Re: How to checkout a revision that contains a deleted submodule?

> Le 20 sept. 2020 à 05:44, Luke Diamand <luke@diamand.org> a écrit :
> 
> On Sat, 19 Sep 2020 at 10:03, Luke Diamand <luke@diamand.org> wrote:
>> 
>> Maybe this is a FAQ, but I couldn't figure it out!
>> 
>> I have a repo which has a couple of submodules.
>> 
>> At some point in the past I deleted one of those submodules:
>> 
>>    git rm sub2
>>    git add -u
>>    git commit -m 'Deleting sub2'
>>    git push origin
>>    ...
>>    ... more commits and pushes...
>> 
>> Now I go and clone the head revision. This gives me a clone which has
>> nothing present in .git/modules/sub2.
>>    login on some other machine
>>    git clone git@my.repo:thing
>>    cd thing
>>    ls .git/modules
>>    <sub2 not present>
>> 
>> So when I go and checkout an old revision where sub2 is still around I get:
>>    git checkout oldrevision
>>    fatal: not a git repository: sub2/../.git/modules/sub2
>> 
>> What am I doing wrong?
>> What set of commands do I need to use to ensure that this will always
>> do the right thing?
>> 
>> Thanks
>> Luke
> 
> Replying to myself, adding Jens who added the section below.
> 
> This is a known bug:
> 
> https://git-scm.com/docs/git-rm
> 
>> BUGS
>> ----
>> Each time a superproject update removes a populated submodule
>> (e.g. when switching between commits before and after the removal) a
>> stale submodule checkout will remain in the old location. Removing the
>> old directory is only safe when it uses a gitfile, as otherwise the
>> history of the submodule will be deleted too. This step will be
>> obsolete when recursive submodule update has been implemented.
> 
> I'm wondering what "recursive submodule update" is. If I do:
> 
>    git submodule update --checkout --force --remote --recursive
> 
> then those stale repos are still left lying around. I guess that's a
> different kind of recursive?
>