正则表达式懒惰的量词表现出贪婪 [英] Regex lazy quantifier behave greedy

问题描述

我有这样的文字；

[Some Text][1][Some Text][2][Some Text][3][Some Text][4]

我想将[Some Text][2]与此正则表达式匹配；

I want to match [Some Text][2] with this regex;

/\[.*?\]\[2\]/

但是它返回[Some Text][1][Some Text][2]

我怎么只能匹配[Some Text][2]?

注意:某些文本中可以包含任何字符，包括[和]，并且方括号中的数字可以是不仅是1和2的任何数字.我要匹配的Some Text可以位于行的开头，可以有多个Some Text

Note : There can be any character in Some Text including [ and ] And the numbers in square brackets can be any number not only 1 and 2. The Some Text that i want to match can be at the beginning of the line and there can be multiple Some Texts

JSFiddle

推荐答案

\[.*?\]\[2\]模式工作原理这个:

• \[-查找最左边的[(正则表达式引擎处理从左到右的字符串输入)
• .*?-匹配除换行符以外的任何0+字符，尽可能少，但为成功匹配所需要的尽可能多，还有后续的模式，请参见下文
• \]\[2\]-][2]子字符串.

• \[ - finds the leftmost [ (as the regex engine processes the string input from left to right)
• .*? - matches any 0+ chars other than line break chars, as few as possible, but as many as needed for a successful match, as there are subsequent patterns, see below
• \]\[2\] - ][2] substring.

因此，.*?在每次失败时都会扩展，直到找到最左边的][2]. 请注意，懒惰的量词不能保证最短"匹配..

So, the .*? gets expanded upon each failure until it finds the leftmost ][2]. Note the lazy quantifiers do not guarantee the "shortest" matches.

解决方案

使用否定字符类(而不是边界char)来代替.*?(或.*)，而使用否定字符类.

Instead of a .*? (or .*) use negated character classes that match any char but the boundary char.

\[[^\]\[]*\]\[2\]

请参见此regex演示.

在这里，.*?替换为[^\]\[]*-除]和[之外的0个或更多字符.

Here, .*? is replaced with [^\]\[]* - 0 or more chars other than ] and [.

其他示例:

• <[^<>]*>匹配内部没有<和>的<...>
• \([^()]*\)匹配内部没有(和)的(...)
• "[^"]*"匹配内部没有"的"..."

• <[^<>]*> matches <...> with no < and > inside
• \([^()]*\) matches (...) with no ( and ) inside
• "[^"]*" matches "..." with no " inside

在其他情况下，当起始模式是多字符字符串或复杂模式时，请使用调和贪婪令牌，(?:(?!start).)*?.要匹配abc 0 abc 1 def中的abc 1 def，请使用abc(?:(?!abc).)*?def.

In other situations, when the starting pattern is a multichar string or complex pattern, use a tempered greedy token, (?:(?!start).)*?. To match abc 1 def in abc 0 abc 1 def, use abc(?:(?!abc).)*?def.

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