Git：在准备真正的合并提交之后，如何创建一个简单的提交？ [英] Git: after preparing a real merge commit, how to create a simple commit?

问题描述

在调用 git merge --no-commit< commit> 之后，执行一次提交将导致一个合并提交更多）父母。而不必用 - squash 选项重新执行合并命令）来调用 创建简单提交 p>

解决方案

根据 git-merge man page ，--squash选项不记录 $ GIT_DIR / MERGE_HEAD 。 $ GIT_DIR / MERGE_HEAD 负责创建合并提交;你可以在Git源文件中看到这个文件 builtin / commit.c ：


in_merge = file_exists（git_path（MERGE_HEAD））;

...

if（in_merge）{

... //执行merge_commit







解决方案：执行正常合并后， <$> $ GIT_DIR / MERGE_HEAD 以避免获得合并提交。您也可以手动清理 $ GIT_DIR / MERGE_MSG 和$ GIT_DIR / MERGE_MODE，或者在成功提交时将此任务保留为Git。


After invoking git merge --no-commit <commit>, performing a commit will result in a merge commit with two (or more) parents. What command to invoke to create a simple commit instead (without having to re-perform the merge command with the --squash option)?

解决方案

According to the git-merge man page, the --squash option does not record $GIT_DIR/MERGE_HEAD. $GIT_DIR/MERGE_HEAD is responsible for creating merge commits; you can see this in the Git sources, file builtin/commit.c:

in_merge = file_exists(git_path("MERGE_HEAD"));
...
if (in_merge) {
... // Perform merge_commit
}

Solution: after having performed a normal merge, simply get rid of $GIT_DIR/MERGE_HEAD to avoid getting a merge commit. You may manually clean up $GIT_DIR/MERGE_MSG and $GIT_DIR/MERGE_MODE as well or leave this task up to Git upon successful commit.

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