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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?

1 Answer

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by (51.2k points)

If you are using a fast-forward merge, then you can simply use

git fetch <remote> <sourceBranch>:<destinationBranch>


# Merge local branch foo into local branch master,

# without having to checkout master first.

# Here `.` means to use the local repository as the "remote":

git fetch . foo:master

# Merge remote branch origin/foo into local branch foo,

# without having to checkout foo first:

git fetch origin foo:foo

And also, you can work on this using git fetch. 

This way is a little safer than just force-moving the branch reference since git fetch will automatically prevent accidental non-fast-forwards as long as you don't use + in the refspec.

Sometimes, you cannot merge a branch B into branch A without checking out A first if it would result in a non-fast-forward merge. 

This is because a working copy is needed to resolve any potential conflicts.

To do this without checking out a branch first, you can use 

git fetch with a refspec.

git fetch upstream master:master

You'll probably want to make an alias for it in your git configuration file, like this one:


    sync = !sh -c 'git checkout --quiet HEAD; git fetch upstream master:master; git checkout --quiet -'

This will help you to push the changes without any checkouts.

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