如何从分支获取所有提交并放入单个提交

Question

I have this

A-----1'--2'--3'--4'  (master)
 \
  \
   D-----1--2--3--4--5--6--7--8--9 (dev)

what i want get all chnages from D - 9 since diverge from master and then put all in one commit and make new branch dev2 which will have all chnages from dev branch

There are 30 commits in there

Answer 1

If you already created dev2 from master branch:

A-----1'--2'--3'--4'  (master, dev2)
 \
  \
   D-----1--2--3--4--5--6--7--8--9 (dev)

Then you can get all commits from dev branch into one commit to dev2 by:

git checkout dev2
git merge dev --squash

Now the commit history will be (commit M is the merge commit which contains all the changes from dev branch):

              (master)
                  |
A-----1'--2'--3'--4'---M (dev2)
 \
  \                          
   D-----1--2--3--4--5--6--7--8--9 (dev)

Answer 2

1.从主节点创建分支dev2。获取最新的dev 3.结帐dev2 4.将dev合并到dev2