Re: best way to fastforward all tracking branches after a fetch

On 12/11/2011 03:22 AM, Sitaram Chamarty wrote:

On Sat, Dec 10, 2011 at 01:26:32PM +0100, Gelonida N wrote:

quoted

Hi,

What is the best way to fastforward all fastforwardable tracking
branches after a git fetch?

I dont think there is a single command to do it for *all*
branches, but for any particular branch, this should work:

A tiny script is fine. I didn't really expect the magic command.

Currently I'm using one script per repository, which hard coded the
branches,
that I want to fast-forward (checking them out and doing a git-pull)

    git merge --ff-only @{u}

Thanks, '--ff=only @{u}'  is already the first improvement for my script.

So what you want would boil down to this script (untested):

    #!/bin/bash
    git status --porcelain -uno | grep . && {echo dirty tree, exiting...; exit 1; }

    for b in `git for-each-ref '--format=%(refname:short)' refs/heads`
    do
        git checkout $b
        git merge --ff-only @{u}
    done

Is there no way to distinguish tracking branches from other branches?
without checking them out?

In order to save time I'd like to avoid checking out local branches.

Ideally I would even like to avoid checking out branches, which don't
need to be forwarded.


I also had to remember on which branch I was in order to avoid, that I
am at a random branch after running the script.

I could imagine something like my snippet below , though I guess,
there's something more elegant.

git stash
mybranch=`git branch | sed -n 's/\* *//p'`
# do_script . . .
git checkout $mybranch
git stash apply