mean class Controller: def __init__(self): self.model = Model() self.view = View() def main(self): data = self.model.filename() self.view.tcpdump(data) class View: def tcpdump(self, command): subprocess.call(command.split(), shell=False) When i run my code i get this error: subprocess.call(command.split(), shell=False) AttributeError: 'NoneType' object has no attribute 'split' My guess means that its not picking up command (look at source for reference) or that its not getting command with variables. But i know the error when variables are not being picked up so i don't think it is that. My question is, from what i have thus far, how do i from "class view" grab "command" for my subprocesses to run. Thanks~ John Riselvato

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• Handling download abort in PHP

  - by Aron Rotteveel

  Is it somehow possible to handle a download abort in PHP? In this specific case I am not speaking of a connection abort, but handling the event that triggers when the 'cancel' button in the browser download dialog button is clicked. Since this dialog already interprets the headers of the file that is to be download but does not actually start the download, it only seems logical there should be some way to catch this. Small (pseudo) code example to clear things up: // set some headers header('...'); // Question: what happens between the part where the headers are sent // and the actual data is being outputted to the client? IE: this is the part // where the download dialog should show up // Logical question that follows is: is there a way to detect a 'cancel'? $filename = '/some/file.txt'; $handle = fopen($filename, 'rb'); // output data to client while (!feof($handle)) { echo fread($handle, 8096); } fclose($handle);

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• How to redirect a live data stream adding to it another header and returning it on demand? (PHP)

  - by Ole Jak

  I have a url like http://localhost:8020/stream.flv On request to my php sctipt I want to return (be something like a proxy) all data I can get from that URL (so I mean my php code should get data from that url and give it to user) and my header and my beginning of file. So I have my header and some data I want to write in the beginning of response like # content headers header("Content-Type: video/x-flv"); header("Content-Disposition: attachment; filename=\"" . $fileName . "\""); header("Content-Length: " . $fileSize); # FLV file format header if($seekPos != 0) { print('FLV'); print(pack('C', 1)); print(pack('C', 1)); print(pack('N', 9)); print(pack('N', 9)); } How to do such thing?

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• MFMailComposeViewController configure mimetype for attachment

  - by user749918

  I need some help. I am trying to attach files to mail using, [mail addAttachmentData:attachmentData mimeType:@"image/png" fileName:fileName]; but the problem is that if i need to send a .jpeg image i need to repeat code just for setting mime type to "mimeType:@"image/jpeg". My question is that is there any general mimeType that can attach any kind of file irrespective of .doc,.ppt,.pdf or an audio or video file. is there any general mimeType: for an kind of attachment. Thanks in advance.

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• regular expression not behaving as expected - Python

  - by philippe

  I have the following function which is supposed to read a .html file and search for <input> tags, and inject a <input type='hidden' > tag into the string to be shown into the page. However, that condition is never met:( e.g the if statement is never executed. ) What's wrong with my regex? def print_choose( params, name ): filename = path + name f = open( filename, 'r' ) records = f.readlines() print "Content-Type: text/html" print page = "" flag = True for record in records: if re.match( '<input*', str(record) ) != None: print record page += record page += "<input type='hidden' name='pagename' value='psychology' />" else: page += record print page Thank you

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• Is there a way to efficiently yield every file in a directory containing millions of files?

  - by Josh Smeaton

  I'm aware of os.listdir, but as far as I can gather, that gets all the filenames in a directory into memory, and then returns the list. What I want, is a way to yield a filename, work on it, and then yield the next one, without reading them all into memory. Is there any way to do this? I worry about the case where filenames change, new files are added, and files are deleted using such a method. Some iterators prevent you from modifying the collection during iteration, essentially by taking a snapshot of the state of the collection at the beginning, and comparing that state on each move operation. If there is an iterator capable of yielding filenames from a path, does it raise an error if there are filesystem changes (add, remove, rename files within the iterated directory) which modify the collection? There could potentially be a few cases that could cause the iterator to fail, and it all depends on how the iterator maintains state. Using S.Lotts example: filea.txt fileb.txt filec.txt Iterator yields filea.txt. During processing, filea.txt is renamed to filey.txt and fileb.txt is renamed to filez.txt. When the iterator attempts to get the next file, if it were to use the filename filea.txt to find it's current position in order to find the next file and filea.txt is not there, what would happen? It may not be able to recover it's position in the collection. Similarly, if the iterator were to fetch fileb.txt when yielding filea.txt, it could look up the position of fileb.txt, fail, and produce an error. If the iterator instead was able to somehow maintain an index dir.get_file(0), then maintaining positional state would not be affected, but some files could be missed, as their indexes could be moved to an index 'behind' the iterator. This is all theoretical of course, since there appears to be no built-in (python) way of iterating over the files in a directory. There are some great answers below, however, that solve the problem by using queues and notifications. Edit: The OS of concern is Redhat. My use case is this: Process A is continuously writing files to a storage location. Process B (the one I'm writing), will be iterating over these files, doing some processing based on the filename, and moving the files to another location. Edit: Definition of valid: Adjective 1. Well grounded or justifiable, pertinent. (Sorry S.Lott, I couldn't resist). I've edited the paragraph in question above.

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• DataSet XML export is empty

  - by Shaine

  I've got in-memory dataset with couple of tables that is populated in code. Data-bound grids on the gui show table contents without a problem. Then I try to export the dataset into XML: ds.WriteXml(fdSave.FileName, XmlWriteMode.WriteSchema); and get empty XML (with couple of lines regarding dataset names but without any tables) If I export table directly I've got all the data but dataset name is obviously wrong: ds.Fields.WriteXml(fdSave.FileName, XmlWriteMode.WriteSchema); What am I missing? Is there any reasonable way to write the whole dataset into file?

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• php most memory efficient way to return files

  - by bumperbox

  so i have a bunch of files, some can be up to 30-40mb and i want to use php to handle security of the files, so i can control who has access to them that means i have a script sort of like this rough example $has_permission = check_database_for_permission($user, filename); if ($has_permission) { header('Content-Type: image/jpeg'); readfile ($filename); exit; } else { // return 401 error } i would hate for every request to load the full file into memory, as it would soon chew up all the memory on my server with a few simultaneous requests so a couple of questions is readfile the most memory efficient way of doing this? is there some better method of achieving the same outcome, that i am overlooking? server: apache/php5 thanks

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• In Eclipse, how do I change the default modifiers in the class/type template?

  - by gustafc

  Eclipse's default template for new types (Window Preferences Code Style Code Templates New Java Files) looks like this: ${filecomment} ${package_declaration} ${typecomment} ${type_declaration} Creating a new class, it'll look something like this: package pkg; import blah.blah; public class FileName { // Class is accessible to everyone, and can be inherited } Now, I'm fervent in my belief that access should be as restricted as possible, and inheritance should be forbidden unless explicitly permitted, so I'd like to change the ${type_declaration} to declare all classes as final rather than public: package pkg; import blah.blah; final class FileName { // Class is only accessible in package, and can't be inherited } That seems easier said than done. The only thing I've found googling is a 2004 question on Eclipse's mailing list which was unanswered. So, the question in short: How can I change the default class/type modifiers in Eclipse? I'm using Eclipse Galileo (3.5) if that matters.

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• Is there a way of getting a file name and inserting into Matlab script?

  - by torr

  In a folder, I have both my .m file that contains the script and an imaging .dcm file that needs to be analyzed. Folder structure: Folder1/analysis.m Folder1/meas_dynamic_123.dcm My script (analysis.m) begins as follows: target =''; <== here should go the full path to the file + filename example: /Volumes/Data/Folder1/meas_dynamic_123.m txt = dir(target); // etc So I'm wondering if there is a way of when running analysis.m it will: automatically search the folder it's in, grab the full path + filename of file containing string dynamic in the name, insert its full path + name into target variable continue running the script Does anyone have any pointers on how to achieve this? Using ffpath?

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• Why does SendMailMAPI rename file attachments to shorter ones?

  - by Tom

  I use the following emailing function with Eudora. For some reason the attachment file name is renamed to be something else. How can I make sure the attachment file name remains intact? function SendMailMAPI(const Subject, Body, FileName, SenderName, SenderEMail, RecepientName, RecepientEMail: String) : Integer; var message: TMapiMessage; lpSender, lpRecepient: TMapiRecipDesc; FileAttach: TMapiFileDesc; SM: TFNMapiSendMail; MAPIModule: HModule; begin FillChar(message, SizeOf(message), 0); with message do begin if (Subject<>'') then begin lpszSubject := PChar(Subject) end; if (Body<>'') then begin lpszNoteText := PChar(Body) end; if (SenderEMail<>'') then begin lpSender.ulRecipClass := MAPI_ORIG; if (SenderName='') then begin lpSender.lpszName := PChar(SenderEMail) end else begin lpSender.lpszName := PChar(SenderName) end; lpSender.lpszAddress := PChar('SMTP:'+SenderEMail); lpSender.ulReserved := 0; lpSender.ulEIDSize := 0; lpSender.lpEntryID := nil; lpOriginator := @lpSender; end; if (RecepientEMail<>'') then begin lpRecepient.ulRecipClass := MAPI_TO; if (RecepientName='') then begin lpRecepient.lpszName := PChar(RecepientEMail) end else begin lpRecepient.lpszName := PChar(RecepientName) end; lpRecepient.lpszAddress := PChar('SMTP:'+RecepientEMail); lpRecepient.ulReserved := 0; lpRecepient.ulEIDSize := 0; lpRecepient.lpEntryID := nil; nRecipCount := 1; lpRecips := @lpRecepient; end else begin lpRecips := nil end; if (FileName='') then begin nFileCount := 0; lpFiles := nil; end else begin FillChar(FileAttach, SizeOf(FileAttach), 0); FileAttach.nPosition := Cardinal($FFFFFFFF); FileAttach.lpszPathName := PChar(FileName); nFileCount := 1; lpFiles := @FileAttach; end; end; MAPIModule := LoadLibrary(PChar(MAPIDLL)); if MAPIModule=0 then begin Result := -1 end else begin try @SM := GetProcAddress(MAPIModule, 'MAPISendMail'); if @SM<>nil then begin Result := SM(0, Application.Handle, message, MAPI_DIALOG or MAPI_LOGON_UI, 0); end else begin Result := 1 end; finally FreeLibrary(MAPIModule); end; end; if Result<>0 then begin MessageDlg('Error sending mail ('+IntToStr(Result)+').', mtError, [mbOk], 0) end;

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• PHP exec() hangs, Error 500

  - by MRX_II

  So, my plan is to make small thumbnails of URL's with PHP and IECapt. IECapt works well, a nice command line tool, gets the full sized image of specified URL in 1 to 4 seconds. But my problem is to execute it trough PHP. This is the code I've trying to get working: exec('IECapt.exe ' . escapeshellarg($URL) . ' ' . escapeshellarg($Filename)) $URL is of course the URL, and $filename is a simplified version of the URL. Sometimes I get the IECapt to snap the image(trough PHP), but it takes awfully long (30-60s), and in the end I always get a 500-error, with no error messages to tell me what's wrong. Both variables are fine, they work manually with commandline: IECapt http://google.com Google.png My server set-up is IIS7 and PHP5.2.9, if relevant. (Windows Vista, all on my personal computer, so full access.) Any ideas?

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• php convert images and upload to amazon s3

  - by faraklit

  I am looking for a best practice while uploading images to amazon s3 server and serving from there. We need four different sizes of an image. So just after image upload we convert the image and scale in 4 different widths and heights. And then we send them to the amazon s3 using official php api. // ... // image conversions, bucket setting, s3 initialization etc. $sizes= array("", "48", "64", "128"); foreach($sizes as $size) { $filename = $upload_path.$dest_file.$size.$ext; $s3->batch()->create_object($bucket, , array( 'fileUpload' => $filename, 'acl' => AmazonS3::ACL_PUBLIC, )); } But for a 1M image the client sometimes wait up to 30 seconds which is a very long time. Instead of sending images immediately to S3, it may be better to add them to a job queue. But the user should see the uploaded image immediately.

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• Function argument treated as undeclared

  - by Mikulas Dite

  I've prepared this simple example which is not working for me #include <stdio.h> #include <stdlib.h> FILE *fp; char filename[] = "damy.txt"; void echo (char[] text) { fp = fopen(filename, "a"); fwrite(text, 1, strlen(text), fp); fclose(fp); printf(text); } int main () { echo("foo bar"); return 0; } It's supposed to write both to command window and to file. However, this gives compilation error - the text used in echo() is not declared. Does c need another declaration of the variable?

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• what is -p in perl

  - by user136104

  perl -p -i.bak -e 's/search_str/replace_str/g' filename here what is -p -i.bak /s and /g

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• What is the universal way to use file I/O API with unicode filenames?

  - by dma_k

  In Windows there is a common problem: the filenames should be converted to local codepage, before they are passed to open(). Of course, there is a possibility to use Win32::API for that, but I don't want my script to be platform-dependent. At the moment I have to write something like: open IN, "<", encode("cp1251", $filename) or die $!; but is there any library, that hides these details? I think the local codepage can be automatically detected, so I just want to pass unicode filename and forget about the details. Why is it still not in the box?

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• ruby on rails photo upload problem

  - by dodo00700

  Hallo rails version 2.3.5 I'm learning rails and I run into a problem. I'm doing some nesting forms from the railscasts tutorials. I changed the text area into a data field to upload photos and everything is working. Now i have to display the uploaded pictures and i simply can't do it. I Tried everything I could find on the net but nothing worked. PROBLEM I have the Article controller which handles the article CRUD. inside the article new form there is nested a form for uploading images. article controller def code_image @image_data = Photo.find(params[:id]) @image = @image_data.binary_data send_data(@image, :type => @image_data.content_type, :filename => @image_data.filename, :disposition => 'inline') end photo model def image_file=(input_data) self.filename = input_data.original_filename self.content_type = input_data.content_type.chomp self.binary_data = input_data.read end articles/show.html.erb <%=h @article.title %> <%=h @article.body %> <% for photos in @article.photos %> <%= image_tag(url_for({:action => 'code_image', :id => @article.photos.id})) -%> <% end %> articles/_formnew.html.erb <% form_for (:article, @article, :url => {:action=>'create'}, :html=> {:multipart=>true}) do |f| %> <%= f.error_messages % <%= f.label :title %><br /> <%= f.text_field :title %><br /><br /> <%= f.label :body %><br /> <%= f.text_area :body, :style => 'width: 600px;' %><br /><br /> <% f.fields_for :photos do |builder|%> <%= builder.label :content, "Photo"%><br /> <%= builder.file_field :image_file %><br /> <% end %> <br /> <%= f.submit "Create" %> <% end % Thanks

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• Serialze an Object to a String

  - by Vaccano

  I have the following method to save an Object to a file: // Save an object out to the disk public static void SerializeObject<T>(this T toSerialize, String filename) { XmlSerializer xmlSerializer = new XmlSerializer(toSerialize.GetType()); TextWriter textWriter = new StreamWriter(filename); xmlSerializer.Serialize(textWriter, toSerialize); textWriter.Close(); } I confess I did not write it (I only converted it to a extension method that took a type parameter). Now I need it to give the xml back to me as a string (rather than save it to a file). I am looking into it, but I have not figured it out yet. I thought this might be really easy for someone familiar with these objects. If not I will figure it out eventually.

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• (PHP) 1)How to genrate Secreate key on User & Client Side ? 3) How to Compare Server side MD5 and Client side Md5 ?

  - by user557994

  /* In Below Code .. My problem is that 1) How to genrate Secreate key on User Side ? 2) How to genrate Secreate key on Client Side ? 3) How to Compare Server side MD5 and Client side Md5 ? Can you solve my problem ? */ $gid = $_GET['id']; if($gid=="") { $filename = "counter.txt"; $fp = fopen( $filename, "r" ) or die("Couldn't Generate Whiteboard"); while ( ! feof( $fp ) ) { $countfile = fgets( $fp); $countfile++; } fclose( $fp ); $fp = fopen( $filename, "w" ) or die("Couldn't generate whiteboard"); fwrite( $fp, $countfile ); fclose( $fp ); $doc = new DOMDocument('1.0', 'UTF-8'); $ele = $doc-createElement( 'root' ); $ele-nodeValue = $uvar; $doc-appendChild( $ele ); $test = $doc-save("$countfile.xml"); genkey($id); echo ""; $uvar=$_POST['msgval']; exit; } else { if($uvar == "") { $xdoc = new DOMDocument( '1.0', 'UTF-8' ); $xdoc-Load("$gid.xml"); $candidate = $xdoc-getElementsByTagName('root')-item(0); $newElement = $xdoc -createElement('root'); $txtNode = $xdoc -createTextNode ($root); $newElement - appendChild($txtNode); $candidate - appendChild($newElement); $msg = $candidate-nodeValue; } } function genkey($id) { $encrypt_key = "GJHsahakst1468464a"; $key = MD5("$id","$$encrypt_key"); return $key; } ? function sendRequest() { var uvar = document.getElementById('txtHint').value; var xmlhttp = new XMLHttpRequest(); xmlhttp.onreadystatechange = function() { if(xmlhttp.readyState == 4 && xmlhttp.status==200) { document.getElementById('txtHint').value = ""; } } xmlhttp.open("POST","post.php?id=",true); xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded"); xmlhttp.send("umsg="+uvar); return; } Msg " /

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• How to connect two files and use the radio button?

  - by Stupefy101

  I have here a set of form from the index.php to upload a zip file, select an option then perform a converter process. <form action="" method="post" accept-charset="utf-8"> <p class="buttons"><input type="file" value="" name="zip_file"/></p> </form> <form action="index.php" method="post" accept-charset="utf-8" name="form1"> <h3><input type="radio" name="option" value="option1"/> Option1 </h3> <h3><input type="radio" name="option" value="option2"/> Option2 </h3> <h3><input type="radio" name="option" value="option3"/> Option3 </h3> <p class="buttons"><input type="submit" value="Convert"/></p> </form> In the other hand, this is my code for the upload.php that will extract the Zip file. <?php if($_FILES["zip_file"]["name"]) { $filename = $_FILES["zip_file"]["name"]; $source = $_FILES["zip_file"]["tmp_name"]; $type = $_FILES["zip_file"]["type"]; $name = explode(".", $filename); $accepted_types = array('application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/x-compressed'); foreach($accepted_types as $mime_type) { if($mime_type == $type) { $okay = true; break; } } $continue = strtolower($name[1]) == 'zip' ? true : false; if(!$continue) { $message = "The file you are trying to upload is not a .zip file. Please try again."; } $target_path = "C:xampp/htdocs/themer/".$filename; // change this to the correct site path if(move_uploaded_file($source, $target_path)) { $zip = new ZipArchive(); $x = $zip->open($target_path); if ($x === true) { $zip->extractTo("C:xampp/htdocs/themer/"); // change this to the correct site path $zip->close(); unlink($target_path); } $message = "Your .zip file was uploaded and unpacked."; } else { $message = "There was a problem with the upload. Please try again."; } } ?> How can i connect both files that will perform the extracting process? And how to include the codes for radio button after submission? Please Help.

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• Image URL has the contentType "text/html"

  - by user1503025

  I want to implement a method to download Image from website to laptop. public static void DownloadRemoteImageFile(string uri, string fileName) { HttpWebRequest request = (HttpWebRequest)WebRequest.Create(uri); HttpWebResponse response = (HttpWebResponse)request.GetResponse(); if ((response.StatusCode == HttpStatusCode.OK || response.StatusCode == HttpStatusCode.Moved || response.StatusCode == HttpStatusCode.Redirect) && response.ContentType.StartsWith("image", StringComparison.OrdinalIgnoreCase)) { //if the remote file was found, download it using (Stream inputStream = response.GetResponseStream()) using (Stream outputStream = File.OpenWrite(fileName)) { byte[] buffer = new byte[4096]; int bytesRead; do { bytesRead = inputStream.Read(buffer, 0, buffer.Length); outputStream.Write(buffer, 0, bytesRead); } while (bytesRead != 0); } } } But the ContentType of request or response is not "image/jpg" or "image/png". They're always "text/html". I think that's why after I save them to local, they has incorrect content and I cannot view them. Can anyone has a solution here? Thanks

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• Rotating images with PHP reduces quality especially over about 10-20 actions

  - by Dylan Cross

  I am using this code below to rotate my jpeg images, the problem is that after around 10-20 times of rotating the image the image is dramatically lower quality, especially blue skies and such, my question is how can I keep these images the same high quality image? There must be a way. I mean, i keep the original image on the server for each image uploaded, and I don't do anything to that, so if need be it, I guess I could always come up with some way of using that over whenever possible.. But would rather not have to. $source = imagecreatefromjpeg($filename); $rotate = imagerotate($source, 90, 0); imagejpeg($rotate, $filename ,100);

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