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When running the following command line to convert multiple .mkv files into .mp4 files, things go awry. It's as if the script tries to run all the commands at once or they get entangled in each other somehow, and the output gets messy. I don't know exactly. When I ran this just now, only the first file was converted. Other times, I seem to get unexpected or random results.

ls -1 *.mkv | while read f; do ffmpeg -i "$f" -codec copy "${f%.*}.mp4"; done

However, if I do the following, things work as expected:

ls -1 *.mkv | while read f; do echo ffmpeg -i \"$f\" -codec copy \"${f%.*}.mp4\"; done > script; bash script

But why? What is it with ffmpeg? This doesn't happen with bash loops and other commands than ffmpeg. I'd like to do it a simple way without escaping quotes, piping to a script file, etc.

forthrin
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    Avoid piping ls output, especially to while read. Use the for f in *.mkv; do... construct. Oh, and btw, you don't need -1 when the output is a pipe. – don_crissti Nov 07 '15 at 21:40
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    What don_crissti said. Don't parse ls for things like this, especially not media files whose names are very likely to contain strange characters and whitespace. That said, please [edit] your question and show us this messy output you speak of. What does it look like? We can't help unless you show us the actual problem. – terdon Nov 07 '15 at 22:31
  • I had this same problem using something like find . -name "*.aifc" | while read f; do ffmpeg -i $f... -- ffmpeg was stripping characters off of the input filename and I couldn't figure out why! Using the for construct solved the problem. – bonh Jan 07 '17 at 17:03

1 Answers1

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ffmpeg reads from standard input as well, consuming data meant for read. Redirect its input from /dev/null:

... | while read f; do ffmpeg -i "$f" -codec copy "${f%.*}.mp4" < /dev/null; done

That said, don't read the output from ls like this. A simple for loop will suffice (and remove the original problem).

for f in *.mkv; do
    ffmpeg -i "$f" -codec copy "${f%.*}.mp4"
done
chepner
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