我有一个Git存储库,看起来像这样:

A <- B <- C <- D <- HEAD

我希望分支的头部指向A,即,我希望B、C、D和head消失,我希望头部与A同义。

听起来我可以尝试重新设置基础(不适用,因为我已经在两者之间进行了更改),也可以恢复。但如何恢复多次提交?我一次回复一个吗?订单重要吗?


当前回答

与Jakub的回答类似,这允许您轻松选择要还原的连续提交。

# Revert all commits from and including B to HEAD, inclusively
git revert --no-commit B^..HEAD
git commit -m 'message'

其他回答

如果你

合并提交和您无法恢复,并且你不介意粉碎你要还原的历史,

那么你可以

git reset --soft HEAD~(number of commits you'd like to revert)
git commit -m "The stuff you didn't like."
git log
# copy the hash of your last commit
git revert <hash of your last (squashed) commit>

然后,当您想要推送更改时,请记住使用-f标志,因为您修改了历史记录

git push <your fork> <your branch> -f

为此,您只需使用revert命令,指定要恢复的提交范围。

考虑到您的示例,您必须这样做(假设您在分支“master”上):

git revert master~3..master

或git revert B…D或git return D C B

这将使用B、C和D的反向提交在本地创建一个新的提交(这意味着它将撤消这些提交带来的更改):

A <- B <- C <- D <- BCD' <- HEAD

如果您想暂时恢复功能的提交,那么可以使用以下一系列命令。

下面是它的工作原理

git log --pretty=oneline | grep 'feature_name' | cut -d ' ' -f1 | xargs -n1 git revert --no-edit

我发现自己需要恢复一系列的提交,然后重新恢复它们,以帮助团队提出明确的拉取请求,而不必强制推送他们的目标分支(直接提交给)

# checkout the branch that should be targeted
git checkout $branch_target

# revert the commits in $branch_target to some $count where
#   $count is the number of commits to revert
#   cut is used to slice just the commit hash field from each line of output
#   xargs runs the command once for each line of input, reversing the commits!
git log --oneline -n $count | cut -d' ' -f1 | xargs git revert

# check out the branch which should be the source of the pull request
git checkout -b $branch_for_pull

# revert the revert commits
# $count is that same number of commits being reverted (again)
git log --oneline -n $count | cut -d' ' -f1 | xargs git revert

# push branches up and go off to create PR in whatever web UI
git push --set-upstream origin $branch_for_pull  # it's new!
git checkout $branch_target
git push  # if this branch wasn't pushed, just fix the issue locally instead..

因为这会以相反的顺序将所有提交从HEAD还原为git-log-n$count,所以它可以很好地、干净地处理任何数量的提交

在此状态下从$branch_target查看

% git log --oneline origin/$branch_target
ffff006 (origin/$branch_target, $branch_target) Revert "first commit"
ffff005 Revert "second commit"
ffff004 Revert "third commit"
ffff003 third commit
ffff002 second commit
ffff001 first commit

在此状态下从$branch_for_pull查看

% git log --oneline origin/$branch_for_pull
ffff009 (origin/$branch_for_pull, $branch_for_pull) Revert "Revert "third commit""
ffff008 Revert "Revert "second commit""
ffff007 Revert "Revert "first commit""
ffff006 (origin/$branch_target, $branch_target) Revert "first commit"
ffff005 Revert "second commit"
ffff004 Revert "third commit"
ffff003 third commit
ffff002 second commit
ffff001 first commit

如果意图是用变更集创建N个分支,但它们都被提交到同一个分支,您仍然可以将它们全部还原为基本提交,然后只还原所需的还原,因为变更集应该按逻辑排序(试着说5倍快)

使用像HEAD~7..HEAD~5这样的语法可能有助于描述范围以精确地分割还原-还原分支

在这里,当恢复最后7次提交(gitlog-n7),但在一个分支中恢复5次(gitlog-n 5),然后在另一个gitlogHEAD~12..HEAD~10(12是7次提交+5次提交,假设新的PR分支基于“之前”分支,或FF(未压缩)的结果,将分支“之前”合并到原始目标分支中时,这是有意义的

我真的想避免硬重置,这就是我想到的。

A -> B -> C -> D -> HEAD

返回A(后退4步):

git pull                  # Get latest changes
git reset --soft HEAD~4   # Set back 4 steps
git stash                 # Stash the reset
git pull                  # Go back to head
git stash pop             # Pop the reset 
git commit -m "Revert"    # Commit the changes