Store Grep Output in an Array
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...
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?
1 Answer
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" .)
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