了解否定字符类 [英] Understanding the negated character class

问题描述

正则表达式:

/''+[^f]/g

应用于字符串时:

don't '''theater'''  but not'''d and not do'''f

还匹配 do'''f 中的三个撇号.为什么 [^f] 不排除它?

also matches the three apostrophes in do'''f. Why is the [^f] not excluding it?

小提琴是这里.

PS:我想找到连续的两个或多个撇号后跟不是 f.

PS: I want to find consecutive two or more apostrophes followed by not an f.

推荐答案

+ 使正则表达式引擎在 2 个或多个 之后找到 f 后回溯' s.您可以在否定前瞻中使用 ' 替代项来防止(以免使用 f 和 ' 以外的字符，当您使用 [^f]，该字符成为匹配的一部分，因为否定字符类是一个消耗模式，而前瞻是零宽度断言).

The + makes the regex engine backtrack once a f is found after 2 or more 's. You may prevent with a ' alternative in a negative lookahead (so as not to consume the character other than f and ', when you use [^f], the character becomes part of the match since a negated character class is a consuming pattern and lookaheads are zero-width assertions).

''+(?!['f])

查看正则表达式演示.这里，(?!['f]) 将阻止匹配，如果 2 个或多个 ' 符号后跟 f 或 '.此外，您可以使用限制量词 {2,}(出现 2 次或更多)来编写它:'{2,}(?!['f]).

See the regex demo. Here, (?!['f]) will prevent a match if 2 or more ' symbols are followed with f or '. Also, you may write it with a limiting quantifier {2,} (2 or more occurrences): '{2,}(?!['f]).

如果您的正则表达式引擎支持防止回溯到量化模式的占有量词，请使用一个:

If your regex engine supports possessive quantifiers that prevent backtracking into the quantified patterns, use one:

''++(?!f)
  ^^

参见另一个演示(另一种编写方式是'{2,}+(?!f)).

See another demo (another way of writing it is '{2,}+(?!f)).

如果您正在使用不支持所有格量​​词的 .NET 正则表达式库，您可以使用原子组代替(其工作方式与所有格量词相同，但适用于整个组):

If you are using a .NET regex library that does not support possessive quantifiers, you may use the atomic group instead (that works the same way as possessive quantifier, but for the entire group):

(?>'{2,})(?!f)

见.NET正则表达式演示.

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