Run commandline in php file

[sarayu]

Hi,
I am converting the video files to .flv using ffmpeg.I run and i converted all the files to .flv format in cmd.But how can i call the cmd commands in my php file.I used exec for conversion like this.But it wont be converted.Pleas e help me about this,

Thanks in advance

[Atli]

What do you mean by "it wont be converted"?

The exec function should allow you to execute any command, assuming the user running PHP has permission to execute it.

[sarayu]

Hi Atli,

I used ffmpeg for video conversion in windows.I converted those files to .flv format in cmd by using this command.

C:\>ffmpeg -i sruthi.wmv -ar 22050 -ab 32 -f flv project/video.flv

It converted to video.flv in my project directory.But when i call this command in my php file using exec like this

$ffmpegPath="C://ffmpeg/ffmpeg.exe";
$in="C://ffmpeg/sruthi.wmv";
$out="C://ffmpeg/video1.flv";
exec('$ffmpegPa th -i $in -ar 22050 -ab 32 -f flv -s 320×240 $out');
When i want to execute like this nothing will be displayed.How can i call exec in my file.

[Markus]

Hi Atli,

I used ffmpeg for video conversion in windows.I converted those files to .flv format in cmd by using this command.

C:\>ffmpeg -i sruthi.wmv -ar 22050 -ab 32 -f flv project/video.flv

It converted to video.flv in my project directory.But when i call this command in my php file using exec like this

$ffmpegPath="C://ffmpeg/ffmpeg.exe";
$in="C://ffmpeg/sruthi.wmv";
$out="C://ffmpeg/video1.flv";
exec('$ffmpegPa th -i $in -ar 22050 -ab 32 -f flv -s 320×240 $out');
When i want to execute like this nothing will be displayed.How can i call exec in my file.

$variables aren't parsed inside single quotes ('). Try using double quotes or concatenating your variables.

[php]
exec($ffmpegPat h . ' -i ' . $in . ' -ar 22050 -ab 32 -f flv -s 320×240 ' . $out);
[/php]

[sarayu]

I tried using double quotes and concatenating also.But it does not work.Any other solution?