From mboxrd@z Thu Jan  1 00:00:00 1970
From: Johan Herland <johan@herland.net>
Subject: Re: Find successor of common ancestor
Date: Fri, 16 Jul 2010 14:19:39 +0200
Message-ID: <201007161419.39469.johan@herland.net>
References: <20100716082937.GA22894@nibiru.local> <AANLkTintIToGZu3fvK4aE5LFL7MgOnlfwKgsZ79Q0-o1@mail.gmail.com> <20100716090613.GB22894@nibiru.local>
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To: weigelt@metux.de
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On Friday 16 July 2010, Enrico Weigelt wrote:
> * Santi B=E9jar <santi@agolina.net> wrote:
> > On Fri, Jul 16, 2010 at 10:29 AM, Enrico Weigelt <weigelt@metux.de>=
=20
wrote:
> > > Hi folks,
> > >
> > >
> > > suppose the following situation:
> > >
> > > I've forked some branch A into B, now A and B have evolved
> > > independently for quite some time into A' and B'. Now I'd like
> > > to rebase B' along A' history line step by step - first on A+1,
> > > then A+3, ... until A' (that's what I'd call zip-rebase).
> >
> > This is just what "git rebase"  does:
> >
> > $ git checkout B'
> > $ git rebase A'
>
> Yes, but I need to find out A+1 for my zip-rebase.

IINM you have the following situation

A0---A1---A2---A3---A4  <-- A'
  \
   B1---B2---B3--B4  <-- B'

=2E..and you want to rebase B' (i.e. B1 through B4), first onto A1, the=
n=20
onto A2, the onto A3, and finally onto A4. I have no idea WHY you would=
=20
want to do this (rebasing directly onto A4 (like Santi suggests) is=20
much cheaper), but you can certainly coax Git into doing it, anyway.

To list commits A1, A2, A3, A4 (in that order), do:

  git rev-list --reverse B'..A'

You can now loop over the results like this:

  git checkout B'
  for a in $(git rev-list --reverse ..A'); do
      git rebase $a
  done


Hope this helps,

=2E..Johan

--=20
Johan Herland, <johan@herland.net>
www.herland.net