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" .)