Bash script that lists files in a directory doesn't work

Question

I made a bash script because I need to convert a lot of files in a directory from .MOV to .mp4 format. I created this script for the purpose:

#!/bin/bash

touch .lista
ls -1 "$1" | grep -i .MOV > .lista
list= `pwd`/.lista
cd "$1"
while read -r line;
    do filename=${line%????}
    ffmpeg -i "$line" -vcodec copy -acodec copy "$filename.mp4"; done < $list
rm .lista

This script is supposed to convert me each .MOV file into the directory indicated by $1, but it doesn't work, it converts me only one file, then it terminates. I can't understand why. What's wrong with that?

PesaThe · Accepted Answer · 2018-07-27 10:59:58Z

2

It's better to simply loop using globs:

for file in "$1"/*.MOV; do
    ffmpeg -i "$file" ... "${file%.*}.mp4" 
done

Why you shouldn't parse the output of ls.

answered Jul 27, 2018 at 10:59

PesaThe

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Comments

Mark Setchell · Accepted Answer · 2018-07-27 13:04:33Z

2

Do them all fast and succinctly in parallel with GNU Parallel like this:

parallel --dry-run ffmpeg -i {} -vcodec copy -acodec copy {.}.mp4 ::: movies/*MOV

Sample Output

ffmpeg -i movies/a.MOV -vcodec copy -acodec copy movies/a.mp4
ffmpeg -i movies/b.MOV -vcodec copy -acodec copy movies/b.mp4

If that looks good, do it again but without --dry-run.

Note how easily GNU Parallel takes care of all the loops, all the quoting and changing the extension for you.

edited Jul 27, 2018 at 13:04

answered Jul 27, 2018 at 11:59

Mark Setchell

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2 Comments

Fabrizio Gabriele Over a year ago

I'm gonna try this way very soon. Unfortunately my router is broken, due to a lighting bolt tempest happened yesterday, so I can't download GNU Parallel at the moment.

Mark Setchell Over a year ago

Cool - I'm sure you'll enjoy the ease of use and especially the performance especially in this world where CPUs are hardly getting much faster but are getting more and more cores, so parallel solutions are increasingly appropriate to make the most of them.

Poshi · Accepted Answer · 2018-07-27 11:33:01Z

1

Your code is working for me. I cannot see any error. But I can suggest you a better approach. Don't use ls to get the filenames, it is not a good idea. Also, you can avoid changing dir.

#!/bin/bash

for line in $(find "$1" -maxdepth 1 -type f -iname "*.mov")
do
    ffmpeg -i "$line" -vcodec copy -acodec copy "${line%????}.mp4"
done

You don't need to start by touching the file. In any case, you don't need a file at all, you can use a for loop to iterate over the files returned by find directly. With find, I'm already selecting all the files in the specified folder that have the expected extension.

Here I add a one-liner that should avoid problems with spaces:

find "$1" -maxdepth 1 -type f -iname "*.mov" -print0 | xargs -0 -n 1 -I{} bash -c "F={}; ffmpeg -i \"\$F\" -vcodec copy -acodec copy \"\${F%.*}\".mp4"

edited Jul 27, 2018 at 11:33

answered Jul 27, 2018 at 10:58

Poshi

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2 Comments

Inian Over a year ago

It would be much better use the globs as in PesaThe's answer, because the words could break if the filenames contain spaces

Fabrizio Gabriele Over a year ago

That's why I use "read -r line"...spaces in the names gave me a lot of problems. Anyway thanks to both for the tips

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