从git diff中排除一个目录 [英] Exclude a directory from git diff

问题描述

我想知道如何从Git diff中排除整个目录。 （在这种情况下/规格）。我正在使用git diff命令为我们的整个软件版本创建diff。然而，对规格的更改与此过程无关，只是令人头痛。现在我知道了我可以做的

  git diff previous_release..current_release app / 
  

这会为app目录中的所有更改创建diff，但不会在lib /目录中创建。有谁知道如何完成这项任务？谢谢！

编辑：我只想说清楚，我知道我可以将所有目录的参数串到最后，减去/ spec。我希望有一种方法可以真正排除命令中的单个目录。 解决方案

假设你使用bash，已启用扩展通配符（ shopt -s extglob ），您可以从shell端处理：

  git diff previous_release current_release！（spec）
  

保存你不得不列出所有其他的东西。

或者shell不可知：

  git diff previous_release current_release --name-only | grep -v'^ spec /'\ 
 | xargs git diff previous_release current_release  -  
  

您可以将其封装在单行shell脚本中以保存你必须重新输入参数。

I'm wondering how I can exclude an entire directory from my Git diff. (In this case /spec). I'm creating a diff for our entire software release using the git diff command. However the changes to the specs are irrelevant for this procedure, and just create headaches. now I know i can do

git diff previous_release..current_release app/

This would create a diff for all the changes in the app directory, but not for instance, in the lib/ directory. Does anyone know how to accomplish this task? Thanks!

Edit: I just want to be clear, I know I can just string the parameters of all of my directories on the end, minus /spec. I was hoping there was a way to truly exclude a single directory in the command.

解决方案

Assuming you use bash, and you've enabled extended globbing (shopt -s extglob), you could handle that from the shell side:

git diff previous_release current_release !(spec)

Saves you having to list all other things.

Or, shell-agnostic:

git diff previous_release current_release --name-only | grep -v '^spec/' \
    | xargs git diff previous_release current_release --

You could wrap that up in a one-liner shell script to save yourself having to retype the arguments.

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