C. Michael Pilato wrote:
> Julian Foad <email@example.com> writes:
>>C. Michael Pilato wrote:
>>>Julian Foad <firstname.lastname@example.org> writes:
>>>>"svn status -u gone-dir" shows children of gone-dir as normal. They
>>>>should have "!" status. (gone-dir exists in repository but has been
>>>>deleted from the disk.)
>>>Closed, WONTFIX. If gone-dir is missing from disk, we have no idea
>>>what revision that directory was at, b) that it even had children, and
>>>c) what those children were. Without that knowledge, we can't presume
>>>that children we learn about from the server also happen to be
>>>children we used to have on disk but not do not (which is what
>>Er, well, in that case, how can it be correct that "svn status -u"
>>lists all those children that it should know nothing about? Should
>>it not either stop recursing or, if it does actually know about
>>them, then show them as missing?
> svn status -u isn't showing children of the missing thing. It's
> showing children of an object in the repository that happens to have
> the same name and children as the missing thing (in your particular
> test-case). I *know* that reads like a cop-out answer, but trust me
> -- I'm not trying to pull one over on you. Let me see if I can
> demonstrate the difference:
> So, what does 'svn status -u' really mean? It means, quite literally,
> show me the union of a) my working copy status and b) what I would
> see modified if I ran 'svn update' right now. And indeed, this is
> what I see. Running just 'svn status A/D/H' shows just this missing
> $ svn st A/D/H
> ! A/D/H
> And what would get touched if I updated right now? Well, let's see:
> $ svn up A/D/H
> A A/D/H
> A A/D/H/psi
> Updated to revision 2.
> As you can see, the union of these two operations is exactly what 'svn
> st -u A/D/H' shows.
> Does that make sense to you?
Yes, it does. Thank you for taking the time to write a detailed explanation.
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Received on Thu Oct 9 16:36:35 2003