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

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))

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'

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 ｝

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.

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。