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

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])
    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."

我修改了本·麦考密克的答案以配合你的新测试用例。 我只是在正则表达式中添加了单词边界:

/\b(cathy|cat|catch)\b/gi

“运行代码片段”可以看到下面的结果:

var str = "我有一只猫,一个catch,和一个cathy."; var mapObj = { 凯茜:“猫”, 猫:“抓”, 抓住:“凯蒂” }; STR = STR .replace(/\b(cathy|cat|catch)\b/gi, function(matched){ 返回mapObj(匹配); }); console.log (str);

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.'

以防有人想知道为什么原来海报上的解决方案不管用:

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."