On Wednesday 08 October 2003 23:54, C. Michael Pilato wrote:
[...]
> ### Now, oops! I removed A/D/H (at revision 1, remember) from disk
> $ rm -rf A/D/H
>
> ### Now I run status.
> $ svn st -u A/D/H
> * A/D/H/psi
> ! * ? A/D/H
> Status against revision: 2
>
> Now, the first thing to notice is that only H is reported as missing.
> Why? Because this is the only thing we actually know about that is
> missing. There is *no record* in the working copy that we had H at
> revision 1 before its untimely demise. It could have been at any
> revision. If this display showed chi and omega as missing, well that
> would just be wrong, because they weren't supposed to be in the
> working copy.
Sorry for jumping in, but I saw something that I didn't understand. You
removed the A/D/H directory completely... so why does your 'svn st -u A/D/H'
command show A/D/H/psi as *not* missing?
> 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
> A/D/H:
>
> $ 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.
Here the update command added both A/D/H and A/D/H/psi. I would've expected
that either the 'svn st -u' command would've shown both as missing or only
A/D/H as missing and not even mention A/D/H/psi. Considering the fact that
you said it was the union of a) and b), I would expect both to be listed as
missing.
Am I missing something here too? :-)
-John
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Received on Thu Oct 9 10:44:19 2003