On Fri, Sep 27, 2013 at 07:09:03AM +0200, Francis Moreau wrote:
> Hi,
>
> On Thu, Sep 26, 2013 at 10:21 PM, John Keeping <john@xxxxxxxxxxxxx> wrote:
> > On Thu, Sep 26, 2013 at 06:35:57PM +0200, Francis Moreau wrote:
> >> I'm trying to use "git log --cherry ..." in order to display new, kept
> >> and removed commits between two branches A and B.
> >>
> >> So commits which are only in B are considered new and should be marked
> >> with '+'. Commits which are in both branches are marked with '=' but
> >> only commit in branch B are shown. Eventually commits which are in A
> >> but not in B anymore should be marked with '-'.
> >>
> >> So far I found this solution:
> >>
> >> $ git log --cherry-mark --right-only A...B
> >> $ git log --cherry-pick --left-only A...B
> >>
> >> but I have to call twice git-log. This can be annoying since depending
> >> on A and B, calling git-log can take time.
> >>
> >> Is there another option that I'm missing which would do the job but
> >> with only one call to git-log ?
> >
> > Does this do what you want?
> >
> > git log --cherry-mark --left-right A...B |
> > sed -e '/^commit / {
> > y/<>/-+/
> > }'
>
> Nope because --left-right shows common commits (with '=' mark) that
> belong to A *and* B, and I'd like to have only the ones in B.
I think the only way you can address this is to post-process the result,
I don't know any way to remove a left side commit only if it is
patch-identical to a right side commit.
It should be relatively easy to filter out any '=' commits that are in
the output of "git rev-list --left-only A...B".
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