在Lua中将路径名称拆分成其组件的最巧妙的方法是什么 [英] What is the neatest way to split out a Path Name into its components in Lua

问题描述

我有一个带路径的标准Windows文件名.我需要从字符串中拆分出文件名，扩展名和路径.

I have a standard Windows Filename with Path. I need to split out the filename, extension and path from the string.

我目前只是从末尾开始向后读取字符串，以寻找.切断扩展名，第一个\以获取路径.

I am currently simply reading the string backwards from the end looking for . to cut off the extension, and the first \ to get the path.

我确信我应该可以使用Lua模式来做到这一点，但是在从字符串的右边进行工作时，我一直失败.

I am sure I should be able to do this using a Lua pattern, but I keep failing when it comes to working from the right of the string.

例如 c:\ temp \ test \ myfile.txt 应该返回

eg. c:\temp\test\myfile.txt should return

• c:\ temp \ test \
• myfile.txt
• txt

如果这是重复的话，谢谢您的道歉，但是我可以找到很多其他语言的示例，但不是Lua的示例.

Thank you in advance apologies if this is a duplicate, but I could find lots of examples for other languages, but not for Lua.

推荐答案

> return string.match([[c:\temp\test\myfile.txt]], "(.-)([^\\]-([^%.]+))$")
c:\temp\test\   myfile.txt  txt

这似乎完全符合您的要求.

This seems to do exactly what you want.

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