使用我的replace-once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.'

可以使用分隔符查找和替换字符串。

Var obj = { “firstname”:“约翰”, “姓”:“母鹿” ｝ var文本= "你好{firstname}，你的名字是{firstname}和姓氏是{lastname}" console.log (mutliStringReplace (obj、文本)) 函数mutliStringReplace(对象，字符串){ Var val =字符串 var entries = object .entries(object); entries.forEach ((para) = > { Var find = '{' + para[0] + '}' var regExp = new regExp (find，'g') val = val.replace(regExp, para[1]) }) 返回val; ｝

我们也可以使用split()和join()方法:

var str = "我有一只猫，一只狗，和一只山羊。"; str = str.split(“猫”)。映射(x =>{返回x.split("dog"))。地图(y = >{返回y.split(“山羊”). join(“猫”);}). join(“山羊”);}). join(“狗”); console.log (str);

所有的解决方案都很好，除了应用于闭包的编程语言(如Coda, Excel，电子表格的REGEXREPLACE)。

我下面的两个原始解决方案只使用1个连接和1个正则表达式。

方法#1:查找替换值

其思想是，如果替换值不在字符串中，则附加替换值。然后，使用一个regex，我们执行所有需要的替换:

var str = "我有一只猫，一只狗，和一只山羊。"; STR = (STR +"||||猫，狗，山羊").replace( /猫(? = [\ s \ s] *(狗))|狗(? = [\ s \ s] *(山羊))|山羊(? = [\ s \ s] *(猫 ))|\|\|\|\|.* $ / gi, " $ 1 $ 2 $ 3”); document.body.innerHTML = str;

解释:

cat(?=[\s\S]*(dog)) means that we look for "cat". If it matches, then a forward lookup will capture "dog" as group 1, and "" otherwise. Same for "dog" that would capture "goat" as group 2, and "goat" that would capture "cat" as group 3. We replace with "$1$2$3" (the concatenation of all three groups), which will always be either "dog", "cat" or "goat" for one of the above cases If we manually appended replacements to the string like str+"||||cat,dog,goat", we remove them by also matching \|\|\|\|.*$, in which case the replacement "$1$2$3" will evaluate to "", the empty string.

方法#2:查找替换对

方法#1的一个问题是它一次不能超过9个替换，这是反向传播组的最大数量。 方法#2声明不只是附加替换值，而是直接替换:

var str = "我有一只猫，一只狗，和一只山羊。"; str = (str + " | | | |,猫= >狗,狗= >山羊,山羊= >猫”).replace ( / (\ b \ w + \ b) (? = [\ s \ s] * \ 1 =>([^,]*))|\|\|\|\|.* $ / gi, " $ 2 "); document.body.innerHTML = str;

解释:

(str+"||||,cat=>dog,dog=>goat,goat=>cat") is how we append a replacement map to the end of the string. (\b\w+\b) states to "capture any word", that could be replaced by "(cat|dog|goat) or anything else. (?=[\s\S]*...) is a forward lookup that will typically go to the end of the document until after the replacement map. ,\1=> means "you should find the matched word between a comma and a right arrow" ([^,]*) means "match anything after this arrow until the next comma or the end of the doc" |\|\|\|\|.*$ is how we remove the replacement map.

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])

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