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• FFMPEG installation & configuration with xampp on fedora

  10 octobre 2013, par UMI

  I have running Fedora 8, where I have installed XAMPP-Linux. First I installed latest version with Xampp-Linux-1.8.3 including PHP 5.5 from "http://www.apachefriends.org/en/xampp-linux.html".

  After setting up everything, tried to install "FFMPEG" from "http://www.nixgurus.com/fedora/installing-ffmpeg-using-yum-on-centos-redhat-fedora/". This is the only guide which gets completed with successful installation and configuration, otherwise all other tried tutorials got leave me middle of nowhere.

  Note : While installing FFMPEG following above provided URL, "phpize" did not work to me so I had to install php-devel by running "yum install php-devel" command. I always got different "Zend Module" and "Zend Extension API" versions.

  When I installed ffmpeg and put extension in "php.ini" with correct path, it never appeared in phpinfo().

  Then after searching, I got a hint that installed "FFMPEG" will not be compatible to the installed PHP version. So I removed complete pack of XAMMP-LINUX and installed old one with Xampp-Linux-1.8.2 including PHP 5.4. But it result were same.

  Then, I decided to use more older version and it was Xampp-Linux-1.8.1 including PHP 5.4. Now currently I have this version but having version results are totally changed that after installation of FFMPEG when I put ffmpeg.so with its correct path in PHP.INI (definitely restart the server on every change) server behaves strange way that PHP files get downloaded by browsers and when I comment same line and restart web server everything works well.

  Just to let you know that, even I have given path "extension=/usr/lib/php/modules/ffmpeg.so" which did not work and even moved files to default extension path "/opt/lampp/lib/php/extensions/no-debug-non-zts-20100525" as well but in vain.

  Looking forward to your help please.

• FFMPEG on Fedora but PHP Compilation confliction

  11 octobre 2013, par UMI

  I have running Xampp with PHP5.5 on Fedora from Apache Friend, with default settings what the installer package do on Linux.

  When I install FFMPEG successfully and try to load from php.ini it always says :

  *

  11-Oct-2013 14:05:51 Europe/Berlin] PHP Warning:  PHP Startup: ffmpeg: Unable to initialize module
  
  Module compiled with module API=20060613
  
  PHP    compiled with module API=20121212
  
  These options need to match
  
  in Unknown on line 0

  *

  The only thing which is confusing me is that when I did phpize even that I already installed xampp server which means I have PHP running. FFMPEG phpize did not work and I had to install php-devel. Does that mean that FFMPEG is configured with PHP, other than the installed with XAMPP ? I am not sure what is happening. Whatever I do I always receive this error message in php_error_log file.

  It is amazing :) that I just run below command to see what version of PHP is, I have and below are the results which are shocking one because I am in a feeling that I have PHP 5.5 installed and running from XAMMP.

  [root@localhost ~]# php -v
  
  PHP 5.2.6 (cli) (built: May  8 2008 08:53:44)
  
  Copyright (c) 1997-2008 The PHP Group
  
  Zend Engine v2.2.0, Copyright (c) 1998-2008 Zend Technologies

  How can I get this covered that if I simple install XAMPP server on linux and want to install and configure FFMPEG along with ?

• ffmpeg command and args execute successfully with child_process.exec() but not child_process.spawn()

  15 octobre 2013, par Zugwalt

  I am trying to use ffmpeg to add text to a video via the drawtext filter from within node.js in a windows environment.

  I have a command and arguments that work in the command line and when executed using the child_process module's exec function, but it encounters an error when the same arguments are used with the spawn function.

  The below code illustrates the problem :

  var child_process = require('child_process');
  
  
  
  var cmd = 'ffmpeg';
  
  
  
  var args = [ '-i',
  
               'c:\\path\\to\\my\\inputfile.mp4',
  
               '-vf',
  
               'drawtext="fontfile=/Windows/Fonts/arial.ttf:text=\'Hello World\':fontcolor=white@0.6:fontsize=70:x=0:y=40"',
  
               '-y',
  
               'c:\\path\\to\\my\\outputfile.mp4' ];
  
  
  
  // Above creates the command line equivalent of:
  
  // ffmpeg -i c:\path\to\my\inputfile.mp4 -vf drawtext="fontfile=/Windows/Fonts/arial.ttf:text='Hello fluent text':fontcolor=white@0.7:fontsize=70:x=0:y=40" -y c:\path\to\my\outputfile.mp4
  
  // this works when run from the command line
  
  
  
  var execCmd = cmd+' '+args.join(' ');
  
  
  
  child_process.exec(execCmd, function (error, stdout, stderr) {
  
          /* ffmpeg runs fine, adding the text to the video */
  
  
  
          var spawn = child_process.spawn(cmd,args);
  
  
  
          spawn.on('close', function (code) {
  
              /* ffmpeg fails, with standard error (obtained via spawn.stderr) reading:
  
                 Could not load font ""fontfile=/Windows/Fonts/arial.ttf": impossible to find a matching font
  
                 Error initializing filter 'drawtext' with args '"fontfile=/Windows/Fonts/arial.ttf:text=Hello fluent text:fontcolor=white@0.6:fontsize=70:x=0:y=40"'
  
              */
  
          });
  
      });

  Based on the error message :

  Could not load font ""fontfile=/Windows/Fonts/arial.ttf" : impossible
  to find a matching font

  And comparing it to giving ffmpeg a bogus font on the command line :

  Could not load font "/Windows/Fonts/bogus.ttf" : impossible to find a
  matching font

  It seems the problem is that when executed from spawn the drawtext argument is incorrectly parsed and "fontfile= is incorrectly making its way into the font's path. This does not happen when the same argument is executed with exec or from the command line. Is there any additional escaping that needs to be done when executing using spawn ?