给定文件的完整路径，如何只获取没有文件名的路径? [英] Given the full path to a file, how do I get just the path without the filename?

问题描述

假设我在名为/home/user/directory/HelloWorld.txt"的字符串中有一个路径.我想删除 HelloWorld.txt，并以/home/user/directory"结尾.我需要什么正则表达式.

Suppose I have a path in a string called '/home/user/directory/HelloWorld.txt'. I would like to remove the HelloWorld.txt, and end up with '/home/user/directory'. What regex would I need.

推荐答案

不要使用正则表达式.相反，使用 File::Basename，可以处理所有特殊情况.

Don't use a regex. Instead, use File::Basename, which can handle all the special cases.

use File::Basename;

dirname("/foo/bar/baz/quux.txt"); --> "/foo/bar/baz"

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