Javascript 如何在特定字符之前和之后提取嵌套括号的内容

发布于03月05日

在以下字符串中:

(10+10)*2*((1+1)*1)√(16)+(12*12)+2

I am trying replace ((1+1)*1)√(16) with nthroot(16,(1+1)*1).
Specifically, I want to extract everything in the first sets of brackets on each side of the √.
The brackets themselves could contain many layers of brackets and many different symbols.
Language is JavaScript.

I tried a couple things like <str>.replace(/$(.+)$√$(.+)$/g, 'nthroot($1,$2)')
but every one of my attempts at learning RegEx fails and I can't figure this out.

推荐答案

我认为目前不能用Javascript中的正则表达式以一般方式解决这个问题，因为不能递归地匹配平衡括号.

就我个人而言，我会将文本拆分为组成字符，构建括号组，然后用一些逻辑将所有内容重新连接起来.例如:

let text = '(10+10)*2*((1+1)*1)√(16)+(12*12)+2';
let changedText = '';
let parts = text.split('');
let parCount = null;
let group = '';
let groups = [];

// Group the original text into nested parentheses and other characters.
for (let i = 0; i < parts.length; i++) {
    // Keep a track of parentheses nesting; if parCount is larger than 0,
    // then there are unclosed parentheses left in the current group.
    if (parts[i] == '(') parCount++;
    if (parts[i] == ')') parCount--;

    group += parts[i];

    // Add every group of balanced parens or single characters.
    if (parCount === 0 && group !== '') {
        groups.push(group);
        group = '';
    }
}

// Join groups, while replacing the root character and surrounding groups
// with the nthroot() syntax.
for (let i = 0; i < groups.length; i++) {
    let isRoot = i < groups.length - 2 && groups[i + 1] == '√';
    let hasParGroups = groups[i][0] == '(' && groups[i + 2][0] == '(';

    // If the next group is a root symbol surrounded by parenthesized groups,
    // join them using the nthroot() syntax.
    if (isRoot && hasParGroups) {
        let stripped = groups[i + 2].replace(/^\(|\)$/g, '');
        changedText += `nthroot(${stripped}, ${groups[i]})`;
        // Skip groups that belong to root.
        i = i + 2;
    } else {
        // Append non-root groups.
        changedText += groups[i]
    }
}

console.log('Before:', text, '\n', 'After:', changedText);

但并不是说它很漂亮