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我有一组这种形式的字符串：

NOOO (2), { AAA (1), BBB (2), CCC-CC (3), DDD (4) }

（括号内的元素可以多于四个）

我需要匹配括号内的内容并仅提取（使用组）'AAA'、'BBB'、...子字符串。所以这个例子的结果将是

group1 : AAA
group2 : BBB
group3 : CCC-CC
group4 : DDD

我试过这个表达式：

\{ (?:(\S+) \(\d+\),?\s?)+ \}

但它只返回最后一个匹配的组（因此，在这种情况下，只返回“DDD”）。我错过了什么？谢谢

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1 回答 1

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如果您使用的是 .NET 正则表达式，那么您的表达式将起作用，因为捕获组将捕获其所有值。否则，您必须使用更棘手的正则表达式或分两步进行匹配，首先匹配{ ... }组，然后匹配其中的元素。

棘手的正则表达式看起来像：

(?:{|\G(?!^),)   # match a { or where the previous match ended followed by a ,
\s+              # space between elements
(\S+)\s+\(\d+\)  # an element
(?=[^{]*})       # make sure it's eventually followed by a }

You can use that expression as it's written if you use the /x flag (can also be set by adding (?x) in the beginning of the expression).

The regex without the comments:

(?:{|\G(?!^),)\s+(\S+)\s+\(\d+\)(?=[^{]*})

This expression uses \G which your regex flavor has to support. Most modern regex flavors have it, including: Perl, PCRE (PHP/etc), .NET.

Note that such an expression isn't perfect. It would capture AAA and BBB in the following string for example:

{ AAA (1), BBB (23), CCC, something invalid here #¤% ))),,,,!! }

Altho that can be fixed if required (except for the counter).

于 2012-06-05T12:58:08.670 回答