正则表达式以匹配两个相同字符串之间的所有字符串 [英] Regex to match all the strings between two identical strings

问题描述

例如我有这串 -- This -- is -- one -- another -- comment -- 我希望匹配的元素是 "This"，"is"，"one"，"another"和"comment"

我正在尝试此正则表达式 --\s+([^--]+)\s+-- 这给了我匹配的元素 这个"，一个"和评论"

我已经搜索了其他问题，它们都提供了类似的解决方案，例如#A#，我会得到A，但是对于#A#B#，我也会得到A，但是在这种情况下，我想要两个元素和B，因为它们都在两个#字符之间.

我正在针对javascript正则表达式进行测试，但是我认为解决方案应该与平台/语言无关.

解决方案

通常，您需要使用类似

的模式

STRING([\s\S]*?)(?=STRING|$)

它将匹配STRING，然后将任何零个或多个字符(尽可能少)捕获到组1中，直到第一个STRING出现*停止在此单词之前**，因为(?=...)是一个正向前进，它是零宽度的断言，不会消耗匹配的文本或字符串结尾.

模式的通用变体是

STRING((?:(?!STRING)[\s\S])*)

它使用脾气暴躁的令牌，(?:(?!STRING)[\s\S])*，该匹配的char等于0个或多个出现，不能启动STRING字符序列.

要获取当前解决方案中的所有子字符串，请使用前瞻性

/--\s+([\s\S]*?)(?=\s+--)/g
                ^^^^^^^^^

请参见 regex演示.

请注意，[^--]+匹配除-之外的1个或多个符号，它不匹配任何不等于--的文本. [...]是与单个字符匹配的字符类.要匹配从一个字符到模式的第一次出现的任何长度的任何文本，您可以依靠[\s\S]*?构造:0个以上的字符，并且尽可能少(由于懒惰的*?量词). /p>

JS演示:

 var s = '-- This -- is -- one -- another -- comment --';
var rx = /--\s+([\s\S]*?)(?=\s+--)/g;
var m, res=[];
while (m = rx.exec(s)) {
  res.push(m[1]);
}
console.log(res); 

E.g. I have this string -- This -- is -- one -- another -- comment -- I want the matched elements to be "This", "is", "one", "another", and "comment"

I was trying this regex --\s+([^--]+)\s+-- which gives me the matched elements as "This", "one" and "comment"

I have searched for other problems, they all provide solution like this i.e. #A# and I will get A but for #A#B# also I get A, but in this case I want both the elements A and B as both of them are between two # chars.

I am testing it for javascript regex, but I think solution should be irrespective of platform/language.

解决方案

In general, you need to use a pattern like

STRING([\s\S]*?)(?=STRING|$)

It will match STRING, then capture into Group 1 any zero or more chars, as few as possible, up to the first occurrence of STRING *stopping right before this word** because the (?=...) is a positive lookahead that, being a zero-width assertion, does not consume matched text or end of string.

A generic variation of the pattern is

STRING((?:(?!STRING)[\s\S])*)

It uses a tempered greedy token, (?:(?!STRING)[\s\S])*, that matches any char, 0 or more occurrences, that does not start a STRING char sequence.

To get all the substrings in the current solution, use a lookahead like

/--\s+([\s\S]*?)(?=\s+--)/g
                ^^^^^^^^^

See the regex demo.

Note that [^--]+ matches 1 or more symbols other than a -, it does not match any text that is not equal to --. [...] is a character class that matches a single character. To match any text of any length from one char up to the first occurrence of a pattern, you can rely on a [\s\S]*? construct: any 0+ chars, as few as possible (due to the lazy *? quantifier).

JS demo:

var s = '-- This -- is -- one -- another -- comment --';
var rx = /--\s+([\s\S]*?)(?=\s+--)/g;
var m, res=[];
while (m = rx.exec(s)) {
  res.push(m[1]);
}
console.log(res);

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