List all local branches without a remote

Question

Problem: I want a way of deleting all the local branches I have that do not have a remote. It's easy enough to pipe the names of the branches into a git branch -D {branch_name}, but how do I get that list in the first place?

For example:

I create a new branch without a remote:

$ git co -b no_upstream

I list all my branches, and there's only one with a remote

$ git branch -a
master
* no_upstream
remotes/origin/HEAD -> origin/master
remotes/origin/master

What command can I run to get no_upstream as an answer?

I can run git rev-parse --abbrev-ref --symbolic-full-name @{u} and that will show that it has no remote:

$ git rev-parse --abbrev-ref --symbolic-full-name @{u}
error: No upstream configured for branch 'no_upstream'
error: No upstream configured for branch 'no_upstream'
fatal: ambiguous argument '@{u}': unknown revision or path not in the working tree.
Use '--' to separate paths from revisions, like this:
'git <command> [<revision>...] -- [<file>...]'

But as this is an error, it won't let me use it or pipe it to other commands. I'm intending to use this as either a shell script alias'd to git-delete-unbranched or maybe make a super simple Gem like git-branch-delete-orphans

Answer 1

I recommend using git branch --format to specify the exact output you want from the git branch command. By doing that, you can pull out just the refname and the upstream, like this:

git branch --format "%(refname:short) %(upstream)"

It outputs the branches along with the remote branches if they exist, in the following format:

25-timeout-error-with-many-targets
31-target-suggestions refs/remotes/origin/31-target-suggestions
54-feedback-link refs/remotes/origin/54-feedback-link
65-digest-double-publish

Once you have this nicely formatted output, it's as easy as piping it through awk to get your list:

git branch --format "%(refname:short) %(upstream)" | awk '{if (!$2) print $1;}'

Results in the following output:

25-timeout-error-with-many-targets
65-digest-double-publish

The awk portion prints the first column if there is no second column.

Bonus: Create an alias

Make it easy to run by creating an alias in your global .gitconfig file (or wherever):

[alias]
  local-branches = "!git branch --format '%(refname:short) %(upstream:short)' | awk '{if (!$2) print $1;}'"

Bonus: Remote Filtering

If for some reason you have multiple tracking remotes for different branches, it's easy enough to specify which remote you want to check against. Just add the remote name to the awk pattern. In my case, it's origin so I can do this:

git branch --format "%(refname:short) %(upstream)" | awk '$2 !~/\/origin\// { print $1 }'

Important: The backslash needs to be escaped in the alias or else you will have an invalid gitconfig file.

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Previous Answer

The previous answer was functionally similar, but used the following as it's starting point. Over time, commenters have pointed out that a regex is unreliable due to the variance possible in a commit message, so I no longer recommend this method. But, here it is for reference:

I recently discovered git branch -vv which is the "very verbose" version of the git branch command.

It outputs the branches along with the remote branches if they exist, in the following format:

  25-timeout-error-with-many-targets    206a5fa WIP: batch insert
  31-target-suggestions                 f5bdce6 [origin/31-target-suggestions] Create target suggestion for team and list on save
* 54-feedback-link                      b98e97c [origin/54-feedback-link] WIP: Feedback link in mail
  65-digest-double-publish              2de4150 WIP: publishing-state

Once you have this nicely formatted output, it's as easy as piping it through cut and awk to get your list:

git branch -vv | cut -c 3- | awk '$3 !~/\[/ { print $1 }'