Save grep result to array

Question

I want to save all filenames that matches the pattern in bash array.

My solution does not work. I think the problem is because of pipe usage, but I don't know how to fix it.

i=0
find . -type f | grep -oP "some pattern" | while read -r line; do
    arr[$i]=$line;
    let i=i+1;
done

Stéphane Chazelas · Accepted Answer · 2017-05-01 15:17:53Z

With bash-4.4 and above, you'd use:

readarray -d '' -t arr < <(
  find . -type f -print0 | grep -zP 'some pattern')

With older bash versions:

arr=()
while IFS= read -rd '' file; do
  arr+=("$file")
done < <(find . -type f -print0 | grep -zP 'some pattern')

Or (to be compatible to even older versions of bash that didn't have the zsh-style arr+=() syntax):

arr=() i=0
while IFS= read -rd '' file; do
  arr[i++]=$line
done < <(find . -type f | grep -zP 'some pattern')

Your approach has several problems:

with -o, grep only prints the parts of the records that match the pattern as opposed to the full record. You don't want it here.
find's default newline-delimited output can't be post processed as the newline character is as valid as any in a file path. You need a NUL-delimited output (so -print0 in find and -z in grep to process NUL-delimited records.
you also forgot to pass IFS= to read.
in bash, and without the lastpipe option, the last part of a pipe line runs in a subshell, so there, you'd only be updating the $arr of that subshell.

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