12

I need to read the output from ffmpeg in order to even try the solution to my question from yesterday. This is a separate issue from my problem there, so I made a new question.

How the heck do I get the output from an ffmpeg -i command in PHP?

This is what I've been trying:

<?PHP
    error_reporting(E_ALL);
    $src = "/var/videos/video1.wmv";
    $command = "/usr/bin/ffmpeg -i " . $src;
    echo "<B>",$command,"</B><br/>";
    $command = escapeshellcmd($command);

    echo "backtick:<br/><pre>";
    `$command`;

    echo "</pre><br/>system:<br/><pre>";
    echo system($command);

    echo "</pre><br/>shell_exec:<br/><pre>";
    echo shell_exec($command);

    echo "</pre><br/>passthru:<br/><pre>";
    passthru($command);

    echo "</pre><br/>exec:<br/><pre>";
    $output = array();
    exec($command,$output,$status);
    foreach($output AS $o)
    {
            echo $o , "<br/>";
    }
    echo "</pre><br/>popen:<br/><pre>";
    $handle = popen($command,'r');
    echo fread($handle,1048576);
    pclose($handle);
    echo "</pre><br/>";
?>

This is my output:

<B>/usr/bin/ffmpeg -i /var/videos/video1.wmv</B><br/>
backtick:<br/>
    <pre></pre><br/>
system:<br/>
    <pre></pre><br/>
shell_exec:<br/>
    <pre></pre><br/>
passthru:<br/>
    <pre></pre><br/>
exec:<br/>
    <pre></pre><br/>
popen:<br/>
    <pre></pre><br/>

I don't get it. safe_mode is off. There's nothing in disable_functions. The directory is owned by www-data (the apache user on my Ubuntu system). I get a valid status back from exec() and system() and running the same command from the command line give me tons of output. I feel like I must be missing something obvious but I have no idea what it is.

Community
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Andrew Ensley
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4 Answers4

29

The problem is you catch only stdout and not stderr (see Standard Streams). Change this line:

$command = "/usr/bin/ffmpeg -i " . $src;

into

$command = "/usr/bin/ffmpeg -i " . $src . " 2>&1";

and give it another try :)

hakre
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hegemon
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  • Ahhh... yeah. That makes sense because my return status is 1 (not 0). I wish that worked, but I still get the same result, just with: "/usr/bin/ffmpeg -i /var/videos/video1.wmv 2>&1" at the beginning. I'll do some research to see if Ubuntu has some differences for redirecting stderr. – Andrew Ensley Jul 10 '09 at 18:16
  • I can confirm that the following should work. I've used this exact line in a hobby project once. exec('ffmpeg -i ' . escapeshellarg($filepath) . ' 2>&1', $output); – Werner Jul 10 '09 at 18:51
  • Ahha! I figured it out. My problem was the escapeshellcmd() function. It was changing "2>&1" to "2\>\&1". I tried your code and it worked. Lesson I've learned: output the command *after* you've made all modifications to it =p. Thanks. – Andrew Ensley Jul 10 '09 at 19:07
8

Use ffprobe instead, it's much quicker and supports JSON output.

$output = shell_exec('ffprobe -v quiet -print_format json -show_format -show_streams "path/to/yourfile.ext"');
$parsed = json_decode($output, true);

And you have all your video info in a php array! This is much faster than ffmpeg -i for some reason.

tweak2
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4

To get output status and output:

exec("ffmpeg -i input.avi output.mp4 2>&1", $output, $returnStatus);

print_r($output);

if($returnStatus === 0){
   // success
}
else {
   //fail
}
Kristian
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0

You can use exec and print_r the output... 

exec("ffmpeg -i input.avi -vcodec h264 -acodec aac -strict -2 output.mp4 2>&1",$output);

echo "<pre>";
print_r($output);
echo "</pre>";
Luke Dobner
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