sh: ffmpeg: command not found when run command thr

Question

I have installed successfully installed FFMpeg on root of my Centos 6 machine (https://trac.ffmpeg.org/wiki/CompilationGuide/Centos).

My workplace of apache/php is /var/www/html

Now I'm running below command successfully on /var/www/html directory to capture frame from the video file. It's capturing a frame.

[root@localhost html]# ffmpeg -i video.mpg -an -ss 30  -t 00:00:01 -r 1 -y -vcodec mjpeg -f mjpeg -s 160x100 frame8  2>&1

I want to run this command through php and using shell_exec() or exec() php functions. My php code for running the command is:

$cmd = "/root/bin/ffmpeg -i /project/app/webroot/videos/video.mpg -ss 00:00:14.435 -f image2 -vframes 1 /project/app/webroot/videothumbnails/example-thumb.jpg";
$locale = 'en_IN.UTF-8';
setlocale(LC_ALL, $locale);
putenv('LC_ALL='.$locale);
echo shell_exec($cmd);

When I'm trying to run command through above php code, I'm getting below error:

sh: ffmpeg: command not found

Please help me to solve out this problem.

Answer 1

Try using which to locate the ffmpeg binary. At least on Ubuntu/Debian based Linuxes the which command should be available by default. which returns the file path to the preferred executable of any available application. E.g. which php should return the PHP installation executable that is used by default for CLI operations. Therefore which ffmpeg should return the system default location for the ffmpeg installation.

<?php

// shell_exec returns the command output or null on error/empty output.
$ffmpeg_path = shell_exec('which ffmpeg');

if ($ffmpeg_path != null) {
    shell_exec($ffmpeg_path . ' -i file.mpg --snip--');
}

Also make sure that the Linux user running the PHP script has access to ffmpeg (e.g. $ chmod +x ffmpeg). If $ sudo which ffmpeg returns a path but $ which ffmpeg doesn't, try reinstalling ffmpeg for the wanted user or play around with the permissions.

EDIT: use care with which though, there's no knowing whether a cracker could replace the ffmpeg executable with something malicious (which leads to the above PHP script running the wrong executable).

Answer 2

For Video Thumbnail creation we use ffmpeg.

In Linux Systems(centos 6.x) Ffmpeg installation process and Php example:-

This process is done in my centos 6 and created thumbnails using php.

Step1:- Please check If any existing ffmpeg available or not in your linux server .If available please remove that files.

Step2:- For New ffmpeg installation follow this link
http://root.uabid.com/compile-ffmpeg-on-centos-6-x/ .

Step3:- After installation complete check whereis your ffmpeg installed.

Command: whereis ffmpeg

type this command in linux command line and check.If for example your ffmpeg path in (/usr/local/bin/ffmpeg). Use this path in your code.

Php Example:-

<?php

    if($extension === 'mp4' OR $extension == 'MP4' )
    {
    $video = $timestamp.$imagename;
    $videoname=substr($imagename,0, -4).$timestamp;
    $image = "sites/default/files/content_images/{$videoname}-thumb.jpg";

    var_dump($video);

    $cmd = "/usr/local/bin/ffmpeg -i    /opt/lampp/htdocs/myproject/sites/default/files/content_videos/".$video." -ss   00:00:01.435 -f image2 -vframes 1       /opt/lampp/htdocs/myproject/sites/default/files/content_images/".$videoname."-  thumb.jpg";

    $cmdstr = $cmd;
    $locale = 'en_IN.UTF-8';
    setlocale(LC_ALL, $locale);
    putenv('LC_ALL='.$locale);
    echo exec($cmd);
?>

Process 2:- This is for static file paths:

<?php    
    $cmd = "/root/bin/ffmpeg -i /var/www/html/project/app/webroot/videos/example.mp4 -ss   00:00:01.435 -f image2 -vframes 1 /var/www/html/project/app/webroot/videothumbnails/example-thumb.jpg";
    $cmdstr = $cmd;
    $locale = 'en_IN.UTF-8';
    setlocale(LC_ALL, $locale);
    putenv('LC_ALL='.$locale);
    echo exec($cmd);
?>

Hope this helps to you & others..

Answer 3

I've seen this problem. For me 'command not found' may be the $PATH variable problem.

When run this command through php and using shell_exec() or exec() php functions, you may use different $PATH variable compare with running command in linux directory.

For example, the /usr/local/php/etc/vhostphp-fpm.conf for my php project is below:

env[PATH] = /usr/local/bin:/usr/bin:/bin
env[TMP] = /tmp
env[TMPDIR] = /tmp
env[TEMP] = /tmp
env[LD_LIBRARY_PATH] = /usr/local/lib

but ffmpeg command locates in /data/bin directory. My Linux's $PATH variable contain the /data/bin.

So add /data/bin to env[PATH] variable above can solve the problem.