为此，您可以使用https://www.npmjs.com/package/union-replacer。它基本上是一个字符串。Replace (regexp，…)对等体，它允许在一次传递中发生多次替换，同时保留string.replace(…)的全部功能。

披露:我是作者。开发这个库是为了支持更复杂的用户可配置替换，它解决了所有有问题的事情，比如捕获组、反向引用和回调函数替换。

上面的解决方案对于精确的字符串替换来说已经足够好了。

使用Array.prototype.reduce ():

更新(更好)答案(使用对象): 此函数将替换所有出现的情况，并且不区分大小写

/**
 * Replaces all occurrences of words in a sentence with new words.
 * @function
 * @param {string} sentence - The sentence to modify.
 * @param {Object} wordsToReplace - An object containing words to be replaced as the keys and their replacements as the values.
 * @returns {string} - The modified sentence.
 */
function replaceAll(sentence, wordsToReplace) {
  return Object.keys(wordsToReplace).reduce(
    (f, s, i) =>
      `${f}`.replace(new RegExp(s, 'ig'), wordsToReplace[s]),
      sentence
  )
}

const americanEnglish = 'I popped the trunk of the car in a hurry and in a hurry I popped the trunk of the car'
const wordsToReplace = {
  'popped': 'opened',
  'trunk': 'boot',
  'car': 'vehicle',
  'hurry': 'rush'
}

const britishEnglish = replaceAll(americanEnglish, wordsToReplace) 
console.log(britishEnglish)
// I opened the boot of the vehicle in a rush and in a rush I opened the boot of the vehicle

原始答案(使用对象数组):

    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)

Example // ///////////// 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

这个解决方案可以只替换整个单词——例如，当搜索“猫”时，“catch”、“ducat”或“locator”将找不到。这可以通过对正则表达式中每个单词前后的单词字符使用负向后查找(?<!\w)和负向前查找(?!\w)来实现:

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

JSFiddle demo： http://jsfiddle.net/mfkv9r8g/1/

试试我的解决方案。请随意改进

函数multiReplace(字符串，regex，替换){ 返回str.replace(regex, function(x) { //检查替换键以防止错误，如果为false则返回原始值 return Object.keys(replace).includes(x) ?替换[x]: x; })； } var str = "我有一只猫，一只狗，和一只山羊。"; //(json)使用value替换键 Var替换= { “猫”:“狗”, “狗”:“山羊”, “山羊”:“猫”, ｝ console.log(multiReplace(str， /Cat|dog|goat/g, replace))

使用编号的物品，防止再次更换。 如

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

then

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

它的工作原理:- %\d+查找跟在%后面的数字。括号表示数字。

这个数字(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以索引宠物数组。

然后将%数字替换为数组下标处的字符串。

/g导致lambda函数被重复调用，每个数字被替换为数组中的字符串。

在现代JavaScript中:-

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

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/