如何去掉不必要的括号? [英] How to remove unnecessary parenthesis?
本文介绍了如何去掉不必要的括号?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
与 this 相同,但 JavaScript.举几个例子来说明我的目标:
Same as this but JavaScript. A few examples to illustrate my goal:
- ((((foo))) => (foo)
- ((foo)) => (foo)
- (foo) => (foo)
- (foo (bar)) => (foo (bar))
- ((foo b)ar) => ((foo b)ar)
- (((a)b(c))) => ((a)b(c))
我制作了一个正则表达式,它应该与我想要更改的那些匹配 /\({2,}[\s\S]*\){2,}/g
但我不能似乎不知道如何删除它们.
I have made a regex which should match the ones I want to change /\({2,}[\s\S]*\){2,}/g
but I can't seem to figure out how to remove them.
有没有类似String.replace(/\({2,}[\s\S]*\){2,}/g, '(${rest})')
?
推荐答案
与其纠结于正则表达式,我建议您尝试经典的标记化解析生成工作流程.想法是将字符串解析为数据结构(在本例中为嵌套数组),简化该结构并从中生成新字符串.
Instead of struggling with regexes I'd suggest you try the classic tokenize-parse-generate workflow. The idea is to parse a string into a data structure (in this case, nested arrays), simplify that structure and generate a new string from it.
示例:
function tokenize(str) {
return str.match(/[()]|[^()]+/g);
}
function parse(toks, depth = 0) {
let ast = [];
while (toks.length) {
let t = toks.shift();
switch (t) {
case '(':
ast.push(parse(toks, depth + 1));
break;
case ')':
if (!depth)
throw new SyntaxError('mismatched )');
return ast;
default:
ast.push(t);
}
}
if (depth) {
throw new SyntaxError('premature EOF');
}
return ast;
}
function generate(el) {
if (!Array.isArray(el))
return el;
while (el.length === 1 && Array.isArray(el[0]))
el = el[0];
return '(' + el.map(generate).join('') + ')';
}
//
let test = `
(((foo)))
((foo))
(foo)
(foo (bar))
((foo b)ar)
((((foo)) bar))
((((((foo))bar))baz))
((((((((foo))))))))
foo
`;
for (let s of test.trim().split('\n'))
console.log(s, '=>', generate(parse(tokenize(s))));
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