git: update a local branch without checking it out?

Question

I work on a project that has 2 branches, A and B. I typically work on branch A, and merge stuff from branch B. For the merging, I would typically do:

git merge origin/branchB

However, I would also like to keep a local copy of branch B, as I may occasionally check out the branch without first merging with my branch A. For this, I would do:

git checkout branchB
git pull
git checkout branchA

Is there a way to do the above in one command, and without having to switch branch back and forth? Should I be using git update-ref for that? How?

Answer 1

You can only do this if the merge is a fast-forward. If it's not, then git needs to have the files checked out so it can merge them!

To do it for a fast-forward only:

 git fetch <branch that would be pulled for branchB>
 git update-ref -m "merge <commit>: Fast forward" refs/heads/<branch> <commit>

where <commit> is the fetched commit, the one you want to fast-forward to. This is basically like using git branch -f to move the branch, except it also records it in the reflog as if you actually did the merge. Please, please, please don't do this for something that's not a fast-forward, or you'll just be resetting your branch to the other commit. (To check, see if git merge-base <branch> <commit> gives the branch's SHA1.)