非贪婪的正则表达式或否定的字符类哪个更好? [英] Which would be better non-greedy regex or negated character class?

问题描述

我需要从字符串@anything_here@dhhhd@shdjhjs@中匹配@anything_here@.因此，我使用了以下正则表达式.

I need to match @anything_here@ from a string @anything_here@dhhhd@shdjhjs@. So I'd used following regex.

^@.*?@

或

^@[^@]*@

两种方法都可以，但是我想知道哪种方法更好.非贪婪重复的正则表达式或字符类否定的正则表达式?

Both way it's work but I would like to know which one would be a better solution. Regex with non-greedy repetition or regex with negated character class?

推荐答案

如果可能的话，与延迟匹配相比，否定的字符类通常更受欢迎.

Negated character classes should usually be prefered over lazy matching, if possible.

如果正则表达式成功，则^@[^@]*@可以单步匹配@之间的内容，而^@.*?@需要扩展@之间的每个字符.

If the regex is successful, ^@[^@]*@ can match the content between @s in a single step, while ^@.*?@ needs to expand for each character between @s.

当失败时(对于没有结尾的@而言)，大多数正则表达式引擎会施加一点魔力并将内部将[^@]*视为[^@]*+，因为在@与非@，因此它将匹配到字符串的末尾，识别出丢失的@而不是回溯，但会立即失败. .*?将照常扩展字符.

When failing (for the case of no ending @) most regex engines will apply a little magic and internally treat [^@]* as [^@]*+, as there is a clear cut border between @ and non-@, thus it will match to the end of the string, recognize the missing @ and not backtrack, but instantly fail. .*? will expand character for character as usual.

在较大的上下文中使用时，[^@]*也将永远不会扩展到结尾@的边界，而对于延迟匹配则非常有可能.例如. ^@[^@]*a[^@]*@与@bbbb@a@不匹配，而^@.*?a.*?@将与之匹配.

When used in larger contexts, [^@]* will also never expand over the borders of the ending @ while this is very well possible for the lazy matching. E.g. ^@[^@]*a[^@]*@ won't match @bbbb@a@ while ^@.*?a.*?@ will.

请注意，[^@]也将匹配换行符，而.则不匹配(在大多数正则表达式引擎中，除非在单行模式下使用).您可以通过在否定词中添加换行符来避免这种情况-如果不需要的话.

Note that [^@] will also match newlines, while . doesn't (in most regex engines and unless used in singleline mode). You can avoid this by adding the newline character to the negation - if it is not wanted.

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