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
<!DOCTYPE html>
<html>
<body>
<p id="demo">Mr Blue
has a blue house and a blue car.</p>
<button onclick="myFunction()">Try it</button>
<script>
function myFunction() {
var str = document.getElementById("demo").innerHTML;
var res = str.replace(/\n| |car/gi, function myFunction(x){
if(x=='\n'){return x='<br>';}
if(x==' '){return x=' ';}
if(x=='car'){return x='BMW'}
else{return x;}//must need
});
document.getElementById("demo").innerHTML = res;
}
</script>
</body>
</html>
其他回答
在这个实例中，这可能不能满足您的确切需求，但我发现这是一种有用的方法，可以替换字符串中的多个参数，作为通用解决方案。它将替换参数的所有实例，无论它们被引用了多少次:
String.prototype.fmt = function (hash) {
var string = this, key; for (key in hash) string = string.replace(new RegExp('\\{' + key + '\\}', 'gm'), hash[key]); return string
}
你可以这样调用它:
var person = '{title} {first} {last}'.fmt({ title: 'Agent', first: 'Jack', last: 'Bauer' });
// person = 'Agent Jack Bauer'
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
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);
具体的解决方案
您可以使用一个函数来替换每一个。
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];
});
jsfiddle例子
概括它
如果您想动态地维护正则表达式，并且只是将未来的交换添加到映射中，您可以这样做
new RegExp(Object.keys(mapObj).join("|"),"gi");
生成正则表达式。就像这样
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];
});
要添加或更改任何替换，您只需编辑地图。
摆弄动态正则表达式
可重复使用
如果你想让它成为一般形式你可以把它变成这样一个函数
function replaceAll(str,mapObj){
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
return str.replace(re, function(matched){
return mapObj[matched.toLowerCase()];
});
}
然后你可以把str和你想要的替换的映射传递给函数它会返回转换后的字符串。
摆弄函数
确保对象。key适用于旧的浏览器，添加一个填充，例如从MDN或Es5。
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; ｝
