Re: Some advanced index playing

From: Linus Torvalds <torvalds@osdl.org>
Date: Dec 3, 2006, 20:40 UTC
Message-ID: <Pine.LNX.4.64.0612031236560.3476@woody.osdl.org>
In-Reply-To: <200612032011.25922.alan@chandlerfamily.org.uk>

On Sun, 3 Dec 2006, Alan Chandler wrote:

Show 9 quoted lines

> >
> > 	git ls-tree HEAD -- that-file | git update-index --index-info
> > 	git commit that-file
> 
> I don't quite understand this - maybe it should be
> 
> git ls-tree HEAD -- that-file | git update-index --index-info
> git commit
> git commit -a

Sure. It depends on which file you want to commit first.

If you want to commit "that-file" first, you do my sequence.

If you want to commit everything _but_ "that-file", you do the second sequence (which basically removes the changes to "that-file" from the index, then commits the index, and then with "git commit -a" commits the remaining dirty state, which is obviously those changes to "that-file" that you still had in the working tree).

> either I want to ONLY commit that file at the working tree state (and index 
> before these commands), or I want to commit ALL except this file (so I can 
> later come and commit just that file)

Right. If you do

	git ls-tree HEAD -- that-file | git update-index --index-info
	git commit that-file

you basically ONLY commit "that-file". You first reset it (in the index) to the old state, but that's just so that "git commit that-file" will now happily commit the current state (in the working tree) of "that-file".

So "git commit that-file" will basically _ignore_ your current index. Because you told "git commit" (by naming "that-file") that you _only_ wanted to commit "that-file". So whatever state you had in your current index doesn't matter at all - it will only look at the HEAD tree _and_ that single file that you specified.