正则表达式单词边界不包括连字符 [英] regex word boundary excluding the hyphen

问题描述

我需要一个正则表达式来匹配以单词边界结尾的表达式，但它不将连字符视为边界.即获取所有匹配的表达式

i need a regex that matches an expression ending with a word boundary, but which does not consider the hyphen as a boundary. i.e. get all expressions matched by

type ([a-z])\b

但不匹配例如

type a-1

重新表述:我想要一个等效的单词边界运算符 \b，它不使用单词字符类 [A-Za-z0-9_]，而是使用扩展类:[A-Za-z0-9_-]

to rephrase: i want an equivalent of the word boundary operator \b which instead of using the word character class [A-Za-z0-9_], uses the extended class: [A-Za-z0-9_-]

推荐答案

您可以为此使用前瞻，最短的是使用否定前瞻:

You can use a lookahead for this, the shortest would be to use a negative lookahead:

type ([a-z])(?![\w-])

(?![\w-]) 表示如果下一个字符在 \w 中或者是 - 则匹配失败>".

(?![\w-]) would mean "fail the match if the next character is in \w or is a -".

这是一个使用正常前瞻的选项:

Here is an option that uses a normal lookahead:

type ([a-z])(?=[^\w-]|$)

您可以将 (?=[^\w-]|$) 读作仅匹配字符类 [\w-]，或者这是字符串的结尾".

You can read (?=[^\w-]|$) as "only match if the next character is not in the character class [\w-], or this is the end of the string".

看看它是否有效:http://www.rubular.com/r/NHYhv72znm

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