您可以使用分隔符查找和替换字符串。

var obj = {
  'firstname': 'John',
  'lastname': 'Doe'
}

var text = "My firstname is {firstname} and my lastname is {lastname}"

alert(mutliStringReplace(obj,text))

function mutliStringReplace(object, string) {
      var val = string
      var entries = Object.entries(object);
      entries.forEach((para)=> {
          var find = '{' + para[0] + '}'
          var regExp = new RegExp(find,'g')
       val = val.replace(regExp, para[1])
    })
  return val;
}

var str = "I have a cat, a dog, and a goat.";

    str = str.replace(/goat/i, "cat");
    // now str = "I have a cat, a dog, and a cat."

    str = str.replace(/dog/i, "goat");
    // now str = "I have a cat, a goat, and a cat."

    str = str.replace(/cat/i, "dog");
    // now str = "I have a dog, a goat, and a cat."

使用常规函数定义要替换的模式，然后使用替换函数处理输入字符串，

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

此解决方案可以调整为仅替换整个单词-因此，例如，“catch”、“ducat”或“locator”在搜索“cat”时不会被找到。这可以通过使用负查找来完成 (?<!\w) 消极前瞻 (?!\w) 关于正则表达式中每个单词前后的单词字符：

(?<!\w)(cathy|cat|ducat|locator|catch)(?!\w)

JSFIDLE演示：http://jsfiddle.net/mfkv9r8g/1/

万一有人想知道为什么原始海报的解决方案不起作用：

var str = "I have a cat, a dog, and a goat.";

str = str.replace(/cat/gi, "dog");
// now str = "I have a dog, a dog, and a goat."

str = str.replace(/dog/gi, "goat");
// now str = "I have a goat, a goat, and a goat."

str = str.replace(/goat/gi, "cat");
// now str = "I have a cat, a cat, and a cat."

使用我的replace once软件包，您可以执行以下操作：

const replaceOnce = require('replace-once')

var str = 'I have a cat, a dog, and a goat.'
var find = ['cat', 'dog', 'goat']
var replace = ['dog', 'goat', 'cat']
replaceOnce(str, find, replace, 'gi')
//=> 'I have a dog, a goat, and a cat.'

这对我很有用：

String.prototype.replaceAll = function(search, replacement) {
    var target = this;
    return target.replace(new RegExp(search, 'g'), replacement);
};

function replaceAll(str, map){
    for(key in map){
        str = str.replaceAll(key, map[key]);
    }
    return str;
}

//testing...
var str = "bat, ball, cat";
var map = {
    'bat' : 'foo',
    'ball' : 'boo',
    'cat' : 'bar'
};
var new = replaceAll(str, map);
//result: "foo, boo, bar"

使用array.prototype.reduce（）：

const arrayOfObjects = [
  { plants: 'men' },
  { smart:'dumb' },
  { peace: 'war' }
]
const sentence = 'plants are smart'

arrayOfObjects.reduce(
  (f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence
)

// as a reusable function
const replaceManyStr = (obj, sentence) => obj.reduce((f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence)

const result = replaceManyStr(arrayOfObjects , sentence1)

例子

// /////////////    1. replacing using reduce and objects

// arrayOfObjects.reduce((f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence)

// replaces the key in object with its value if found in the sentence
// doesn't break if words aren't found

// Example

const arrayOfObjects = [
  { plants: 'men' },
  { smart:'dumb' },
  { peace: 'war' }
]
const sentence1 = 'plants are smart'
const result1 = arrayOfObjects.reduce((f, s) => `${f}`.replace(Object.keys(s)[0], s[Object.keys(s)[0]]), sentence1)

console.log(result1)

// result1: 
// men are dumb

// Extra: string insertion python style with an array of words and indexes

// usage

// arrayOfWords.reduce((f, s, i) => `${f}`.replace(`{${i}}`, s), sentence)

// where arrayOfWords has words you want to insert in sentence

// Example

// replaces as many words in the sentence as are defined in the arrayOfWords
// use python type {0}, {1} etc notation

// five to replace
const sentence2 = '{0} is {1} and {2} are {3} every {5}'

// but four in array? doesn't break
const words2 = ['man','dumb','plants','smart']

// what happens ?
const result2 = words2.reduce((f, s, i) => `${f}`.replace(`{${i}}`, s), sentence2)

console.log(result2)

// result2: 
// man is dumb and plants are smart every {5}

// replaces as many words as are defined in the array
// three to replace
const sentence3 = '{0} is {1} and {2}'

// but five in array
const words3 = ['man','dumb','plant','smart']

// what happens ? doesn't break
const result3 = words3.reduce((f, s, i) => `${f}`.replace(`{${i}}`, s), sentence3)

console.log(result3)

// result3: 
// man is dumb and plants

使用编号的项目以防止再次更换。如

let str = "I have a %1, a %2, and a %3";
let pets = ["dog","cat", "goat"];

然后

str.replace(/%(\d+)/g, (_, n) => pets[+n-1])

工作原理：-%\d+查找%后面的数字。括号表示数字。
该数字（作为字符串）是lambda函数的第二个参数n。
+n-1将字符串转换为数字，然后减去1以索引pets数组。
然后将%数字替换为数组索引处的字符串。
/g导致使用每个数字重复调用lambda函数，然后用数组中的字符串替换这些数字。
在现代javascript中：-

replace_n=(str,...ns)=>str.replace(/%(\d+)/g,(_,n)=>ns[n-1])

在本例中，这可能无法满足您的确切需求，但我发现这是替换字符串中多个参数的一种有用方法，是一种通用解决方案。它将替换参数的所有示例，无论引用了多少次：

String.prototype.fmt = function (hash) {
        var string = this, key; for (key in hash) string = string.replace(new RegExp('\\{' + key + '\\}', 'gm'), hash[key]); return string
}

您可以按如下方式调用它：

var person = '{title} {first} {last}'.fmt({ title: 'Agent', first: 'Jack', last: 'Bauer' });
// person = 'Agent Jack Bauer'

特定溶液

您可以使用一个函数替换每个函数。

var str = "I have a cat, a dog, and a goat.";
var mapObj = {
   cat:"dog",
   dog:"goat",
   goat:"cat"
};
str = str.replace(/cat|dog|goat/gi, function(matched){
  return mapObj[matched];
});

JSFIDLE示例

概括起来

如果您想动态地维护regex，并且只需将未来的交换添加到Map中，您可以这样做

new RegExp(Object.keys(mapObj).join("|"),"gi");

生成正则表达式。那么看起来是这样的

var mapObj = {cat:"dog",dog:"goat",goat:"cat"};

var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
str = str.replace(re, function(matched){
  return mapObj[matched];
});

要添加或更改任何其他替换，您只需编辑Map即可
摆弄动态正则表达式

使其可重复使用

如果你想让它成为一个通用模式，你可以把它拉出来，变成这样一个函数

function replaceAll(str,mapObj){
    var re = new RegExp(Object.keys(mapObj).join("|"),"gi");

    return str.replace(re, function(matched){
        return mapObj[matched.toLowerCase()];
    });
}

因此，您只需将str和替换的Map传递给函数，它就会返回转换后的字符串。
玩弄功能
为了确保object.keys在较旧的浏览器中工作，请从mdn或es5添加polyfill。