ffmpeg converting using a cronjob

oMIKEo · 05-13-2008, 09:46 PM

Hi,

I'm making some progress with ffmpeg but need some help working out why i cant run the same code using php as when i run in ssh.

Here is the php:

PHP Code:


			
$dir_root = "../../var/www/html/uploads/media/";

$temp_new = substr($o_newname, 0, strripos($o_newname, '.'));



$output = exec ("ffmpeg -i ".$dir_root.'o_'.$o_newname." -s qvga -y ".$dir_root.'n_'.$temp_new.".flv");

echo($output);

its a bit of a mess but its basically getting the input video, setting the size to 320x240 (qvga) and then outputting it as a .flv. When i use that output in ssh it works fine but when i run the above script it doesnt seem to do anything.

ffmpeg is located: /usr/bin/ffmpeg
website is location: /var/www/html/

Here is an example of what i paste into ssh:

Code:

ffmpeg -i ../../var/www/html/uploads/media/o_l2mWwMYOmDhuieDpNnb.avi -s qvga -y ../../var/www/html/uploads/media/n_l2mWwMYOmDhuieDpNnb.flv

is there any way to get some error checking in there or get an output back from the exec/ffmpeg command?

Regards,
Michael

drewbee · 05-14-2008, 02:03 PM

exec will only return the last line which in this case might be a blank line.

Is safemode on? If safemode is on, exec will not work.

You could also try PHP: shell_exec - Manual

I believe shell_exec returns all output as a string.

drewbee · 05-14-2008, 02:10 PM

Also, after reading the manual, exec can return values, you just need to specify it;

PHP Code:


			
exec('whoami', $array, $returnStatus);
echo print_r($array);

oMIKEo · 05-14-2008, 04:24 PM

ok i changed the script to do this:

PHP Code:


			
$output = exec ("ffmpeg -i ".$dir_root.'o_'.$o_newname." -s qvga -y ".$dir_root.'n_'.$temp_new.".flv",$arrayOutput, $returnOutput);



echo print_r($arrayOutput); 

echo '<br />';

echo print_r($returnOutput);

and got the following:

PHP Code:


			
Array ( ) 1

11

Any idea what that means?

I then tried this:

PHP Code:


			
$output = shell_exec("ffmpeg -i ".$dir_root.'o_'.$o_newname." -s qvga -y ".$dir_root.'n_'.$temp_new.".flv");

echo "<pre>$output</pre>";

i didnt get an onscreen output but the code read <pre></pre> thats was it.

Thanks for your help!

PS: safe mode is not turned on.

drewbee · 05-14-2008, 07:23 PM

Hmm, not sure.

Does the PHP user have access to run ffmpeg? There may be a potential permissions problem.

That does tell me that nothing is being returned from the command line though;

What happens when you do just exec 'whoami'? See if that returns any information in the array.