我如何通过编程快速转发单个git commit？ [英] How can I fast-forward a single git commit, programmatically?

问题描述

我定期从git获取消息，如下所示：

 您的分支位于跟踪的远程分支'local-master / master'
提交3次，并且可以快进。 
  

我希望能够在可以执行以下操作的shell脚本中编写命令： p>

1. 如何判断当前分支是否可以从正在跟踪的远程分支进行快速转发？




2. 我怎么知道我的分支后面有多少提交？




3. 如何快速转发通过一个 提交，例如，我的本地分支将从后面3次提交到后面2次提交？

（对于那些有兴趣的人，我正在尝试组合一个高质量的git / darcs镜像。）

解决方案

如果当前提交是远程分支头的祖先，远程分支可以被快速转发到本地分支。换句话说，如果远程分支的一分支历史包含当前提交（因为如果它确实如此，那么确定新提交已经提交到当前提交上）。

因此，一个安全的方式来确定远程分支是否可以被快速转发：

 ＃转换参考提交ID的名称
 current_commit = $（git rev-parse HEAD）
 remote_commit = $（git rev -parse remote_name / remote_branch_name）
 
＃调用git log，只提交ID 
 log = $（git log --topo-order --format ='％H'$ remote_commit | grep $ current_commit）
 
＃检查当前提交的存在日志
 if [！ -z$ log] 
然后echo'远程分支可以快进！'
 fi 
  

请注意，没有使用--all参数（它会列出所有分支）调用git log，所以当前提交不可能在side branch上并且仍然打印在输出上。

当前提交之前的提交数量等于$ current_commit之前的$ log中的行数。

如果只想快速转发一个提交，则将当前提交的前一行（例如grep -B 1）取出，并将本地分支重置为此提交。

UPDATE ：您可以使用 git log commit1..commit2 来确定快速转发提交的数量：

  if [！ -z$ log] 
然后
＃打印当前提交之前的提交数
 ff_commits = $（git log --topo-order --format ='％H' \ 
 $ current_commit .. $ remote_commit | wc -l）
 echo快进提交的数量：$ ff_commits
 
＃快进只有一个提交
 if [$ ff_commits -gt 1] 
然后
 next_commit = $（git log --topo-order --format ='％H'\ 
 $ current_commit .. $ remote_commit | tail -1）
 git reset --hard $ next_commit 
 fi 
 fi 
  

当然，如果您将第一次调用的结果保存到文件中，您可以使用一次git日志调用完成此操作。

I periodically get message from git that look like this:

Your branch is behind the tracked remote branch 'local-master/master' 
by 3 commits, and can be fast-forwarded.

I would like to be able to write commands in a shell script that can do the following:

1. How can I tell if my current branch can be fast-forwarded from the remote branch it is tracking?

2. How can I tell how many commits "behind" my branch is?

3. How can I fast-forward by just one commit, so that for example, my local branch would go from "behind by 3 commits" to "behind by 2 commits"?

(For those who are interested, I am trying to put together a quality git/darcs mirror.)

解决方案

The remote branch can be fast-forwarded to the local branch if the current commit is the ancestor of the remote branch head. In other words, if the "one-branch history" of the remote branch contains the current commit (because if it does, it is sure that the new commits were committed "onto" the current commit)

So a safe way to determine whether the remote branch can be fast-forwarded:

# Convert reference names to commit IDs
current_commit=$(git rev-parse HEAD)
remote_commit=$(git rev-parse remote_name/remote_branch_name)

# Call git log so that it prints only commit IDs
log=$(git log --topo-order --format='%H' $remote_commit | grep $current_commit)

# Check the existence of the current commit in the log
if [ ! -z "$log" ]
  then echo 'Remote branch can be fast-forwarded!'
fi

Note that git log was called without the --all parameter (which would list all branches), so it is not possible that the current commit is on a "side branch" and is still printed on the output.

The number of commits ahead of the current commit equals the number of rows in $log before $current_commit.

If you want to fast-forward only one commit, you take the row previous to the current commit (with grep -B 1, for example), and reset the local branch to this commit.

UPDATE: you can use git log commit1..commit2 to determine the number of fast-forwarding commits:

if [ ! -z "$log" ]
then
  # print the number of commits ahead of the current commit
  ff_commits=$(git log --topo-order --format='%H' \
    $current_commit..$remote_commit | wc -l)
  echo "Number of fast-forwarding commits: $ff_commits"

  # fast-forward only one commit
  if [ $ff_commits -gt 1 ]
  then
    next_commit=$(git log --topo-order --format='%H' \
      $current_commit..$remote_commit | tail -1)
    git reset --hard $next_commit
  fi
fi

Of course, you can do this with one git log call if you save the result of the first call into a file.

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