“[^][]"是什么意思?正则表达式是什么意思? [英] What does the "[^][]" regex mean?

问题描述

我在以下正则表达式中找到了它:

I found it in the following regex:

[(?:[^][]|(?R))*]

它匹配方括号(及其内容)和嵌套方括号.

It matches square brackets (with their content) together with nested square brackets.

推荐答案

[^][] 是一个字符类，表示除 [ 和 之外的所有字符].

[^][] is a character class that means all characters except [ and ].

您可以避免转义 [ 和 ] 特殊字符，因为它对于 PCRE(preg_ 函数中使用的正则表达式引擎)来说没有歧义.

You can avoid escaping [ and ] special characters since it is not ambiguous for the PCRE, the regex engine used in preg_ functions.

由于 [^] 在 PCRE 中不正确，正则表达式解析的唯一方法是 ] 位于稍后将关闭的字符类中.与后面的 [ 相同.它不能重新打开字符类内的字符类(POSIX 字符类 [:alnum:] 除外).那么最后一个]就清晰了；这是字符类的结束.但是，字符类之外的 [ 必须被转义，因为它被解析为字符类的开头.

Since [^] is incorrect in PCRE, the only way for the regex to parse is that ] is inside the character class which will be closed later. The same with the [ that follows. It can not reopen a character class (except a POSIX character class [:alnum:]) inside a character class. Then the last ] is clear; it is the end of the character class. However, a [ outside a character class must be escaped since it is parsed as the beginning of a character class.

同样的，你可以写[]] or [[] or [^[] 不用转义[ 或 ] 在字符类中.

In the same way, you can write []] or [[] or [^[] without escaping the [ or ] in the character class.

注意:自 PHP 7.3 起，您可以使用内联 xx 修饰符，即使在字符类中也允许忽略空白字符.通过这种方式，您可以以不那么模棱两可的方式编写这些类:(?xx) [^ ][ ] [ ] ] [ [ ] [^ [ ].

Note: since PHP 7.3, you can use the inline xx modifier that allows blank characters to be ignored even inside character classes. This way you can write these classes in a less ambigous way like that: (?xx) [^ ][ ] [ ] ] [ [ ] [^ [ ].

您可以将此语法与多种正则表达式一起使用:PCRE(PHP、R)、Perl、Python、Java、.NET、GO、awk、Tcl(如果您用大括号分隔您的模式，感谢 Donal Fellows), ...

You can use this syntax with several regex flavour: PCRE (PHP, R), Perl, Python, Java, .NET, GO, awk, Tcl (if you delimit your pattern with curly brackets, thanks Donal Fellows), ...

但不适用于:Ruby、JavaScript(IE < 9 除外)、...

But not with: Ruby, JavaScript (except for IE < 9), ...

正如 m.buettner 指出的那样，[^]] 不是二义性的，因为 ] 是 第一个 字符，[^a]] 被视为所有不是 a 后跟 ] 的东西.要拥有 a 和 ]，你必须写:[^a]] 或 [^]a]

As m.buettner noted, [^]] is not ambiguous because ] is the first character, [^a]] is seen as all that is not a a followed by a ]. To have a and ], you must write: [^a]] or [^]a]

在 JavaScript 的特殊情况下，规范允许 [] 作为 永远 匹配的正则表达式标记(换句话说，[] 将总是失败)和 [^] 作为匹配任何字符的正则表达式.那么[^]] 被视为任何字符后跟一个].实际实现各不相同，但现代浏览器一般都遵循规范中的定义.

In particular case of JavaScript, the specification allow [] as a regex token that never matches (in other words, [] will always fail) and [^] as a regex that matches any character. Then [^]] is seen as any character followed by a ]. The actual implementation varies, but modern browser generally sticks to the definition in the specification.

模式详情:

[          # literal [
(?:         # open a non capturing group
    [^][]   # a character that is not a ] or a [
  |         # OR
    (?R)    # the whole pattern (here is the recursion)
)*          # repeat zero or more time
]          # a literal ]

在您的模式示例中，您不需要转义最后一个 ]

In your pattern example, you don't need to escape the last ]

但是你可以用这个模式做同样的事情，稍微优化一下，更有用的原因可重用为子模式 (with the (?-1)): ([(?:[^][]+|(?-1))*+])

But you can do the same with this pattern a little bit optimized, and more useful cause reusable as subpattern (with the (?-1)): ([(?:[^][]+|(?-1))*+])

(                     # open the capturing group
    [                # a literal [
        (?:           # open a non-capturing group
            [^][]+    # all characters but ] or [ one or more time
          |           # OR
            (?-1)     # the last opened capturing group (recursion)
                      # (the capture group where you are)
        )*+           # repeat the group zero or more time (possessive)
    ]                 # literal ] (no need to escape)
)                     # close the capturing group

或更好:([[^][]*(?:(?-1)[^][]*)*+]) 避免了交替的成本.

or better: ([[^][]*(?:(?-1)[^][]*)*+]) that avoids the cost of an alternation.

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