我试图用多个其他单词替换字符串中的多个单词。字符串是“我有一只猫,一只狗和一只山羊。”

然而,这并不会产生“我有一只狗、一只山羊和一只猫”,而是产生“我有一只猫、一只猫和一只猫”。是否有可能在JavaScript中同时用多个其他字符串替换多个字符串,以便产生正确的结果?

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

//this produces "I have a cat, a cat, and a cat"
//but I wanted to produce the string "I have a dog, a goat, and a cat".

当前回答

这招对我很管用:

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"

其他回答

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);

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/

这招对我很管用:

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"

作为对以下问题的回答:

寻找最新的答案

如果在当前示例中使用“words”,则可以使用非捕获组扩展Ben McCormick的答案,并在左侧和右侧添加单词边界\b以防止部分匹配。

\b(?:cathy|cat|catch)\b

防止部分匹配的单词边界 (?:非捕获组 Cathy |cat|catch匹配其中一个选项 )关闭非捕获组 防止部分匹配的单词边界

原问题的例子:

let str = "我有一只猫,一只狗和一只山羊。"; const mapObj = { 猫:“狗”, 狗:“山羊”, 山羊:“猫” }; str = str.replace(/\b(?:猫|狗|山羊)\b/gi, matched => mapObj[matched]); console.log (str);

评论中的例子似乎并没有很好地工作:

let str = "I have a cat, a catch and a cathy."; const mapObj = { 凯茜:“猫”, 猫:“抓”, 抓住:“凯蒂” }; str = str.replace(/\b(?:cathy|cat|catch)\b/gi, matched => mapObj[matched]); console.log (str);

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