Git diff - how to get different files (only name) between two tags by 'git diff' and order by the file commit time

Question

Through the command:

$git diff --name-status tag1 tag2 -- target-src

The output is like:

M       src/config.h
M       src/createTable.sql
A       src/gameserver/BattleGround_10v10.h
M       src/gameserver/CMakeLists.txt
M       src/gameserver/achieve.cpp
M       src/gameserver/achieve.h
A       src/gameserver/action.cpp
A       src/gameserver/action.h
A       src/gameserver/activity.cpp
A       src/gameserver/activity.h
M       src/gameserver/admin.cpp

I got the files that has modified between the two tags. But I want the list order by committed time. How can I do that?

---

Thanks to ilius's answer , I added awk for my request:

git diff --name-status tag1 tag2 | while read line ; do
    status=${line:0:1}
    path=${line:2}
    date=$(git log -1 '--pretty=format:%ci' -- "$path")
    echo "$date    $status   $path"
done | sort -r | awk '{print $4" "$5}'

But I think it is too complicated. Can it be simpler?

Answer 1

You can try the same thing, but with git log :

git log --name-status tag1 tag2 -- target-src

That way, the files are sorted in the order the commits were made. Also git log has various sorting options you can try out.

Answer 2

Using mart1n's idea ,

git log --name-status '--pretty=format:' tag1 tag2 -- target-src | grep -vxh '\s*'

gives a clean output.

Also try this script:

git diff --name-status tag1 tag2 | while read line ; do
    status=${line:0:1}
    path=${line:2}
    date=$(git log -1 '--pretty=format:%ci' -- "$path")
    echo "$date    $status   $path"
done | sort -r

You can remove the dates (used for sorting) later, I think dates are useful though.

You can also remove -r option from sort if you want the older changed files to be on top.