正则表达式中最短的匹配结束 [英] Shortest match in regex from end

问题描述

给定一个输入字符串 fooxxxxxxfooxxxboo 我正在尝试编写一个匹配 fooxxxboo 的正则表达式，即从第二个 foo 开始到最后一个 boo.

Given an input string fooxxxxxxfooxxxboo I am trying to write a regex that matches fooxxxboo i.e. starting from the second foo till the last boo.

我尝试了以下

foo.*?boo 匹配完整的字符串 fooxxxxxxfooxxxboo

foo.*boo 也匹配完整的字符串 fooxxxxxxfooxxxboo

foo.*boo also matches the complete string fooxxxxxxfooxxxboo

我读了这篇贪婪与不情愿与占有量词 并且我理解它们的区别，但我试图从匹配正则表达式的末尾匹配最短的字符串，即类似于要从后面评估的正则表达式的东西.有什么办法可以只匹配最后一部分吗?

I read this Greedy vs. Reluctant vs. Possessive Quantifiers and I understand their difference, but I am trying to match the shortest string from the end which matches the regex i.e. something like the regex to be evaluated from back. Is there any way I can match only the last portion?

推荐答案

使用 否定前瞻断言.

foo(?:(?!foo).)*?boo

演示

(?:(?!foo).)*? - 任何字符的非贪婪匹配，但不是 foo 零次或多次.也就是说，在匹配每个字符之前，它会检查字符是否不是字母f 后跟两个o.如果是，则只匹配相应的字符.

(?:(?!foo).)*? - Non-greedy match of any character but not of foo zero or more times. That is, before matching each character, it would check that the character is not the letter f followed by two o's. If yes, then only the corresponding character will be matched.

为什么正则表达式 foo.*?boo 匹配完整的字符串 fooxxxxxxfooxxxboo?

Why the regex foo.*?boo matches the complete string fooxxxxxxfooxxxboo?

因为您的正则表达式中的第一个 foo 匹配 foo 字符串和以下 .*? 将进行非贪婪匹配字符串 boo，所以我们得到了两个匹配项 fooxxxxxxfooxxxboo 和 fooxxxboo.因为第二个匹配出现在第一个匹配中，正则表达式引擎只显示第一个.

Because the first foo in your regex matches both the foo strings and the following .*? will do a non-greedy match upto the string boo, so we got two matches fooxxxxxxfooxxxboo and fooxxxboo. Because the second match present within the first match, regex engine displays only the first.

这篇关于正则表达式中最短的匹配结束的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！