使用正则函数定义要替换的模式，然后使用replace函数处理输入字符串，

var i = new RegExp('"{','g'),
    j = new RegExp('}"','g'),
    k = data.replace(i,'{').replace(j,'}');

我扩展了一下@本麦考密克斯。他的工作规则字符串，但不如果我转义字符或通配符。我是这么做的

str = "[curl] 6: blah blah 234433 blah blah";
mapObj = {'\\[curl] *': '', '\\d: *': ''};


function replaceAll (str, mapObj) {

    var arr = Object.keys(mapObj),
        re;

    $.each(arr, function (key, value) {
        re = new RegExp(value, "g");
        str = str.replace(re, function (matched) {
            return mapObj[value];
        });
    });

    return str;

}
replaceAll(str, mapObj)

返回"blah blah 234433 blah blah"

这样它将匹配mapObj中的键，而不是匹配的单词'

一种可能的解决方案是使用mapper表达式函数。

const regex = /(?:cat|dog|goat)/gmi;
const str = `I have a cat, a dog, and a goat.`;

let mapper = (key) => {
  switch (key) {
    case "cat":
      return "dog"
    case "dog":
      return "goat";
    case "goat":
      return "cat"
  }
}
let result = str.replace(regex, mapper);

console.log('Substitution result: ', result);
//Substitution result1:  I have a dog, a goat, and a cat.

const str = '感谢为Stack Overflow贡献一个答案!' Const substr = ['for'， 'to'] 函数boldString(str, substr) { 让boldStr boldStr = str 字符串的子串。映射(e => { const strRegExp = new RegExp(e， 'g'); boldStr = boldStr。替换(strRegExp ' <强> $ {e} < / >强'); ｝ ） 返回boldStr ｝

String.prototype.replaceSome = function() {
    var replaceWith = Array.prototype.pop.apply(arguments),
        i = 0,
        r = this,
        l = arguments.length;
    for (;i<l;i++) {
        r = r.replace(arguments[i],replaceWith);
    }
    return r;
}

／* 字符串的replaceSome方法 它需要尽可能多的参数，然后替换所有参数 我们指定的最后一个参数 2013年版权保存:Max Ahmed 这是一个例子:

var string = "[hello i want to 'replace x' with eat]";
var replaced = string.replaceSome("]","[","'replace x' with","");
document.write(string + "<br>" + replaced); // returns hello i want to eat (without brackets)

*/

jsFiddle: http://jsfiddle.net/CPj89/

注意!

如果您正在使用动态提供的映射，这里的解决方案都不够!

在这种情况下，有两种解决方法:(1)使用分割连接技术，(2)使用正则表达式和特殊字符转义技术。

这是一个分割连接技术，它比另一个快得多(至少快50%):

var str = "I have {abc} a c|at, a d(og, and a g[oat] {1} {7} {11." var mapObj = { 'c|at': "d(og", 'd(og': "g[oat", 'g[oat]': "c|at", }; var entries = Object.entries(mapObj); console.log( entries .reduce( // Replace all the occurrences of the keys in the text into an index placholder using split-join (_str, [key], i) => _str.split(key).join(`{${i}}`), // Manipulate all exisitng index placeholder -like formats, in order to prevent confusion str.replace(/\{(?=\d+\})/g, '{-') ) // Replace all index placeholders to the desired replacement values .replace(/\{(\d+)\}/g, (_,i) => entries[i][1]) // Undo the manipulation of index placeholder -like formats .replace(/\{-(?=\d+\})/g, '{') );

这一个，是Regex特殊字符转义技术，它也有用，但慢得多:

var str = "I have a c|at, a d(og, and a g[oat]." var mapObj = { 'c|at': "d(og", 'd(og': "g[oat", 'g[oat]': "c|at", }; console.log( str.replace( new RegExp( // Convert the object to array of keys Object.keys(mapObj) // Escape any special characters in the search key .map(key => key.replace(/[-[\]{}()*+?.,\\^$|#\s]/g, '\\$&')) // Create the Regex pattern .join('|'), // Additional flags can be used. Like `i` - case-insensitive search 'g' ), // For each key found, replace with the appropriate value match => mapObj[match] ) );

后者的优点是，它也可以用于不区分大小写的搜索。