Store grep output in an array

Question

I need to search a pattern in a directory and save the names of the files which contain it in an array.

Searching for pattern:

grep -HR "pattern" . | cut -d: -f1

This prints me all filenames that contain "pattern".

If I try:

targets=$(grep  -HR "pattern" . | cut -d: -f1)
length=${#targets[@]}
for ((i = 0; i != length; i++)); do
   echo "target $i: '${targets[i]}'"
done

This prints only one element that contains a string with all filnames.

output: target 0: 'file0 file1 .. fileN'

But I need:

 output: target 0: 'file0'
 output: target 1: 'file1'
 .....
 output: target N: 'fileN'

How can I achieve the result without doing a boring split operation on targets?

Answer 1

You can use:

targets=($(grep -HRl "pattern" .))

Note use of (...) for array creation in BASH.

Also you can use grep -l to get only file names in grep's output (as shown in my command).

---

Above answer (written 7 years ago) made an assumption that output filenames won't contain special characters like whitespaces or globs. Here is a safe way to read those special filenames into an array: (will work with older bash versions)

while IFS= read -rd ''; do
   targets+=("$REPLY")
done < <(grep --null -HRl "pattern" .)

# check content of array
declare -p targets

On BASH 4+ you can use readarray instead of a loop:

readarray -d '' -t targets < <(grep --null -HRl "pattern" .)