巴什 - 获得最后目录名/文件名的文件路径参数 [英] Bash - get last dirname/filename in a file path argument

问题描述

我试图写一个post-commit钩子为SVN，这是我们开发服务器上托管。我的目标是尝试自动签承诺项目的副本，在那里它在服务器上托管的目录。不过，我需要能够传递到脚本，以结帐到我们的项目托管在同一子目录的目录串只读最后一个目录。

I'm trying to write a post-commit hook for SVN, which is hosted on our development server. My goal is to try to automatically checkout a copy of the committed project to the directory where it is hosted on the server. However I need to be able to read only the last directory in the directory string passed to the script in order to checkout to the same sub-directory where our projects are hosted.

例如，如果我做一个SVN提交到项目的榜样，我的剧本获得在/ usr /本地/ SVN /回购/例如，作为第一个参数。我需要得到的只是榜样的字符串的结尾，然后用另一个字符串连接，所以我可以检出到/服务器/根/例子，并立即看到变更后居住。

For example if I make an SVN commit to the project "example", my script gets "/usr/local/svn/repos/example" as its first argument. I need to get just "example" off the end of the string and then concat it with another string so I can checkout to "/server/root/example" and see the changes live immediately.

推荐答案

基本名 确实删除路径的目录preFIX：

basename does remove the directory prefix of a path:

$ basename /usr/local/svn/repos/example
example
$ echo "/server/root/$(basename /usr/local/svn/repos/example)"
/server/root/example

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