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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.

05-14-2008, 04:24 PM	#4 (permalink)
oMIKEo The Contributor Join Date: Jan 2008 Location: Leeds Posts: 52 Thanks: 7	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.