I have a clone of a central repo at rev 2048. I want to remove the last 10 changesets on my local repo as if I was back in time two weeks ago. I suppose I could delete my local repo and do "hg clone -rev 2038"
but that would be long (cloning the repo takes several minutes). Is there a way to just "unpull" some changesets?
Notes:
- I'm not trying to backout the changesets. I'll eventually pull those changesets again from the central repo.
- I'm not trying to update the working directory to an earlier version; I really want to affect the repository.
- I don't have any outgoing changesets or pending modif开发者_运维技巧ications in my current repo and working directory.
Use the strip command:
hg strip -r 2039
This command is provided by the StripExtension. It is distributed as part of Mercurial 2.8 and later, but you do need to enable it first by adding the following lines to your .hgrc or Mercurial.ini:
[extensions]
strip =
Before Mercurial 2.8, it was part of the MqExtension.
To prevent you from accidentally destroying history, the command will generate a backup bundle in .hg/strip-backup/
which you can hg unbundle
again if desired.
Cloning your local repo should be fast. I assume "several minutes" refers to a remote repo?
You can use hg clone <local repo> <new repo> -r <revision>
to only clone up to a certain revision.
To remove a changeset that was already committed and pushed use :
hg backout -r (changeset number)
To remove a changeset that was committed but not pushed use :
hg strip -r (changeset number)
For versions previous to Mercurial 2.8, the Strip was part of the MqExtension.
In case you need to Enable the old MQ Extensions,
you can do it by adding this:
[extensions]
hgext.mq =
to your ~/.hgrc (or mercurial.ini) file.
The Strip information used to be here but now it can be found here.
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