Go back N commits in Git to find commit that causes test regressions

Question

Is there a command that will let me checkout a commit based on its distance from the current commit instead of using commit IDs?

Use Case

Basically I am thinking of setting up a cron job type script to do the following on a build server:

• Pull down the latest of a specific git branch (git pull dev).
• Build it, run the tests.
• If the pass percentage is lower than the last stored percentage:
  • Recursively go back a commit, build, run tests, until it finds the commit where the percentage changed.
  • Log the commits that introduced regressions.

I have a rough idea for how this would hinge together but it won't work unless I can go back one commit periodically.

If there is no specific command I suppose I could grep the commit log and take the first one each time?

I appreciate any ideas or help!

Unlike: How to undo last commit(s) in Git?

I want to go back "N" number of commits.

Answer 1

git checkout HEAD~N

will checkout the Nth commit before your current commit.

If you have merge commits in your history, you may be interested in

git checkout HEAD^N

which will checkout the Nth parent of a merge commit (most merge commits have 2 parents).

So, to always go back one commit following the first parent of any merge commits:

git checkout HEAD^1

You may also be interested in git bisect - Find by binary search the change that introduced a bug.