Git如何拉取某个分支的某段提交

Question

现在有两个分支，分别为分支A、分支B。每个分支上都有若干次提交；
分支A的提交commits有 a1, a2, a3, a4, a5，一共有五次提交；
分支B的提交commits有 b1, b2, b3, b4, b5，一共有五次提交；
当前位于分支A。

注：以上10次提交的hash值不一样。

请问：怎么只拉取分支B的 b2, b3, b4 提交节点到 分支A？

要求，在拉取到分支A后，必须保留原来分支B上的提交信息。

求破 ╮(￣▽ ￣)╭ ~

@junnplus 说的cherry-pick方法，是对的，而且我也会；但是有没有更好的回答呢？

Answer 1

On branch A, execute

git cherry-pick <commit_id>

Answer 2

If automation is not required, the most convenient way is this:

$ git rebase -i HEAD $name_of_branch_b

The editor will display five lines from b1 to b5, delete the two lines b1 and b5, save and exit.

Answer 3

It may be difficult to have a better answer, right? According to the git-flow的原则，dev-feature一般不会有只合并中间几次提交到devdivided situation, maybe the subject's need is based on unreasonable version management strategy or very special scenarios.

Answer 4

Should only be chrry pick

Answer 5

Everyone is right, there is no good way; but I still have this special need in my actual work process.
Tonight I used my poor shell programming skills to write a program that can achieve my needs, and it can be achieved through cherry-pick.

This small program can fulfill my needs as long as it is executed as follows:

# b5省略时取到最新的节点
cherry-picks B b2 b5

Next is my debut: github link

#!/usr/bin/env bash

# 初始化
targetBranch=
start=
end= #如果没有这读到最新
currentBranch=$(git symbolic-ref HEAD 2>/dev/null | cut -d"/" -f 3)

# 切换到目标分支
git checkout $targetBranch

echo start proccess commit message...

rawList=$(git log | grep '^commit' | sed '1,$s/commit//')

startIndex=$(echo "$rawList" | sed -n "/$start/=" )

# 如果end不为空，取得结束的偏移量，否则默认为1，也就是最新
if [ -n "$end" ]; then
    endIndex=$(echo "$rawList" | sed -n "/$end/=")
else
    endIndex=1
fi

# 取得需要cherry-pick的区间
list=$(echo "$rawList" | head -$startIndex | tail +$endIndex)

## 倒序
list=$(echo "$list" |sed '1!G;h;$!d')


echo '待cherry-pick的有：'
echo list:
echo "$list"
echo

# 切换回当前分支
git checkout $currentBranch

for i in "$list"; do
    git cherry-pick $i
done

echo '完成'
exit 0