如何在括号

Question

我有一个例如文字：

  我想吃一块馅饼，我想带狗出去，我想游泳（馅饼，狗，游泳）

我需要一个JavaScript RegEx，用,之外的新行替换() 现在，如果我使用.replace(/,/g, "\n")，我会得到：

I would like to eat a pie
I would like to take the dog out
I would like to swim(pie
dog
out)

我需要的是：

I would like to eat a pie
I would like to take the dog out 
I would like to swim(pie,dog,swim)

Answer 1

你可以使用带有负前瞻的正则表达式（假设括号是平衡的和非转义的）：

str = str.replace(/,\s*(?![^()]*\))/g, '\n');

RegEx Demo

(?![^()]*\))是一个消极的先行者，断言我们前面没有)个字符，中间没有(或)个字符。

<强>代码：

var str = 'I would like to eat a pie,I would like to take the dog out, I would like to swim(pie,dog,swim)';

console.log(str.replace(/,\s*(?![^()]*\))/g, '\n'));

Answer 2

匹配(...)子字符串并匹配并捕获其他上下文中的,，以便在检查组1是否匹配后，稍后用换行符替换：

var rx = /\([^)]*\)|(,)\s*/g;
var s = "I would like to eat a pie,I would like to take the dog out, I would like to swim(pie,dog,swim)";
var result = s.replace(rx, function($0,$1) {
  return $1 ? "\n" : $1;
});
console.log(result);

模式匹配：

• \( - (
• [^)]* - 除)
以外的0个字符
• \) - )
• | - 或
• (,) - 第1组：,字符
• \s* - 0+空格（以“修剪”线前面的空格）。

请参阅regex demo。