CODEWITHSUNDEEP
gitgrep
Clumsy cat
Clumsy cat

Reputation: 299

List files in which a word has disappeared

So, say I have two branches old_release (containing an old release) and new_release (containing a new release). In the old release there was a tool called cook, I think it's still in the new release, but it's been renamed. I need to work out what it's been renamed to.

I would like a list of files which must satisfy the criteria;

  1. File exists in both old_release and new_release.
  2. String cook is found in the version from old_release.
  3. String cook is not found in the version from new_release. I need this condition because most of the code is not updated and will contain defunct references to cook.
  4. My repository is big, and checking out a branch takes a long time. I'd like a solution that avoids that.

My current solution looks like;

git checkout old_release
grep cook . -R -l -I > old_release_cooks.txt
sort -o old_release_cooks_sorted.txt old_release_cooks.txt
git checkout new_release
grep cook . -R -l -I > new_release_cooks.txt
sort -o new_release_cooks_sorted.txt new_release_cooks.txt
vim -d old_release_cooks_sorted.txt new_release_cooks_sorted.txt

This meets all my requirements apart from point 4. It requires I do at least one checkout. Strictly I guess it doesn't create a list of files that differ, but the diff is close enough.

Is there a way to get this list without checking out?

Upvotes: 1

Views: 44

Answers (1)

Jay
Jay

Reputation: 3950

Checkout-free one-liner in bash that uses git grep and evaluates its results using comm:

word='cook' rel1='old_release' rel2='new_release'; comm -1 -2 <(git grep -l '\b'"$word"'\b' "$rel1" | sed 's/^[^:]*://' | sort;) <(git grep -L '\b'"$word"'\b' "$rel2" | sed 's/^[^:]*://' | sort;)

EDIT: adjustments so that also your condition 1. is satisfied (file must exist in both)

EDIT2: simplified expression

Upvotes: 1

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