以给定模式开头时的递归匹配 [英] Recursive matching when preceded by a given pattern

问题描述

考虑以下两个字符串:

a (b (c) d) e
f (g (h) i) j

我想递归匹配括号中的内容.我可以使用以下模式来做到这一点:

I would like to match the content in parentheses recursively. I can do so by using the following pattern:

\((?>[^()]|(?R))*\)

这将匹配 (b (c) d) 和 (g (h) i).

现在，假设我想递归匹配括号中的内容，仅当前面有字母 a 时.我怎样才能做到这一点?

Now, let's say I want to match recursively the content in parentheses only when preceded by the letter a. How can I do that?

使用 (?<=a\s)\((?>[^()]|(?R))*\) 似乎不起作用.

Using (?<=a\s)\((?>[^()]|(?R))*\) does not seem to work.

推荐答案

你可以重复创建一个捕获组并重复第一个子模式，而不是重复整个模式:

Instead of repeating the whole pattern, you could repeat create a capturing group and repeat the first sub pattern instead:

(?<=a\s)(\((?>[^()]|(?1))*\))

• (?<=a\s) 断言左边是 a 和一个空格字符
• ( 捕获组 1
  • \( 匹配(
  • (?> 原子组
    • [^()] 匹配除 ( 或 ) 之外的任何字符(使用 [^()]+> 以提高效率)
    • | 或
    • (?1) 递归第一个子模式
    • (?<=a\s) Assert what is on the left is a and a whitespace char
    • ( Capture group 1
      • \( Match (
      • (?> Atomic group
        • [^()] Match any char except ( or ) (Use [^()]+ to make it more efficient)
        • | Or
        • (?1) Recurse the first subpattern

        正则表达式演示

        添加捕获组时，我们可以省略lookbehind并匹配a和一个空格字符:

        When adding a capturing group, we can omit the lookbehind and match a and a whitespace char:

        a\s(\((?>[^()]|(?1))*\))
        

        正则表达式演示

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