对于你想要的行为，我想你必须从 Master 推送， t知道如何从 Slave 中引入子树更改。

推送更改：

 （在Master中）
 $ git subtree split --prefix = lib / slave -b split-branch 
 $ git push&lt ;裸机回购> split-branch：master 
 $ git branch -d split-branch 
 $ cd / path / to / Slave / working / copy 
 $ git pull＃（现在在Slave中）
 

第一个命令根据目录创建一个新的子树，然后将其推送到另一个项目。

从理论上讲，您应该能够直接推送到工作副本，但在实践中，这对我不起作用。

The short of what's below: I want to push commit messages to a subtree, but only the messages that apply to that tree. How do I do this?

---

I have two projects, Master and Slave. Slave is checked out as a subtree of Master checked out to lib/slave via git subtree merge --prefix=lib/slave --squash projects/slave where projects/slave is a branch that Slave is checked out into:

Branches:

master projects/slave → slave

Now I'm working on my master branch, making commits to files that are part of both projects, everything is going smoothly. Now I want to push changes back to Slave:

• git checkout slave
• git merge ??? master

If I do a normal merge I get commits for every commit to master, whether or not any files in lib/slave were modified. Or I can do a --squash and only get a single commit, but I lose the log messages.

So how do I get the appropriate log messages? e.g. if my master log history is:

• added images to master
• modified files in slave only
• more changes to master only
• modified files in master and slave

I'd want this added to Slave:

• modified files in slave only
• modified files in master and slave

For the behavior you want, I think you'll have to push from Master, I don't know of a way to pull subtree changes from Slave.

To push changes:

(in Master)
$ git subtree split --prefix=lib/slave -b split-branch
$ git push <bare Slave repo> split-branch:master
$ git branch -d split-branch
$ cd /path/to/Slave/working/copy
$ git pull  # (now in Slave)

The first command creates a new subtree based on the directory, which is then pushed to the other project.

In theory, you should be able to push directly to a working copy, but in practice that didn't work for me.