How to get part of the filename using (basename "${video%.*}") in bash script?

I have a set of videos scalled xxxyyys.mp4, each filename ends with s. I want to execute a script that takes only the name upto xxxyyy for the argument basefile.

So far, what I have is

for video in folder/*
do python script.py -i "${video%*.mp4}" --basefile $(basename "${video%.*}")
done

How do I edit this to get only the name upto xxxyyy?

3

1 Answer

You can use:

"${video%s.*}"

This will delete everything from (and including) the last s. until the end of the string.

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