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DeMoN064

Posts: 1308

Posted: Tue, 24th Jun 2008 17:08 Post subject: PHP shell_exec not working correctly...

I doubt im going to get a response since i've never even heard of this problem before and im not too sure how this problem occurred :

As far as i know, the entire script broke when I decided it would be a good idea to recompile PHP (Bad idea!) I then decided to downgrade from PHP 5 to PHP 4.4.7 (Orig version)

and.. nothing!

Code:

<?php
$convert = "ffmpeg -i /home/test/test.mov -ar 22050 -vb 524288 -f mp4 -s 320x240 /home/test/test.mp4";
$download=shell_exec($convert);
?>

The problem seems to me, is that it will exec a command and the php script will die right away therefore the shell dies - beforehand it would wait until the shell script had completed. Why is it not doing that?

Rinze
Site Admin

Posts: 2343

Posted: Tue, 24th Jun 2008 18:09 Post subject:	Quick Quote Quote

Any error messages? Permissions? Compiled with safe mode?

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DeMoN064

Posts: 1308

Posted: Tue, 24th Jun 2008 19:09 Post subject:

Rinze wrote:

Any error messages?

Nope, the code is fine.. its just PHP handling it differently :-/

Quote:

Permissions?

have not changed since compliing

Quote:

Compiled with safe mode?

Nope

DeMoN064

Posts: 1308

Posted: Tue, 24th Jun 2008 20:11 Post subject:

Breakthough! it seems there is a workaround to shell_exec :

Code:

<?php
function runExternal($cmd,&$code) {
$descriptorspec = array(
0 => array("pipe", "r"), // stdin is a pipe that the child will read from
1 => array("pipe", "w"), // stdout is a pipe that the child will write to
2 => array("pipe", "w") // stderr is a file to write to
);

$pipes= array();
$process = proc_open($cmd, $descriptorspec, $pipes);

$output= "";

if (!is_resource($process)) return false;

#close child’s input imidiately
fclose($pipes[0]);

stream_set_blocking($pipes[1],false);
stream_set_blocking($pipes[2],false);

$todo= array($pipes[1],$pipes[2]);

while( true ) {
$read= array();
if( !feof($pipes[1]) ) $read[]= $pipes[1];
if( !feof($pipes[2]) ) $read[]= $pipes[2];

if (!$read) break;

$ready= stream_select($read, $write=NULL, $ex= NULL, 2);

if ($ready === false) {
break; #should never happen - something died
}

foreach ($read as $r) {
$s= fread($r,1024);
$output.= $s;
}
}

fclose($pipes[1]);
fclose($pipes[2]);

$code= proc_close($process);

return $output;
}

But the problem now is that when you run :

Code:

$result= runExternal("ffmpeg -i /test/test.mov -ar 22050 -vb 524288 -f mp4 -s 320x240 /test/test.mp4",$code);
print " ";
print $result;
//print "\n";
print "code: $code\n";

It outputs :

[code]sh: ffmpeg: command not found code: 127[/code]

So i'm not sure what its looking for here, since i dont know enough about the linux file system? any ideas?

See the thing is it runs ffmpeg fine when you type in it in shell, so im kinda confused on this one lol

Rinze
Site Admin

Posts: 2343

Posted: Tue, 24th Jun 2008 22:24 Post subject:	Quick Quote Quote

Are you running PHP as a different user than before, with different path settings? Try using the full path to ffmpeg.

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DeMoN064

Posts: 1308

Posted: Wed, 25th Jun 2008 00:01 Post subject:

Used this :

Code:

ln -s /usr/local/bin/ffmpeg /usr/bin/ffmpeg

Outputs fine now Very Happy

thanks!

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