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  • Steganography : Encoded audio and video file not being played, getting corrupted. What is the issue

    - by Shantanu Gupta
    I have made a steganography program to encrypt/Decrypt some text under image audio and video. I used image as bmp(54 byte header) file, audio as wav(44 byte header) file and video as avi(56 byte header) file formats. When I tries to encrypt text under all these file then it gets encrypted successfully and are also getting decrypted correctly. But it is creating a problem with audio and video i.e these files are not being played after encrypted result. What can be the problem. I am working on Turbo C++ compiler. I know it is super outdated compiler but I have to do it in this only. Here is my code to encrypt. int Binary_encode(char *txtSourceFileName, char *binarySourceFileName, char *binaryTargetFileName,const short headerSize) { long BinarySourceSize=0,TextSourceSize=0; char *Buffer; long BlockSize=10240, i=0; ifstream ReadTxt, ReadBinary; //reads ReadTxt.open(txtSourceFileName,ios::binary|ios::in);//file name, mode of open, here input mode i.e. read only if(!ReadTxt) { cprintf("\nFile can not be opened."); return 0; } ReadBinary.open(binarySourceFileName,ios::binary|ios::in);//file name, mode of open, here input mode i.e. read only if(!ReadBinary) { ReadTxt.close();//closing opened file cprintf("\nFile can not be opened."); return 0; } ReadBinary.seekg(0,ios::end);//setting pointer to a file at the end of file. ReadTxt.seekg(0,ios::end); BinarySourceSize=(long )ReadBinary.tellg(); //returns the position of pointer TextSourceSize=(long )ReadTxt.tellg(); //returns the position of pointer ReadBinary.seekg(0,ios::beg); //sets the pointer to the begining of file ReadTxt.seekg(0,ios::beg); //sets the pointer to the begining of file if(BinarySourceSize<TextSourceSize*50) //Minimum size of an image should be 50 times the size of file to be encrypted { cout<<"\n\n"; cprintf("Binary File size should be bigger than text file size."); ReadBinary.close(); ReadTxt.close(); return 0; } cout<<"\n"; cprintf("\n\nSize of Source Image/Audio File is : "); cout<<(float)BinarySourceSize/1024; cprintf("KB"); cout<<"\n"; cprintf("Size of Text File is "); cout<<TextSourceSize; cprintf(" Bytes"); cout<<"\n"; getch(); //write header to file without changing else file will not open //bmp image's header size is 53 bytes Buffer=new char[headerSize]; ofstream WriteBinary; // writes to file WriteBinary.open(binaryTargetFileName,ios::binary|ios::out|ios::trunc);//file will be created or truncated if already exists ReadBinary.read(Buffer,headerSize);//reads no of bytes and stores them into mem, size contains no of bytes in a file WriteBinary.write(Buffer,headerSize);//writes header to 2nd image delete[] Buffer;//deallocate memory /* Buffer = new char[sizeof(long)]; Buffer = (char *)(&TextSourceSize); cout<<Buffer; */ WriteBinary.write((char *)(&TextSourceSize),sizeof(long)); //writes no of byte to be written in image immediate after header ends //to decrypt file if(!(Buffer=new char[TextSourceSize])) { cprintf("Enough Memory could not be assigned."); return 0; } ReadTxt.read(Buffer,TextSourceSize);//read all data from text file ReadTxt.close();//file no more needed WriteBinary.write(Buffer,TextSourceSize);//writes all text file data into image delete[] Buffer;//deallocate memory //replace Tsize+1 below with Tsize and run the program to see the change //this is due to the reason that 50-54 byte no are of colors which we will be changing ReadBinary.seekg(TextSourceSize+1,ios::cur);//move pointer to the location-current loc i.e. 53+content of text file //write remaining image content to image file while(i<BinarySourceSize-headerSize-TextSourceSize+1) { i=i+BlockSize; Buffer=new char[BlockSize]; ReadBinary.read(Buffer,BlockSize);//reads no of bytes and stores them into mem, size contains no of bytes in a file WriteBinary.write(Buffer,BlockSize); delete[] Buffer; //clear memory, else program can fail giving correct output } ReadBinary.close(); WriteBinary.close(); //Encoding Completed return 0; } Code to decrypt int Binary_decode(char *binarySourceFileName, char *txtTargetFileName, const short headerSize) { long TextDestinationSize=0; char *Buffer; long BlockSize=10240; ifstream ReadBinary; ofstream WriteText; ReadBinary.open(binarySourceFileName,ios::binary|ios::in);//file will be appended if(!ReadBinary) { cprintf("File can not be opened"); return 0; } ReadBinary.seekg(headerSize,ios::beg); Buffer=new char[4]; ReadBinary.read(Buffer,4); TextDestinationSize=*((long *)Buffer); delete[] Buffer; cout<<"\n\n"; cprintf("Size of the File that will be created is : "); cout<<TextDestinationSize; cprintf(" Bytes"); cout<<"\n\n"; sleep(1); WriteText.open(txtTargetFileName,ios::binary|ios::out|ios::trunc);//file will be created if not exists else truncate its data while(TextDestinationSize>0) { if(TextDestinationSize<BlockSize) BlockSize=TextDestinationSize; Buffer= new char[BlockSize]; ReadBinary.read(Buffer,BlockSize); WriteText.write(Buffer,BlockSize); delete[] Buffer; TextDestinationSize=TextDestinationSize-BlockSize; } ReadBinary.close(); WriteText.close(); return 0; } int text_encode(char *SourcefileName, char *DestinationfileName) { ifstream fr; //reads ofstream fw; // writes to file char c; int random; clrscr(); fr.open(SourcefileName,ios::binary);//file name, mode of open, here input mode i.e. read only if(!fr) { cprintf("File can not be opened."); getch(); return 0; } fw.open(DestinationfileName,ios::binary|ios::out|ios::trunc);//file will be created or truncated if already exists while(fr) { int i; while(fr!=0) { fr.get(c); //reads a character from file and increments its pointer char ch; ch=c; ch=ch+1; fw<<ch; //appends character in c to a file } } fr.close(); fw.close(); return 0; } int text_decode(char *SourcefileName, char *DestinationName) { ifstream fr; //reads ofstream fw; // wrrites to file char c; int random; clrscr(); fr.open(SourcefileName,ios::binary);//file name, mode of open, here input mode i.e. read only if(!fr) { cprintf("File can not be opened."); return 0; } fw.open(DestinationName,ios::binary|ios::out|ios::trunc);//file will be created or truncated if already exists while(fr) { int i; while(fr!=0) { fr.get(c); //reads a character from file and increments its pointer char ch; ch=c; ch=ch-1; fw<<ch; //appends character in c to a file } } fr.close(); fw.close(); return 0; }

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  • $_GET['page'] loading content incorrectly

    - by s32ialx
    OK so here is my previous post PHP Templated Site w/ file_get_content links now i got that issue resolved BUT the problem is now that the content loads it displays UNDER the div i placed #CONTENT# inside so the styles are being ignored and it's posting #CONTENT# outside the divs at positions 0,0 any suggestions? Found out whats happening by using "View Source" seems that it's putting all of the #CONTENT#, content that's being loaded in front of the tag. Like this <doctype...> <div class="home"> blah blah </div> <head> <script src=""></script> </head> <body> <div class="header"></div> <div class="contents"> #CONTENT# < where content SHOULD load </div> <div class="footer"></div> </body> so anyone got a fix? OK so a better description I'll add relevant screen-shots Whats happening is /* file.class.php */ <?php $file = new file(); class file{ var $path = "templates/clean"; var $ext = "tpl"; function loadfile($filename){ return file_get_contents($this->path . "/" . $filename . "." . $this->ext); } function css($val,$content='',$contentvar='#CSS#') { if(is_array($val)) { $css = 'style="'; foreach($val as $p) { $css .= $p . ";"; } $css .= '"'; } else { $css = 'style="' . $val . '"'; } if($content!='') { return str_replace($contentvar,' ' . $css,$content); } else { return $css; } } function setsize($content,$width='-1',$height='-1',$border='-1'){ $css = ''; if($width!='-1') { $css = $css . "width=\"".$width."\""; } if($height!='-1') { $css = $css . "height=\"".$height."\""; } if($border!='-1') { $css = $css . "border=\"" . $border . "\""; } return str_replace('#SIZE#',' ' . $css,$content); } function setcontent($content,$newcontent,$vartoreplace='#CONTENT#'){ $val = str_replace($vartoreplace,$newcontent,$content); return $val; } function p($content) { $v = $content; $v = str_replace('#CONTENT#','',$v); $v = str_replace('#SIZE#','',$v); print $v; } } if (isset($_GET['page'])) { $content = $_GET['page'].'.php'; } else { $content = 'main.php'; } ?> is calling for a file_get_contents at the bottom which I use in /* index.php */ <?php include('classes/file.class.php'); // load the templates $header = $file->loadfile('header'); $body = $file->loadfile('body'); $footer = $file->loadfile('footer'); // fill body.tpl #CONTENT# slot with $content $body = $file->setcontent($body, $content); // cleanup and output the full page $file->p($header . $body . $footer); ?> and loads into /* body.tpl */ <div id="bodys"> <div id="bodt"></div> <div id="bodm"> <div id="contents"> #CONTENT# </div> </div> <div id="bodb"></div> </div> but the issue is as follows the $content loads properly img tags etc <h2> tags etc but CSS styling is TOTALY ignored for position width z-index etc. and as follows here's the screen-shot JUST incase you require the css for where $content is being loaded #bodys { top:91px; position:absolute; width:100%; } #bodt { margin-left:auto; margin-right:auto; top:3px; position:relative; width:820px; height:42px; background-image:url('images/pagetop.png'); background-repeat:no-repeat; z-index: 0; } #bodm { margin-left:auto; margin-right:auto; top:3px; position:relative; width:820px; background-image:url('images/pagemid.png'); background-repeat:repeat-y; z-index: 0; } #bodb { margin-left:auto; margin-right:auto; bottom:-42px; position:relative; width:820px; height:42px; background-image:url('images/pagebot.png'); background-repeat:no-repeat; z-index:-1; } #menuo { position:absolute; bottom:-2px; z-index:199; } #contents { position:relative; top:5px; left:25px; width:770px; z-index:10; overflow: auto; color: #000000; line-height: 1.3em; font-size: 12px; } #content { position:absolute; top:5px; left:25px; width:760px; z-index:1; color: #000000; line-height: 1.3em; font-size: 12px; } #contents p{ margin-bottom: 0.7em; } #contents a{ font-weight:bold; color: #6fa5fd; border-bottom: 1px dotted #6fa5fd; }

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  • How to convert Markdown files to Dokuwiki, on a PC

    - by Clare Macrae
    I'm looking for a tool or script to convert Markdown files to Dokuwiki format, that will run on a PC. This is so that I can use MarkdownPad on a PC to create initial drafts of documents, and then convert them to Dokuwiki format, to upload to a Dokuwiki installation that I have no control over. (This means that the Markdown plugin is no use to me.) I could spend time writing a Python script to do the conversion myself, but I'd like to avoid spending time on this, if such a thing exists already. The Markdown tags I'd like to have supported/converted are: Heading levels 1 - 5 Bold, italic, underline, fixed width font Numbered and unnumbered lists Hyperlinks Horizontal rules Does such a tool exist, or is there a good starting point available? Things I've found and considered I initially thought that txt2tags would be helpful, but although it can write both markdown and Dokuwiki, it is very tied to its own specific input format I've also seen Markdown2Dokuwiki, and although I'd certainly be willing to use a sed script, even on a PC, this only supports a tiny, tiny part of Markdown's syntax. python-markdown2 also sounded promising, but it only writes out HTML. pandoc - but it doesn't support Dokuwiki output MultiMarkdown - does not appear to support Dokuwiki output

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  • Static file download from browser breaking in varnish but works fine in Apache

    - by Ron
    I would at first like to thank everyone at serverfault for this great website and I also come to this site while searching in google for various server related issues and setups. I also have an issue today and so I am posting here and hope that the seniors would help me out. I had setup a website on a dedicated server a few days ago and I used Varnish 3 as the frontend to Apache2 on a Debian Lenny server as the traffic was a bit high. There are several static file downloads of around 10-20 MB in size in the website. The website looked fine in the last few days after I setup. I was checking from a 5mbps + broadband connection and the file downloads were also completed in seconds and working fine. But today I realized that on a slow internet connection the file downloads were breaking off. When I tried to download the files from the website using a browser then it broke off after a minute or so. It kept on happening again and again and so it had nothing to do with the internet connection. The internet connection was around 512 kbps and so it was not dial up level speed too but decent speed where files should easily download though not that fast. Then I thought of trying out with the apache backend port and used the port number to check out if the problem occurs. But then on adding the apache port in the static file download url, the files got downloaded easily and did not break even once. I tried it several times to make sure that it was not a coincidence but every time I was using the apache port in the file download url then it was downloading fine while it was breaking each time with the normal link which was routed through Varnish I suppose. So, it seems Varnish has somehow resulted in the broken file downloads. Could anyone give any idea as to why it is happening and how to fix the problem. For more clarification, take this example: Apache backend set on port 8008, Varnish frontend set on port 80 Now when I download say http://mywebsite.com/directory/filename.extension Then the download breaks off after a minute or so. I cannot be sure it is due to the time or size though and I am just assuming. May be some other reason too. But when I download using: http://mywebsite.com:8008/directory/filename.extension Then the file download does not break at all and it gets download fine. So, it seems that varnish is somehow creating the file download breaking and not apache. Does anybody have any idea as to why it is happening and how can it be fixed. Any help would be highly appreciated. And my varnish default.vcl is backend apache { set backend.host = "127.0.0.1"; set backend.port = "8008"; } sub vcl_deliver { remove resp.http.X-Varnish; remove resp.http.Via; remove resp.http.Age; remove resp.http.Server; remove resp.http.X-Powered-By; }

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  • User permissions linux. (proftpd / nginx)

    - by user55745
    I've been having a complete nightmare trying to configure proftpd. I've got proftp server working with an sql database. However I want to have any files uploaded able to viewed by the webserver running on the same box. The folders get created in /var/tmp/ as rwx------ 2 ftpuser ftpgroup 4096 Oct 8 20:35 50730c4346512 drwx------ 2 ftpuser ftpgroup 4096 Oct 8 20:38 50730f3a811ca I've tried adding www-data to group with the following usermod -g www-data ftpuser But this doesn't allow the web server access. In proftpd.conf I have the following umask set Umask 0022 It doesn't seem to make a difference what I set that value to. /etc/group (sure I've messed up one of these two but I'm getting desperate) ftpgroup:x:2001:www-data www-data:x:33:ftpgroup /etc/passwd www-data:x:33:33:www-data:/var/www:/bin/sh proftpd:x:108:65534::/var/run/proftpd:/bin/false ftp:x:109:65534::/srv/ftp:/bin/false ftpuser:x:2001:33:proftpd user www-data:/bin/null:/bin/false The ftpuser table in the database has uid / gid set to 2oo1 for both. I'm going absolutely crazy trying to solve this any help would be greatly appreciated. p.s Also, although if I manually connect to the ftp server I can upload files via FileZilla. Although this isn't working for the web-camera, although there is talky talky going on between the server and the camera.

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  • How to perform this Windows 7 permissions change on many files via GUI or command line

    - by hippietrail
    After using my external hard drive on another Windows 7 computer to tweak photos with Windows Live Photo Gallery then upload them to Facebook I found the modified images were now not visible on the original Windows 7 computer. I'm not sure if the things I tried to get it working subsequently changed anything, but I do know this is the sequence of actions that makes the permissions of the modified files match those of the unmodified files: Right click on broken image file, select "Properties" On the "Security" tab press the "Advanced" button In the "Permissions" tab press the "Continue" button with the shield icon on it Tick the box marked "Include inheritable permissions from this object's parent Click the "Remove" button to remove the only current entry "Type: Allow, Name: Administrators (XYZ\Administrators), Permission: Full control, Inherited From: OK on the "Permissions" tab. OK on the "Security" tab. Now this same procedure does not work at the folder level. It results in "access denied" dialogs. I'm looking for some way to perform this exact modification on all the images I edited on the other computer. I'm happy to use the Windows GUI in Explorer or any other included tools. I'm happy to use the Windows command line. I'd prefer not to use a third-party tool since I'd have to be satisfied it's not doing anything else. I'm not looking for a different way to change permissions to other settings to make an external drive full of photos editable on multiple computers. At least not in this question.

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  • Why would Copying a Large Image to the Clipboard Freeze a Computer?

    - by Akemi Iwaya
    Sometimes, something really odd happens when using our computers that makes no sense at all…such as copying a simple image to the clipboard and the computer freezing up because of it. An image is an image, right? Today’s SuperUser post has the answer to a puzzled reader’s dilemna. Today’s Question & Answer session comes to us courtesy of SuperUser—a subdivision of Stack Exchange, a community-driven grouping of Q&A web sites. Original image courtesy of Wikimedia. The Question SuperUser reader Joban Dhillon wants to know why copying an image to the clipboard on his computer freezes it up: I was messing around with some height map images and found this one: (http://upload.wikimedia.org/wikipedia/commons/1/15/Srtm_ramp2.world.21600×10800.jpg) The image is 21,600*10,800 pixels in size. When I right click and select “Copy Image” in my browser (I am using Google Chrome), it slows down my computer until it freezes. After that I must restart. I am curious about why this happens. I presume it is the size of the image, although it is only about 6 MB when saved to my computer. I am also using Windows 8.1 Why would a simple image freeze Joban’s computer up after copying it to the clipboard? The Answer SuperUser contributor Mokubai has the answer for us: “Copy Image” is copying the raw image data, rather than the image file itself, to your clipboard. The raw image data will be 21,600 x 10,800 x 3 (24 bit image) = 699,840,000 bytes of data. That is approximately 700 MB of data your browser is trying to copy to the clipboard. JPEG compresses the raw data using a lossy algorithm and can get pretty good compression. Hence the compressed file is only 6 MB. The reason it makes your computer slow is that it is probably filling your memory up with at least the 700 MB of image data that your browser is using to show you the image, another 700 MB (along with whatever overhead the clipboard incurs) to store it on the clipboard, and a not insignificant amount of processing power to convert the image into a format that can be stored on the clipboard. Chances are that if you have less than 4 GB of physical RAM, then those copies of the image data are forcing your computer to page memory out to the swap file in an attempt to fulfil both memory demands at the same time. This will cause programs and disk access to be sluggish as they use the disk and try to use the data that may have just been paged out. In short: Do not use the clipboard for huge images unless you have a lot of memory and a bit of time to spare. Like pretty graphs? This is what happens when I load that image in Google Chrome, then copy it to the clipboard on my machine with 12 GB of RAM: It starts off at the lower point using 2.8 GB of RAM, loading the image punches it up to 3.6 GB (approximately the 700 MB), then copying it to the clipboard spikes way up there at 6.3 GB of RAM before settling back down at the 4.5-ish you would expect to see for a program and two copies of a rather large image. That is a whopping 3.7 GB of image data being worked on at the peak, which is probably the initial image, a reserved quantity for the clipboard, and perhaps a couple of conversion buffers. That is enough to bring any machine with less than 8 GB of RAM to its knees. Strangely, doing the same thing in Firefox just copies the image file rather than the image data (without the scary memory surge). Have something to add to the explanation? Sound off in the comments. Want to read more answers from other tech-savvy Stack Exchange users? Check out the full discussion thread here.

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  • How to post a file via HTTP post in vb.net

    - by Worz
    Hi all! Having a problem with sending a file via HTTP post in vb.net. I am trying to mimic the following HTML so the vb.net does the same thing. <form enctype="multipart/form-data" method="post" action="/cgi-bin/upload.cgi"> File to Upload: <input type="file" name="filename"/> <input type="submit" value="Upload" name="Submit"/> </form> Hope someone can help!

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  • Read and write into a file using VBScript

    - by Maddy
    How can we read and write some string into a text file using VBScript? I mean I have a text file which is already present so when I use this code below:- Set fso = CreateObject("Scripting.FileSystemObject" ) Set file = fso.OpenTextFile("C:\New\maddy.txt",1,1) This opens the file only for reading but I am unable to write anything and when I use this code:- Set fso = CreateObject("Scripting.FileSystemObject" ) Set file = fso.OpenTextFile("C:\New\maddy.txt",2,1) I can just use this file for writing but unable to read anything. Is there anyway by which we can open the file for reading and writing by just calling the OpenTextFile method only once. I am really new to VBScript. I am only familiar with C concepts. Is there any link to really get me started with VBScript? I guess I need to have a good knowledge of the objects and properties concepts.

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  • How to retrieve path for a file embedded in Resources (Resource Manager) - .net C#

    - by curiousone
    Hi, I am trying to retrieve file path for a html file that is embedded in resource (resx file) in VS2008 C# project. I want to give path of this file to native webbrowser control (PIEHtml) to be able to navigate (DTM_NAVIGATE) in my application. I know I can pass the string to this control using DTM_ADDTEXTW but since html text size is so big, I dont want to pass string to the control. I need to somehow extract the file path for this html file embedded inside resource manager. I tried using but this does not give the file path of html inside assembly: private ResourceManager resManager = new ResourceManager("AppName.FolderName.FileName", System.Reflection.Assembly.GetExecutingAssembly()); this.lbl.Text = resManager.GetString("StringInResources"); and also read Retrieving Resources in Satellite Assemblies but it did not solve my problem. Can somebody please provide info as to how to achieve this ? thanks,

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  • Perl - Internal File (create and execute)

    - by drewrockshard
    I have a quick question about creating files with perl and executing them. I wanted to know if it was possible to generate a file using perl (I actually need a .bat script) and then execute this file internally to the program. I know I can create files, and I have with perl, however, I'm wanting to do this internally to the program. So, what I want it to do is actually create a batch script internally to the program (no file is actually written to the disk, everything remains in memory, or the perl program), and then once it completes the writing of the file, I'd like to be able to actually execute this file, and then discard the file it just wrote. I'm basically trying to have it create a batch script on the fly, so that I can just have output text files from the output of the script, rather than creating the batch script on disk, then executing it, and then deleting the batch file from disk when its done. Can this be done and how would I go about doing this? Regards, Drew

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  • Imagemagick - File Naming

    - by Josh Crowder
    I am using the convert command to convert a pdf to multiple pngs, I need the naming conventions to be slide-##.png at the moment they come out like slide-1.png but because there is 20+ slides when I loop through them to add them into the model the order comes up wrong, so it looks like slide-1.png slide-10.png slide-11.png and so on, how can I force convert to use double numbers like 01 02 03 and so forth or is there a better way to loop through them, this is the code I have at the moment def convert_keynote_to_slides system('convert -size 640x300 ' + keynote.queued_for_write[:original].path + ' ~/rails/arcticfox/public/system/keynotes/slides/'+File.basename( self.keynote_file_name )+'0%d.png') slide_basename = File.basename( self.keynote_file_name ) files = Dir.entries('/Users/joshcrowder/rails/arcticfox/public/system/keynotes/slides') for file in files #puts file if file.include?(slide_basename +'-') self.slides.build("slide" => "#{file}") if file.include?(slide_basename) end end

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  • Reading an XML File with .NET

    - by Daywalker21
    I am new to xml and unable to find a way to get content in between tags. My XML file is <?xml version="1.0" encoding="utf-8"?> <block1> <file name="c:\w0.xml"> <word>Text</word> <number>67</number> </file> <file name="c:\w1.xml"> <word>Text</word> <number>67</number> </file> <file name="c:\w2.xml"> <word>Text</word> <number>67</number> </file> </block1>

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  • Magic behind R.java file

    - by LambergaR
    Hi! Recently I have been having quite some problems with R.java file. Now I have decided to do a backup and delete the file to see what happens. Nothing happened, so I created an empty R.java file and hopped for the best. Now Eclipse seems to figure out that the file was tempered with and even issues a warning: R.java was modified manually! Reverting to generated version! And that's all there is. I tried building it manually but got no results. So, I have two questions: 1. what should I do to force Eclipse to generate the file 2. what is happening here? How is the file created, where is the code that is generating the file? I would appreciate any help. As usual the problem occurred just a few days before the deadline :)

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  • Django file uploads - Just can't work it out

    - by phoebebright
    OK I give up - after 5 solid hours trying to get a django form to upload a file, I've checked out all the links in stackoverflow and googled and googled. Why is it so hard, I just want it to work like the admin file upload? So I get that I need code like: if submitForm.is_valid(): handle_uploaded_file(request.FILES['attachment']) obj = submitForm.save() and I can see my file in request.FILES['attachment'] (yes I have enctype set) but what am I supposed to do in handle_uploaded_file? The examples all have a fixed file name but obviously I want to upload the file to the directory I defined in the model, but I can't see where I can find that. def handle_uploaded_file(f): destination = open('fyi.xml', 'wb+') for chunk in f.chunks(): destination.write(chunk) destination.close() Bet I'm going to feel really stupid when someone points out the obvious!

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  • MySQL INTO OUTFILE overide existing file?

    - by Derek Organ
    I've written a big sql script that creates a CSV file. I want to call a cronjob every night to create a fresh CSV file and have it available on the website. Say for example I'm store my file in '/home/sites/example.com/www/files/backup.csv' and my SQL is SELECT * INTO OUTFILE '/home/sites/example.com/www/files/backup.csv' FIELDS TERMINATED BY ',' OPTIONALLY ENCLOSED BY '"' LINES TERMINATED BY '\n' FROM ( .... MySQL gives me an error when the file already exists File '/home/sites/example.com/www/files/backup.csv' already exists Is there a way to make MySQL overwrite the file? I could have PHP detect if the file exists and delete it before creating it again but it would be more succinct if I can do it directly in MySQL.

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  • decompress .gz file in batch

    - by kapildalwani
    I have 100 of .gz files which I need to de-compress. I have couple of questions a) I am using the code given at http://www.roseindia.net/java/beginners/JavaUncompress.shtml to decompress the .gz file. Its working fine. Quest:- is there a way to get the file name of the zipped file. I know that Zip class of Java gives of enumeration of entery file to work upon. This can give me the filename, size etc stored in .zip file. But, do we have the same for .gz files or does the file name is same as filename.gz with .gz removed. b) is there another elegant way to decompress .gz file by calling the utility function in the java code. Like calling 7-zip application from your java class. Then, I don't have to worry about input/output stream. Thanks in advance. Kapil

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  • Maven2: Best practise for Enterprise Project (EAR file)

    - by Maik
    Hello everyone, I am just switching from Ant to Maven and am trying to figure out the best practice to set up a EAR file based Enterprise project? Lets say I want to create a pretty standard project with a jar file for the EJBs, a WAR file for the Web tier and the encapsulating EAR file, with the corresponding deployment descriptors. How would I go about it? Create the project with archetypeArtifactId=maven-archetype-webapp liek a war file and extend from there? What is the best project structure (and POM file example) for this? Where do you stick the ear file related deployment descriptors, etc? Thanks for any help.

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  • Creating an SQL Compact file: Template or script?

    - by David Veeneman
    I am writing an application that writes to SQL Compact files that have a specific schema, and I am now implementing the New File use case. The simplest approach seems to be to use a Template pattern: first, create a template file that lives in the application directory. Then, when the user selects New File, the template is copied to the name and destination specified by the user in a New File dialog. The alternative is a scripted approach: Use the same New File dialog, but dispense with the template file. Instead, create an empty SQL Compact file using the name/destination specified by the user, and then execute a T-SQL script on it from managed code. At this point, I am leaning toward the Template approach, because it is simpler. Is there any reason I should not use that approach? Thanks for your help.

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  • using Applescript to retrieve movie file duration via Finder properties

    - by Matt
    When using Get Info on movie files in the Finder, I've noticed that many movie files have an accessible "Duration" property under the More Info tab. Is there an easy way to access this data using Applescript? I've tried lines like "set theTime to (duration of aFile)", but it doesn't seem to be as simple as that. Ideally I'd like to access the movie duration via the Finder properties without having to actually open the file in QuickTime. Is this possible? Edit: I seem to have figured this out... you need to use the "System Events" application, specifically the Movie File Suite of commands (and, importantly, you need to set the movie variable as a string, rather than an alias): set theMovie to (choose file with prompt "Select a movie file:") as string tell application "System Events" set theName to name of movie file theMovie set theDuration to (duration of contents of movie file theMovie) / (time scale of contents of movie file theMovie) end tell

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  • getting the path of a file from its grandparent folder

    - by Saswat
    i have a php file which has the followng path Shubhmangalam/admin/welcome_image_edition/delete_image.php and an image file with the follwing path Shubhmangalam/welcome_images/image_1.jpg i want to delete the image_1.jpg file which i know can be done by using unlink() method.. but the prob is that the parent folder of the .php file and .jpg file is different, and so is their level of file-system...and i cant find the proper way to get the path to delete the image_1.jpg file. now the code on the delete_image.php is accordingly <?php $image=$_REQUEST['image']; if(unlink("./../welcome_images/".$image)) echo "Successfully Deleted"; else echo "Wrong"; ?> now the above is server-scripting code, i want to delete the image by getting appropriate path.. i dnt want the actual path, but the path from the project folder that is Shubhmangalam thanks in advance

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  • Android file-creation fails.

    - by Alxandr
    I use the following code to create a folder "mymir" and a file ".nomedia" (in the mymir-folder) on the sdcard of an android unit. However, somehow it fails with the exception that the folder the ".nomedia"-file is to be placed in dosn't exist. Here's the code: private String EnsureRootDir() throws IOException { File sdcard = Environment.getExternalStorageDirectory(); File mymirFolder = new File(sdcard.getAbsolutePath() + "/mymir/"); if(!mymirFolder.exists()) { File noMedia = new File(mymirFolder.getAbsolutePath() + "/.nomedia"); noMedia.mkdirs(); noMedia.createNewFile(); } return mymirFolder.getAbsolutePath(); }

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  • Use a grepped file as an included source in bash

    - by Andrew
    I'm on a shared webhost where I don't have permission to edit the global bash configuration file at /ect/bashrc. Unfortunately there is one line in the global file, mesg y, which puts the terminal in tty mode and makes scp and similar commands unavailable. My local ~./bashrc includes the global file as a source, like so: # Source global definitions if [ -f /etc/bashrc ]; then . /etc/bashrc fi My current workaround uses grep to output the global file, sans offending line, into a local file and use that as a source. # Source global definitions if [ -f /etc/bashrc ]; then grep -v mesg /etc/bashrc > ~/.bash_global . ~/.bash_global fi Is there a way to do include a grepped file like this without the intermediate step of creating an actual file? Something like this? . grep -v mesg /etc/bashrc > ~/.bash_global

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  • django handling file uploads - target different than media folder

    - by Tom Tom
    Hi, I want to enable the user to upload media which will not be saved in the media folder. When I use the following line of code data will be uploaded to media/upload/logo . logo_img = models.FileField(upload_to='upload/logo', blank=True) I'm wondering how I can change this behaviour. I would try to write a custom FileField and a view that serves the data based on the database entries. I do not want to place the data the user uploads to the media folder, since it is no public data. Is this approach correct? Are there solutions out there which do exactly what I want and I would reinvent the wheel with implementing this by myself? Would appreciate any help!

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  • FileInputStream for a generic file System

    - by Akhil
    I have a file that contains java serialized objects like "Vector". I have stored this file over Hadoop Distributed File System(HDFS). Now I intend to read this file (using method readObject) in one of the map task. I suppose FileInputStream in = new FileInputStream("hdfs/path/to/file"); wont' work as the file is stored over HDFS. So I thought of using org.apache.hadoop.fs.FileSystem class. But Unfortunately it does not have any method that returns FileInputStream. All it has is a method that returns FSDataInputStream but I want a inputstream that can read serialized java objects like vector from a file rather than just primitive data types that FSDataInputStream would do. Please help!

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