I want to remove multiple stashed states at a time.
But "git stash drop <stash>" removes only one stashed state at a time and "git stash clear" remove all.
Can I do that?
threads / discuss / 46962
Subject: Can I remove multiple stashed states at a time?
I want to remove multiple stashed states at a time.
But "git stash drop <stash>" removes only one stashed state at a time and "git stash clear" remove all.
Can I do that?
Re: Can I remove multiple stashed states at a time?
On Fri, Oct 13, 2017 at 11:58:12AM +0900, 小川恭史 wrote:
> I want to remove multiple stashed states at a time. > > But "git stash drop <stash>" removes only one stashed state at a time > and "git stash clear" remove all. > > Can I do that?
There isn't a way to do it through "git stash", I don't think. The stash feature is backed by a reflog, and the underlying machinery is capable of it. E.g.:
git reflog delete --updateref --rewrite refs/stash@{1} refs/stash@{2} refs/stash@{3}works. But that's getting a bit more familiar with the innards of stash than users should be (both because they shouldn't need to, and because the underlying storage may be subject to change).
Probably "git stash drop" should learn to take multiple stash arguments and pass them all along to "reflog delete".
You can also just do:
for i in 1 2 3; do
git stash drop $i
doneof course. It's less efficient than a single invocation (since it has to rewrite the whole reflog each time), but unless you have thousands of stashes, I doubt the difference is all that noticeable.
-Peff
Re: Can I remove multiple stashed states at a time?
On Fri, Oct 13, 2017 at 01:35:22PM -0400, Jeff King wrote:
> On Fri, Oct 13, 2017 at 11:58:12AM +0900, 小川恭史 wrote: > > You can also just do: > > for i in 1 2 3; do > git stash drop $i > done >
Doesn't stash 2 become stash 1 after the first drop, and the same for 3, resulting in dropping stash 1, 3 and 5?
So something like
for i in 3 2 1; do
git stash drop $i;
doneOr leave $i out altogether.
Re: Can I remove multiple stashed states at a time?
On Sat, Oct 14, 2017 at 05:01:23PM +0200, Kevin Daudt wrote:
> On Fri, Oct 13, 2017 at 01:35:22PM -0400, Jeff King wrote: > > On Fri, Oct 13, 2017 at 11:58:12AM +0900, 小川恭史 wrote: > > > > You can also just do: > > > > for i in 1 2 3; do > > git stash drop $i > > done > > > > Doesn't stash 2 become stash 1 after the first drop, and the same for 3, > resulting in dropping stash 1, 3 and 5? > > So something like > > for i in 3 2 1; do > git stash drop $i; > done > > Or leave $i out altogether.
Oops, you're right.
For that matter, I didn't double check that a single "reflog delete" handles this case correctly. That should be easy to test, though:
git init echo base >file git add file git commit -m base
for i in $(seq 10); do
echo $i >file
git stash push -m $i
done git reflog delete --updateref --rewrite refs/stash@{1} refs/stash@{2} refs/stash@{3}That seems to leave:
$ git stash list
stash@{0}: On master: 10
stash@{1}: On master: 8
stash@{2}: On master: 6
stash@{3}: On master: 4
stash@{4}: On master: 3
stash@{5}: On master: 2
stash@{6}: On master: 1which is not what we wanted (I'd guess also that it rewrites the reflog 3 times, negating any efficiency I claimed earlier).
-Peff