Execute "ffmpeg" command in a loop [duplicate]

Question

I have three .wav files in my folder and I want to convert them into .mp3 with ffmpeg.

I wrote this bash script, but when I execute it, only the first one is converted to mp3.

What should I do to make script keep going through my files?

This is the script:

#!/bin/bash
find . -name '*.wav' | while read f; do
    ffmpeg -i "$f" -ab 320k -ac 2 "${f%.*}.mp3"
done

Answer 1

Use the -nostdin flag to disable interactive mode,

ffmpeg -nostdin -i "$f" -ab 320k -ac 2 "${f%.*}.mp3"

or have ffmpeg read its input from the controlling terminal instead of stdin.

ffmpeg -i "$f" -ab 320k -ac 2 "${f%.*}.mp3" </dev/tty

See the -stdin/-nostdin flags in the ffmpeg documentation, as well as the FAQ How do I run ffmpeg as a background task?.

Answer 2

If you do need find (for looking in subdirectories or performing more advanced filtering), try this:

find ./ -name "*.wav" -exec sh -c 'ffmpeg -i "$1" -ab 320k -ac 2 "$(basename "$1" wav).mp3"' _ {} \;

Piping the output of find to the while loop has two drawbacks:

1. It fails in the (probably rare) situation where a matched filename contains a newline character.
2. ffmpeg, for some reason unknown to me, will read from standard input, which interferes with the read command. This is easy to fix, by simply redirecting standard input from /dev/null, i.e. find ... | while read f; do ffmpeg ... < /dev/null; done.

In any case, don't store commands in variable names and evaluate them using eval. It's dangerous and a bad habit to get into. Use a shell function if you really need to factor out the actual command line.