Return list using git diff and grep

Question

I'm using git diff to return the file names of files recently changed and i'm trying to pipe the returned file names into a grep expression that searches each file and returns the files that have "*.yml" in the file name and "- name:" in the actual file. However it seems like my code only returns all the files in the directory that match the conditions, basically ignoring the git diff.

files=($(git -C ${dir} diff --name-only HEAD^ HEAD | grep -rlw --include="*.yml" --exclude-dir=$excluded_paths-e "- name:"))

Any help would appreciated!

Answer 1

git diff --name-only produces a list of files; you need to give that list to grep in a form it will understand, i.e. as command-line arguments. One way to do this is

git -C "${dir}" diff -z --name-only HEAD^ HEAD | xargs -0 grep -lw --include="*.yml" --exclude-dir="$excluded_paths" -e "- name:"

I added -z so that the file names are separated by null bytes; the -0 xargs option specifies the same delimiter for use in xargs’ input. I removed the -r option from grep — that is used for recursive searches, which is counter-productive here since the exact list of files to check is given.