FFMPEG running in Command Line but not PHP

Question

I am using ffmpeg build for windows to make video thumbnails . The command works well in command line but not from PHP exec method. am using PHP 5.2.11

Here is the c

Answer 1

I use it this way::

exec("C:/wamp/bin/ffmpeg -i ./output4.mp4 -sameq -acodec libmp3lame -ar 22050 -ab 32 -f flv -s 320x240 ./output8.flv -vcodec mjpeg -vframes 4 -an -f rawvideo -s 320x240 ./pic008.jpg 2>&1");

Directly connected from WAMP SERVER.

Notice the:

./output4.mp4

That tells PHP that I am dealing with the current directory.

--All the Best

Answer 2

exec("\"E:\\Documents and Settings\\x\\WINDOWS\\ffmpeg\" -i <inputfile> <options> <outfile>");

Here's one of mine I've used in the past (granted I'm on a LAMP stack):

$cmd = "/usr/bin/ffmpeg -i ".$in." -y -an -sameq -vframes 1 -s 100x56 -ss 3 -t 0.001 ".$out;

You may also consider: http://ffmpeg-php.sourceforge.net/

Answer 3

The error message says it doesn't recognize it as a command. Its most probably your quoting. Check your quoting of white spaces. Escape the white spaces when necessary using slash "\ ". And where is your code snippet that calls exec()?

Answer 4

You need to escape your command properly:

exec(escapeshellcmd($cmd), $thumb_stdout, $retval);

Also do you have PHP safe mode on? You should check that $in is a real file before trying to encode too.

Answer 5

are you properly escaping your backslashes, quotes etc.? Is there any error message?

Answer 6

Maybe try this:

$cmd = "\"$path\" -itsoffset -4 -i \"$in\" -vcodec mjpeg -vframes 1 -an -f rawvideo -s 320x240 \"$out\" 2>&1";