Push new local branch to remote using Gitpython

Question

I looked at a few references but I am still having problems:

I want to clone a remote repo, create a new branch, and push the new branch back to remote using GitPython.

This seems to work:

import git
import subprocess

nm_brnch = 'new_branch'

# Clone    
repo_url = r'my_remote.git'
repo = git.Repo.clone_from(repo_url, dnm_wrk, branch=r'some_branch')

# Create new branch
git = repo.git
git.checkout('HEAD', b=nm_brnch)

# Push new branch to remote
subprocess.call(f'git push -u origin {nm_brnch}')

But it's ugly, since it uses subprocess, instead of using GitPython.

I tried using GitPython, but without success:

repo.head.set_reference(nm_brnch)
repo.git.push("origin", nm_brnch)

I have consulted the following references:

score 3 · Answer 1 · answered Jan 15 '19 at 02:11

I'm using gitpython==2.1.11 with Python 3.7. Below is my push function in which I first try a high-level push, and then a low-level push as necessary. Note how I check the return value of either command. I also log the push actions, and this explains what's happening at every step.

class GitCommandError(Exception):
    pass

class Git:
    def _commit_and_push_repo(self) -> None:
        repo = self._repo
        remote = repo.remote()
        remote_name = remote.name
        branch_name = repo.active_branch.name

        # Note: repo.index.entries was observed to also include unpushed files in addition to uncommitted files.
        log.debug('Committing repository index in active branch "%s".', branch_name)
        self._repo.index.commit('')
        log.info('Committed repository index in active branch "%s".', branch_name)

        def _is_pushed(push_info: git.remote.PushInfo) -> bool:
            valid_flags = {push_info.FAST_FORWARD, push_info.NEW_HEAD}  # UP_TO_DATE flag is intentionally skipped.
            return push_info.flags in valid_flags  # This check can require the use of & instead.

        push_desc = f'active branch "{branch_name}" to repository remote "{remote_name}"'
        log.debug('Pushing %s.', push_desc)
        try:
            push_info = remote.push()[0]
        except git.exc.GitCommandError:  # Could be due to no upstream branch.
            log.warning('Failed to push %s. This could be due to no matching upstream branch.', push_desc)
            log.info('Reattempting to push %s using a lower-level command which also sets upstream branch.', push_desc)
            push_output = repo.git.push('--set-upstream', remote_name, branch_name)
            log.info('Push output was: %s', push_output)
            expected_msg = f"Branch '{branch_name}' set up to track remote branch '{branch_name}' from '{remote_name}'."
            if push_output != expected_msg:
                raise RepoPushError(f'Failed to push {push_desc}.')
        else:
            is_pushed = _is_pushed(push_info)
            logger = log.debug if is_pushed else log.warning
            logger('Push flags were %s and message was "%s".', push_info.flags, push_info.summary.strip())
            if not is_pushed:
                log.warning('Failed first attempt at pushing %s. A pull will be performed.', push_desc)
                self._pull_repo()
                log.info('Reattempting to push %s.', push_desc)
                push_info = remote.push()[0]
                is_pushed = _is_pushed(push_info)
                logger = log.debug if is_pushed else log.error
                logger('Push flags were %s and message was "%s".', push_info.flags, push_info.summary.strip())
                if not is_pushed:
                    raise RepoPushError(f'Failed to push {push_desc} despite a pull.')
        log.info('Pushed %s.', push_desc)

The key line of this code `repo.git.push('--set-upstream', remote_name, branch_name)` doesn't really use the `gitpython` API. It's just directly calling `git`. Is there no way to use the API to get a better formatted return value? — Seanny123, Feb 24 '20 at 21:57
Is there a reason to not set the upstream in the first place? If setting it twice doesn't change the affect, might as well do that and eliminate the try-except — Addison Klinke, Mar 10 '21 at 19:22
@Seanny123 The GitPython API seems to be slightly confusing on this point. AFAICT, the trick is to use the `refspec` argument in `remote.push()`: https://stackoverflow.com/a/66820830/9219425 . This answer has a bit more detail and links to a GitHub issue where this is discussed with the package author: https://stackoverflow.com/a/61034176/9219425 — Tim, Jan 01 '23 at 18:07

score 1 · Answer 2 · answered Feb 28 '20 at 17:46

1

You have to define a remote repo, then push to it. e.g.

origin = repo.remote(name='origin')
origin.push()

See the Handling Remotes documentation for more examples of push/pull

answered Feb 28 '20 at 17:46

Fraser

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score 0 · Answer 3 · answered Jun 04 '20 at 21:04

Expanding on @Fraser's answer, here is the full code I used to successfully create a new branch:

from pathlib import Path

# initialize repo and remote origin
repo_path = Path("~/git/sandboxes/git-sandbox").expanduser()
repo = git.Repo(repo_path)
origin = repo.remote(name="origin")

# create new head and get it tracked in the origin
repo.head.reference = repo.create_head(branch_name)
repo.head.reference.set_tracking_branch(origin.refs.master).checkout()

# create a file for the purposes of this example
touch[f"{repo_path}/tmp1.txt"] & plumbum.FG

# stage the changed file and commit it
repo.index.add("tmp1.txt")
repo.index.commit("mod tmp1.txt")

# push the staged commits
push_res = origin.push(branch_name)[0]
print(push_res.summary)

score 0 · Answer 4 · answered Oct 28 '21 at 16:11

Assuming it's the push that this is failing on in GitPython (as it was for me), just using GitPython I was able to solve this problem like this:

import git
repo = git.Repo('<your repo path>')
repo.git.checkout('HEAD', b=<your branch name>)
# -u fixed it for me
repo.git.push('origin', '-u', branch_name)

Push new local branch to remote using Gitpython

4 Answers4