Determining whether you have a commit locally, in a partial clone?
- From
Tao Klerks <tao@klerks.biz>
- Date
- Jun 20, 2023, 11:26 UTC
- Message-ID
- <CAPMMpohiTzANyhzL-mS-gg2kzbOEOiDktNbdwEXBKy9uL0-JgA@mail.gmail.com>
Hi folks,
I "discovered" today that when you're in a partial clone, naive tests to check for whether you have a commit locally no longer work - they fetch the commit on-demand:
git cat-file -t SOME_HASH_NOT_IN_REFSPEC git rev-list SOME_HASH_NOT_IN_REFSPEC
I didn't realize this until today: even commits can be "filtered out" by partial clone, so any reference to a commit that is not found locally must be resolved transparently via jit-fetch.
I'm optimizing some stuff for users, so I need to know whether a given commit exists locally or not... but I can't seem to figure out how!
I tried using "git rev-list"'s "--exclude-promisor-objects" option, but I guess I don't understand what that's supposed to do. In my case it just made a simple check like "git rev-list --exclude-promisor-objects SOME_HASH_NOT_IN_REFSPEC" take forever (10 mins and counting).
I confirmed that removing (commenting out) the "remote.origin.promisor" and "remote.origin.partialclonefilter" config keys achieves my objective, but I can't figure out how to do it safely; "-c remote.origin.promisor=false -c remote.origin.partialclonefilter=" does *not* seem to work. The existence of a "remote.origin.partialclonefilter" value, even if it is empty, appears to override the "remote.origin.promisor=false" setting.
As far as I can tell, config values cannot be unset with "-c" - in fact I see that credential.helper was granted special support for empty string as a way of signalling "no credential helper" by Jeff King in 2016.
So I guess I have two questions: * Is there any way to run a single git command in a "don't use promisors" context? * Is the fact that "-c remote.origin.partialclonefilter=" doesn't work for temporarily unsetting the filter a bug/issue to be resolved?
Thanks, Tao