Re: The --no-commit blues
Brian L. Troutwine <goofyheadedpunk <at> gmail.com> writes:
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>
> I've got two branches, one which I commit to quite frequently and another only
> periodically, call them inward and outward. Inward is where I do my work,
> outward tracks an SVN repository. I'd like to merge inward to outward without
> committing the merge so that I may provide a commit message appropriate for
> checking in to the SVN repo.
>
> `git merge --no-commit inward' from branch outward, I thought, should do it.
> Performing a `git status' and a `git log' directly afterward seem to indicate
> that the merge was committed. `git commit' insists the branch is now up to
> date.
>
If you want the "whole" set of changes from inward to appear in outward as a /single/ commit with an appropriate commit message, probably you want to use also the --squash option of git merge.
Otherwise, git will preserve the "structuring" of the history in many commits. This means that:
- if outward does not need any real merge (because it only needs to be advanced
to a newer status and not to join some changes of its own with changes from
inward), git merge will make a "fast-forward". In other words, you will see
outward becoming up to date without the need of any merge-commit at all. BTW
that is why --no-commit appears to have no effect in this case: there wouldn't
be any merge commit in any case.
- if outward needs a real merge, then git merge will set up everything up to the
point of making the merge commit. Then, depending on the presence of the
--no-commit option it will or will not make the actual commit.
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> Am I going about this the wrong way? What does --no-commit mean, if I am?
>
> (Also, I asked this on #git earlier. If anyone happens to idle there, pardon
> me for re-posting this so quickly. I've got terrible lag this morning, enough
> to time out frequently.)
