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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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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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  • 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 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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  • 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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  • ZendFramework Zend_Form_Element_File setDestination vs rename filter

    - by bibstha
    The code says Zend_Form_element_File::setDestination() is depricated and to use the rename filter. However the rename filter is currently codes such that when path is set, only temporary name is given. Original filename is lost. <?php $file = new Zend_Form_Element_File(); $file->setDestination('/var/www/project/public'); ?> vs <?php $file = new Zend_Form_Element_File(); $file->addFilter('Rename', array('target' => '/var/www/project/public')); ?> Any solution to upload files so that it preserves original filename structure but checks for existing file and appends _1.ext or _2.ext?

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  • I can I separate multiple logical pages in a text file I create in Perl?

    - by Micah
    So far, I've been successful with generating output to individual files by opening a file for output as part of outer loop and closing it after all output is written. I had used a counting variable ($x) and appended .txt onto it to create a filename, and had written it to the same directory as my perl script. I want to step the code up a bit, prompt for a file name from the user, open that file once and only once, and write my output one "printed letter" per page. Is this possible in plain text? From what I understand, chr(12) is an ascii line feed character and will get me close to what I want, but is there a better way? Thanks in advance, guys. :) sub PersonalizeLetters{ print "\n\n Beginning finalization of letters..."; print "\n\n I need a filename to save these letters to."; print "\n Filename > "; $OutFileName = <stdin>; chomp ($OutFileName); open(OutFile, ">$OutFileName"); for ($x=0; $x<$NumRecords; $x++){ $xIndex = (6 * $x); $clTitle = @ClientAoA[$xIndex]; $clName = @ClientAoA[$xIndex+1]; #I use this 6x multiplier because my records have 6 elements. #For this routine I'm only interested in name and title. #Reset OutLetter array #Midletter has other merged fields that aren't specific to who's receiving the letter. @OutLetter = @MiddleLetter; for ($y=0; $y<=$ifLength; $y++){ #Step through line by line and insert the name. $WorkLine = @OutLetter[$y]; $WorkLine =~ s/\[ClientTitle\]/$clTitle/; $WorkLine =~ s/\[ClientName\]/$clName/; @OutLetter[$y] = $WorkLine; } print OutFile "@OutLetter"; #Will chr(12) work here, or is there something better? print OutFile chr(12); $StatusX = $x+1; print "Writing output $StatusX of $NumRecords... \n\n"; } close(OutFile); }

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  • How to access the calling source line from interactive shell

    - by TJD
    I want to make a function that can determine the source code of how it was called. I'm aware of how to do this generally with the inspect module. For example, this question, works well and provides my desired output in the lines variable as shown below: def hello(x): frame,filename,line_number,function_name,lines,index=\ inspect.getouterframes(inspect.currentframe())[1] print(frame,filename,line_number,function_name,lines,index) The problem is that this solution doesn't work in an interactive command line session. For example, from a command line, the result looks like: >>> y = hello(7) (<frame object at 0x01ECA9E8>, '<stdin>', 1, '<module>', None, None) The problem is that the source file is '<stdin>', so the lines variable is None. How can I access the calling line to find the result containing the string y = hello(7) during an interactive session?

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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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  • 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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  • ASP SaveToDisk method takes an incredible amount of time

    - by burnt_hand
    This is a method in ASP Classic that saves a file to disk. It takes a very long time but I'm not sure why. Normally, I wouldn't mind so much, but the files it handles are pretty large so need this needs to faster than 100kB a second save. Seriously slow. (old legacy system, band aid fix till it gets replaced...) Public Sub SaveToDisk(sPath) Dim oFS, oFile Dim nIndex If sPath = "" Or FileName = "" Then Exit Sub If Mid(sPath, Len(sPath)) <> "\" Then sPath = sPath & "\" '" Set oFS = Server.CreateObject("Scripting.FileSystemObject") If Not oFS.FolderExists(sPath) Then Exit Sub Set oFile = oFS.CreateTextFile(sPath & FileName, True) For nIndex = 1 to LenB(FileData) oFile.Write Chr(AscB(MidB(FileData,nIndex,1))) Next oFile.Close End Sub I'm asking because there are plenty of WTF's in this code so I'm fighting those fires while getting some help on these ones.

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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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  • 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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  • Removing response.End() includes the html of the page.

    - by vinay_rockin
    HttpResponse response = Context.Http.Response; response.Headers.Clear(); response.Clear(); response.ContentType = Component.OverrideMimeType ? Component.MimeType : "application/download"; response.AppendHeader("Content-Disposition", String.Format("attachment; filename=\"{0}\"", HttpUtility.HtmlEncode(Path.GetFileName(file.FileName)))); response.OutputStream.Write(file.Contents, 0, file.Contents.Length); response.Flush(); // response.End(); if I use response.End() it throws exception and if I comment response.End() it includes the html of the page on which the download link resides. how want to download file with introducing this exta html. Any Idea how to fix this?

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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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  • SVN always getting errors when I commit (Subclipse)

    - by jax
    I have setup svn on my server and Subclipse at home. I am the only developer and am mainly using it for the backup and versioning features. Everytime I commit my changes I get eighter: Out Of date errors or Tree conflicts Sometimes I even delete files and they don't delete on svn, in a directory hierarchy only the very last item will delete so I have to delete each folder one at a time. How do I avoid these errors in the future? Update: Another problem I am having is that sometimes eclipse seems to sync with the server so that when I refactor a filename it goes off to the server and does something and makes me wait, which is annoying. And for clarity, this is a typical operation: I might change a filename, move a file to a different folder then change the contents of a file. I select the 'Team menu' and click 'commit'. Then I get all these errors above.

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  • Allow users to pull temporary data then delete table?

    - by JM4
    I don't know the best way to title this question but am trying to accomplish the following goal: When a client logs into their profile, they are presented with a link to download data from an existing database in CSV format. The process works, however, I would like for this data to be 'fresh' each time they click the link so my plan was - once a user has clicked the link and downloaded the CSV file, the database table would 'erase' all of its data and start fresh (be empty) until the next set of data populated it. My EXISTING CSV creation code: <?php $host = 'localhost'; $user = 'username'; $pass = 'password'; $db = 'database'; $table = 'tablename'; $file = 'export'; $link = mysql_connect($host, $user, $pass) or die("Can not connect." . mysql_error()); mysql_select_db($db) or die("Can not connect."); $result = mysql_query("SHOW COLUMNS FROM ".$table.""); $i = 0; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_assoc($result)) { $csv_output .= $row['Field'].", "; $i++; } } $csv_output .= "\n"; $values = mysql_query("SELECT * FROM ".$table.""); while ($rowr = mysql_fetch_row($values)) { for ($j=0;$j<$i;$j++) { $csv_output .= '"'.$rowr[$j].'",'; } $csv_output .= "\n"; } $filename = $file."_".date("Y-m-d",time()); header("Content-type: application/vnd.ms-excel"); header("Content-disposition: csv" . date("Y-m-d") . ".csv"); header( "Content-disposition: filename=".$filename.".csv"); print $csv_output; exit; ?> any ideas?

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