On Fri, Sep 27, 2013 at 10:11 AM, John Keeping <john@xxxxxxxxxxxxx> wrote:
> 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".
yes that's what I'm doing but I was wondering if that's possible to do
that with only one run of git-log/git-rev-list.
Thanks
--
Francis
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