返回Git中的N次提交以查找导致测试回归的提交 [英] Go back N commits in Git to find commit that causes test regressions

问题描述

是否有一条命令可以让我根据其与当前提交的距离而不是使用提交ID来检出提交?

Is there a command that will let me checkout a commit based on its distance from the current commit instead of using commit IDs?

基本上，我正在考虑设置cron作业类型脚本以在构建服务器上执行以下操作:

Basically I am thinking of setting up a cron job type script to do the following on a build server:

• 下拉特定git分支的最新版本(git pull dev).
• 构建它，运行测试.
• 如果通过百分比低于上次存储的百分比:
  • 递归地返回一次提交，构建，运行测试，直到找到百分比发生变化的提交.
  • 记录引入回归的提交.
  • Pull down the latest of a specific git branch (git pull dev).
  • Build it, run the tests.
  • If the pass percentage is lower than the last stored percentage:
    • Recursively go back a commit, build, run tests, until it finds the commit where the percentage changed.
    • Log the commits that introduced regressions.

    对于如何将它们铰接在一起，我有一个粗略的想法，但是除非我可以定期回溯一次提交，否则它将无法正常工作.

    I have a rough idea for how this would hinge together but it won't work unless I can go back one commit periodically.

    如果没有特定的命令，我想我可以grep提交日志并每次获取第一个?

    If there is no specific command I suppose I could grep the commit log and take the first one each time?

    我感谢任何想法或帮助！

    I appreciate any ideas or help!

    与以下不同: 如何撤消Git中的最后一次提交?

    我想返回"N"次提交.

    I want to go back "N" number of commits.

    推荐答案

    git checkout HEAD~N
    

    将在当前提交之前检出第N个提交.

    will checkout the Nth commit before your current commit.

    如果您的历史记录中有合并提交，那么您可能会感兴趣

    If you have merge commits in your history, you may be interested in

    git checkout HEAD^N
    

    这将检出合并提交的第N个父级(大多数合并提交有2个父级).

    which will checkout the Nth parent of a merge commit (most merge commits have 2 parents).

    因此，要始终在所有合并提交的第一个父级之后返回一个提交:

    So, to always go back one commit following the first parent of any merge commits:

    git checkout HEAD^1
    

    您可能还对git bisect感兴趣-通过二进制搜索查找引入错误的更改.

    You may also be interested in git bisect - Find by binary search the change that introduced a bug.

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