Using variable in PHP exec [closed]

Answer 1

Replace with double quotes:

exec("ffmpeg -i $audio_input $audio_output");

Check the manual for Strings here.

Answer 2

Use double quotes instead of single quotes:

exec("ffmpeg -i $audio_input $audio_output");

But please, go read this page: http://php.net/manual/en/language.types.string.php

Answer 3

Use double quotes like mentioned before or

exec('ffmpeg -i '.$audio_input.' '.$audio_output);