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bashshellgnu-findutils

How to find all files that match a regex and store the result in a variable?

I have this code so far:

#!/bin/bash
list=$(find "$(dirname "$0")/../temp/" -regex "(part.[a-z])")
echo "$list"

I want to find all files in the temp directory that have a file name like this:

part.aa part.ab part.aaa part.qqk

current path is:

./scripts/script.sh

path where files are in:

../temp

Ideally, the result stored in list should be in a form that allow looping over it line by line.

Solution

  • The regex must match the whole path.

    -regex '.*/part\.[a-z]+'

    Note that . is special in regexes, it matches any character. To match a literal dot, you need to backslash it (or use [.]).