Bash/sed - use sed in variable

Question

I would like to use sed to delete and replace some characters in a bash script.

#!/bin/bash
DIR="."
file_extension=".mkv|.avi|.mp4"
files= `find $DIR -maxdepth 1 -type f -regex ".*\.\(mkv\|avi\|mp4\)" -printf "%f\n"`

In order to simplify $files, I would like to use $file_extension in it, i.e. change .mkv|.avi|.mp4 to mkv\|avi\|mp4

How can I do that with sed ? Or maybe an easier alternative ?

Please add a shebang and then paste your script at shellcheck.net and try to implement the recommendations made there. — Cyrus
– Cyrus, Commented Feb 21, 2021 at 11:51
I'd do files=( "$DIR"/*.{mkv,avi,mp4} ) instead. Storing find's output in a scalar variable is not a good idea, especially when there is no need to use it at all. — oguz ismail
– oguz ismail, Commented Feb 21, 2021 at 12:36

Thomas · Accepted Answer · 2021-02-21 12:00:23Z

1

No need for sed; bash has basic substitution operators built in. The basic syntax for a replace-all operation is ${variable//pattern/replacement}, but unfortunately it can't be nested so you need a helper variable. For clarity, I'll even use two:

file_extension_without_dot="${file_extension//./}"           # mkv|avi|mp4
file_extension_regex="${file_extension_without_dot//|/\\|}"  # mkv\|avi\|mp4
files= `find $DIR -maxdepth 1 -type f -regex ".*\.\($file_extension_regex\)" -printf "%f\n"`

If your find supports it, you could also consider using a different -regextype (see find -regextype help) so you don't need quite so many backslashes anymore.

answered Feb 21, 2021 at 12:00

Thomas

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