用正则表达式计算括号 [英] Count parentheses with regular expression

问题描述

我的字符串是:(as(dh(kshd)kj)ad)...()()

如何用正则表达式计算括号?我想选择一个字符串，该字符串从第一个开括号开始，到 ...

How is it possible to count the parentheses with a regular expression? I would like to select the string which begins at the first opening bracket and ends before the ...

将其应用到上面的示例中，这意味着我想获取以下字符串:(as(dh(k(hshd)kj)ad)

Applying that to the above example, that means I would like to get this string: (as(dh(kshd)kj)ad)

我尝试编写它，但这不起作用:

I tried to write it, but this doesn't work:

var str = "(as(dh(kshd)kj)ad)... ()()";
document.write(str.match(/(.*)/m));

推荐答案

正如我在评论中所说，与流行的信念(不要相信人们所说的一切)匹配嵌套括号 可能相反正则表达式.

As I said in the comments, contrary to popular belief (don't believe everything people say) matching nested brackets is possible with regex.

使用它的不利之处在于，您只能将其执行到固定的嵌套级别.对于您希望支持的每个其他级别，您的正则表达式将越来越大.

The downside of using it is that you can only do it up to a fixed level of nesting. And for every additional level you wish to support, your regex will be bigger and bigger.

但是不要相信我.让我演示给你看.正则表达式 \([^()] * \) 匹配一个级别.对于最多两个级别，在此处查看正则表达式.要匹配您的情况，您需要:

But don't take my word for it. Let me show you. The regex \([^()]*\) matches one level. For up to two levels see the regex here. To match your case, you'd need:

\(([^()]*|\(([^()]*|\([^()]*\))*\))*\)

它将匹配粗体部分: (as(dh(kshd)kj)ad) ...()()

It would match the bold part: (as(dh(kshd)kj)ad)... ()()

检查 此处的演示 ，了解固定嵌套级别的含义.

Check the DEMO HERE and see what I mean by fixed level of nesting.

以此类推.为了继续添加级别，您要做的就是将最后一个 [^()] * 部分更改为([^()] * | \([^()] * \))* (在此处检查三个级别).正如我所说，它将越来越大.

And so on. To keep adding levels, all you have to do is change the last [^()]* part to ([^()]*|\([^()]*\))* (check three levels here). As I said, it will get bigger and bigger.

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