git：所有更改文件的列表，包括子模块中的文件
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git: list of all changed files including those in submodules
提问by 4eyes
I would like to get a list of all files, which have changed betweet two commits including those in submodules.
我想获得所有文件的列表，这些文件在两个提交之间发生了变化，包括子模块中的提交。
I know I can do this:
我知道我可以这样做：
git diff --name-only --diff-filter=ACMR ${revision} HEAD
It returns a list of files, including the submodule-path, but not the files within.
它返回一个文件列表，包括子模块路径，但不返回其中的文件。
Example: I've updated a submodule. I commited the super-project. Now I want to get a list of all files which have been modified.
示例：我更新了一个子模块。我提交了超级项目。现在我想获取所有已修改文件的列表。
Do you know a way to get this done?
你知道有什么方法可以完成这项工作吗？
回答by VonC
Update 2017: as I mentioned in "see diff of commit on submodule in gitlab",
2017 年更新：正如我在“查看 gitlab 子模块上提交的差异”中提到的，
Git 2.11 (Nov. 2016) introduces
git diff --submodule=diffGit 2.14 (Q3 2017) will improve that by recursing into nested submodules.
See commit 5a52214(04 May 2017) by Stefan Beller (stefanbeller).
(Merged by Junio C Hamano --gitster--in commit a531ecf, 29 May 2017)
Git 2.11（2016 年 11 月）引入
git diff --submodule=diffGit 2.14（2017 年第三季度）将通过递归到嵌套子模块来改进这一点。
请参阅Stefan Beller ( ) 的提交 5a52214（2017 年 5 月 4 日）。（由Junio C Hamano合并-- --in commit a531ecf，2017 年 5 月 29 日）stefanbellergitster
Original 2012 answer (pre 2017 Git 2.14)
2012 年原始答案（2017 年 Git 2.14 之前）
Maybe a simple line would be enough:
也许一条简单的线就足够了：
git submodule foreach --recursive git diff --name-status
That would actually list files in submodules withinsubmodules.
(The --recursiveoption comes from git1.7.3+)
子模块中，将实际列出的文件中的子模块。
（该--recursive选项来自git1.7.3+）
回答by Jamey Sharp
You can find out what version a submodule was at, as of a given parent module commit, using git ls-tree:
您可以使用以下命令找出子模块在给定父模块提交时所处的版本git ls-tree：
subcommit=$(git ls-tree $parentcommit $submodulepath | awk '{print }')
So here's a script that should get you much of the way there, up to output formatting and such:
因此，这里有一个脚本，它应该可以帮助您完成大部分工作，包括输出格式等：
#!/bin/sh
function compare {
if [[ -z "" ]];
then git diff --name-only --ignore-submodules=all --diff-filter=ACMR "" ""
else git diff --name-only --ignore-submodules=all --diff-filter=ACMR "" "" | awk -v r= '{ print "" r "/" #!/bin/sh
echo "Listing changes for super module"
git diff --name-only
subs=(`git submodule | awk '{print }'`)
for sub in ${subs[*]}; do
lastrevision=`git diff $sub | fgrep "Subproject" | head -n1 | awk '{print }'`
cd $sub
echo "Listing changes for $sub"
git diff $lastrevision --name-only
cd ..
done
}'
fi
for submodule in `git submodule | awk '{print }'`
do
old=$(git ls-tree $submodule | awk '{print }')
new=$(git ls-tree $submodule | awk '{print }')
(cd $submodule; compare $old $new $submodule)
done
}
compare "" ""
This will output all files like this (although Base is a submodule): HtmlTemplates/Css/Screen.css Base/Php/Classes/Helper.php
这将输出这样的所有文件（尽管 Base 是一个子模块）： HtmlTemplates/Css/Screen.css Base/Php/Classes/Helper.php
回答by BlacKow
So, the very straightforward script that lists all changes compared to a revision
因此，与修订版相比，列出所有更改的非常简单的脚本
@echo off
if NOT %1.==. goto has_rev1
@echo git diff --name-only including submodules
@echo usage:
@echo ^ ^ %~n0 ^<revision1^> [^<revision2^>^|HEAD]
@exit /b 1
:has_rev1
setlocal
set rev1=%1
if %2.==. (set rev2=HEAD) else (set rev2=%2)
call :show_diff %rev1% %rev2%
exit /b
::eof
:show_diff
setlocal ENABLEDELAYEDEXPANSION
for /f "tokens=*" %%l in ('git --no-pager diff --name-only --ignore-submodules^=all --line-prefix^=%3 %1 %2') do set fwds=%%l & set bcks=!fwds:/=\! & echo !bcks!
endlocal
::git submodule is too slow for this
::for /f "tokens=2" %%d in ('git submodule') do call :subm %1 %2 %%d %3
if exist .gitmodules for /f "tokens=1,3*" %%p in (.gitmodules) do if %%p.==path. call :subm %1 %2 %%q %3
exit /b
::show_diff
:subm
setlocal
for /f "tokens=3" %%r in ('git ls-tree %1 %3') do set rev1=%%r
for /f "tokens=3" %%r in ('git ls-tree %2 %3') do set rev2=%%r
set fwdslash=%3
set bckslash=%fwdslash:/=\%
pushd %bckslash%
call :show_diff %rev1% %rev2% %4%bckslash%\
popd
endlocal
exit /b
::subm
it takes one argument - revision you want to compare with.
Make sure that there is fgrep "Subproject", not fgrep "Submodule".
它需要一个参数 - 您想要比较的修订版。确保有fgrep "Subproject"，没有fgrep "Submodule"。
回答by Pavel Studeny
A Windows variant of https://stackoverflow.com/a/13169898/5438298by Jamey Sharpwould be
Jamey Sharp的https://stackoverflow.com/a/13169898/5438298的Windows 变体将是
git diff --submodule=diff
(note that it might need a few more double quotes for spaces in the submodule names to work).
（请注意，子模块名称中的空格可能需要更多双引号才能工作）。
回答by Naman
July 8,2017
2017 年 7 月 8 日
Now to get a diff including that of a submodule, you can use the command -
现在要获得包括 a 的差异submodule，您可以使用命令 -
Note - This has been introduced with Git 2.14.0
注意 - 这已在Git 2.14.0 中引入
"git diff --submodule=diff" now recurses into nested submodules.
“git diff --submodule=diff”现在递归到嵌套的子模块中。
