正则表达式获取所有空格，只要它们不包含在括号中

Question

正则表达式获取所有空格，只要它们不包含在大括号中

这是针对 javascript 提及系统

例如：“说@::{Joseph Empyre}{b0268efc-0002-485b-b3b0-174fad6b87fc}，好吗？”

需要获得：

[ "说 ", "@::{Joseph Empyre}{b0268efc-0002-485b-b3b0-174fad6b87fc}", ",", "所有", "对吗？" ]

[编辑]

解决于： https://codesandbox.io/s/rough-http-8sgk2

对不起，我的英语不好

Answer 1

我将您的问题解释为您所说to fetch all spaces as long as they are not enclosed in braces ，尽管您的结果示例不是我所期望的。 您的示例结果在 speak 之后包含一个空格，以及在{}组之后为,单独匹配。 下面我的 output 显示了我对您所要求的内容的期望，即仅在大括号外的空格上拆分的字符串列表。

const str =
    "Speak @::{Joseph Empyre}{b0268efc-0002-485b-b3b0-174fad6b87fc}, all right?";

// This regex matches both pairs of {} with things inside and spaces
// It will not properly handle nested {{}}
// It does this such that instead of capturing the spaces inside the {},
// it instead captures the whole of the {} group, spaces and all,
// so we can discard those later
var re = /(?:\{[^}]*?\})|( )/g;
var match;
var matches = [];
while ((match = re.exec(str)) != null) {
  matches.push(match);
}

var cutString = str;
var splitPieces = [];
for (var len=matches.length, i=len - 1; i>=0; i--) {
  match = matches[i];
  // Since we have matched both groups of {} and spaces, ignore the {} matches
  // just look at the matches that are exactly a space
  if(match[0] == ' ') {
    // Note that if there is a trailing space at the end of the string,
    // we will still treat it as delimiter and give an empty string
    // after it as a split element
    // If this is undesirable, check if match.index + 1 >= cutString.length first
    splitPieces.unshift(cutString.slice(match.index + 1));
    cutString = cutString.slice(0, match.index);
  }
}
splitPieces.unshift(cutString);
console.log(splitPieces)

安慰：

["Speak", "@::{Joseph Empyre}{b0268efc-0002-485b-b3b0-174fad6b87fc},", "all", "right?"]