我正在尝试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”。是否可以在 JavaScript 中同时用多个其他字符串替换多个字符串，从而产生正确的结果？
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".
您可以使用一个函数来替换每个函数。
var str = "I have a cat, a dog, and a goat.";
var mapObj = {
cat:"dog",
dog:"goat",
goat:"cat"
};
str = str.replace(/cat|dog|goat/gi, function(matched){
return mapObj[matched];
});
如果您想动态维护正则表达式并只将未来的交换添加到地图中，您可以这样做
new RegExp(Object.keys(mapObj).join("|"),"gi");
生成正则表达式。那么它看起来像这样
var mapObj = {cat:"dog",dog:"goat",goat:"cat"};
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
str = str.replace(re, function(matched){
return mapObj[matched];
});
要添加或更改更多替代品，您只需编辑地图即可。
如果您希望这是一个通用模式，您可以将其提取到这样的函数中
function replaceAll(str,mapObj){
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
return str.replace(re, function(matched){
return mapObj[matched.toLowerCase()];
});
}
因此，您只需将 str 和您想要的替换映射传递给函数，它就会返回转换后的字符串。
作为对以下问题的回答：
寻找最新的答案
如果您在当前示例中使用“单词”，您可以使用非捕获组扩展Ben McCormick的答案，并\b在左侧和右侧添加单词边界以防止部分匹配。
\b(?:cathy|cat|catch)\b
\b 防止部分匹配的单词边界(?: 非捕获组
cathy|cat|catch 匹配其中一种选择) 关闭非捕获组\b 防止部分匹配的单词边界原始问题的示例：
let str = "I have a cat, a dog, and a goat.";
const mapObj = {
cat: "dog",
dog: "goat",
goat: "cat"
};
str = str.replace(/\b(?:cat|dog|goat)\b/gi, matched => mapObj[matched]);
console.log(str);评论中的示例似乎效果不佳：
let str = "I have a cat, a catch, and a cathy.";
const mapObj = {
cathy: "cat",
cat: "catch",
catch: "cathy"
};
str = str.replace(/\b(?:cathy|cat|catch)\b/gi, matched => mapObj[matched]);
console.log(str);使用编号的项目以防止再次更换。例如
let str = "I have a %1, a %2, and a %3";
let pets = ["dog","cat", "goat"];
然后
str.replace(/%(\d+)/g, (_, n) => pets[+n-1])
它是如何工作的：- %\d+ 查找 % 之后的数字。括号捕获数字。
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 pets 数组。
然后将 %number 替换为数组索引处的字符串。
/g 导致对每个数字重复调用 lambda 函数，然后将其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])
在这种情况下，这可能无法满足您的确切需求，但作为通用解决方案，我发现这是一种替换字符串中多个参数的有用方法。它将替换参数的所有实例，无论它们被引用多少次：
String.prototype.fmt = function (hash) {
var string = this, key; for (key in hash) string = string.replace(new RegExp('\\{' + key + '\\}', 'gm'), hash[key]); return string
}
您将按如下方式调用它：
var person = '{title} {first} {last}'.fmt({ title: 'Agent', first: 'Jack', last: 'Bauer' });
// person = 'Agent Jack Bauer'
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)
例子
// ///////////// 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