有没有办法获取 git commit 节点的子节点数？

Question

I'd like to find the nth node in the commit graph from where it has 2 or more children. So if the history was:

A -- B -- C -- D
      \
       E -- F*

And I am at F , I'd like to find B if I was searching for the 1st node with more than one child.

How would I accomplish this?

Answer 1

All of Git's internal linkages point backwards, from child to parent:

A <-B <-C <-D
     ↖︎            [this is an up-and-left arrow, but it may not work in some fonts]
      E <-F

That said, git rev-list has the ability to reverse the links while walking the graph. So you can walk the graph from F backwards, reversing each link so as to build a list of children. The problem here is that Git only sees the links it walks:

A-> B <-C <-D
     ↘︎            [this is a down-and-right arrow]
      E-> F

This means that even using the git rev-list option that does this ( --children ), Git will claim that B has one child, namely E . To get Git to claim that B has two children, E and C , you must start from both F and D so that git rev-list walks back from D to C to B :

A-> B-> C-> D
     ↘︎            [this is a down-and-right arrow]
      E-> F

You can get Git to start at all branch names using --branches , or all tag names using --tags , or all references (including things like refs/stash ) using --all . Add --topo-order to the git rev-list command and you'll get a list of commits and their children. From this list you will want the last commit that leads to F but also has multiple children. Unfortunately the output from git rev-list here is still not terribly useful. To do this job quickly, you might want to write your own graph reader, in which case you can do your own internal marking and can just use git rev-list --parents --all or similar.

(This job is much easier in Mercurial, although even there it may still be nontrivial.)