在Bash中的文件路径参数中获取最后一个目录名/文件名 [英] Get last dirname/filename in a file path argument in Bash

问题描述

我正在尝试为SVN编写一个提交后钩子，该钩子驻留在我们的开发服务器上.我的目标是尝试自动将已提交项目的副本检出到服务器上托管该目录的目录中.但是，我需要能够仅读取传递给脚本的目录字符串中的最后一个目录，以便签出到托管我们项目的同一子目录.

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，如果我对项目"example"进行SVN提交，则我的脚本将"/usr/local/svn/repos/example"作为其第一个参数.我只需要从字符串末尾得到"example"，然后将其与另一个字符串连接起来，这样我就可以签出到"/server/root/example"并立即看到更改.

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

basename 确实删除路径的目录前缀:

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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