node.js 用多个其他字符串替换多个字符串
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Replace multiple strings with multiple other strings
提问by Anderson Green
I'm trying to replace multiple words in a string with multiple other words. The string is "I have a cat, a dog, and a goat."
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”。
However, this does not produce "I have a dog, a goat, and a cat", but instead it produces "I have a cat, a cat, and a cat". Is it possible to replace multiple strings with multiple other strings at the same time in JavaScript, so that the correct result will be produced?
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".
回答by Ben McCormick
Specific Solution
具体解决方案
You can use a function to replace each one.
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];
});
Generalizing it
概括它
If you want to dynamically maintain the regex and just add future exchanges to the map, you can do this
如果您想动态维护正则表达式并仅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");
to generate the regex. So then it would look like this
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];
});
And to add or change any more replacements you could just edit the map.?
要添加或更改更多替代品，您只需编辑地图即可。？
Making it Reusable
使其可重用
If you want this to be a general pattern you could pull this out to a function like this
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()];
});
}
So then you could just pass the str and a map of the replacements you want to the function and it would return the transformed string.
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。
To ensure Object.keys works in older browsers, add a polyfill eg from MDNor Es5.
回答by Iian
This may not meet your exact need in this instance, but I've found this a useful way to replace multiple parameters in strings, as a general solution. It will replace all instances of the parameters, no matter how many times they are referenced:
在这种情况下，这可能无法满足您的确切需求，但我发现这是一种替换字符串中多个参数的有用方法，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
}
You would invoke it as follows:
您将按如下方式调用它：
var person = '{title} {first} {last}'.fmt({ title: 'Agent', first: 'Hyman', last: 'Bauer' });
// person = 'Agent Hyman Bauer'
回答by Quentin 2
Use numbered items to prevent replacing again. eg
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"];
then
然后
str.replace(/%(\d+)/g, (_, n) => pets[+n-1])
How it works:- %\d+ finds the numbers which come after a %. The brackets capture the number.
它是如何工作的：- %\d+ 查找 % 之后的数字。括号捕获数字。
This number (as a string) is the 2nd parameter, n, to the lambda function.
这个数字（作为字符串）是 lambda 函数的第二个参数 n。
The +n-1 converts the string to the number then 1 is subtracted to index the pets array.
+n-1 将字符串转换为数字，然后减去 1 以索引 pets 数组。
The %number is then replaced with the string at the array index.
然后将 %number 替换为数组索引处的字符串。
The /g causes the lambda function to be called repeatedly with each number which is then replaced with a string from the array.
/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。
In modern JavaScript:-
在现代 JavaScript 中：-
replace_n=(str,...ns)=>str.replace(/%(\d+)/g,(_,n)=>ns[n-1])
回答by Jo?o Paulo
This worked for me:
这对我有用：
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"
回答by Emmanuel N K
using Array.prototype.reduce():
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
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回答by Hyman
Just in case someone is wondering why the original poster's solution is not working:
以防万一有人想知道为什么原始海报的解决方案不起作用：
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."
回答by Naresh Kumar
You can find and replace string using delimiters.
您可以使用分隔符查找和替换字符串。
var obj = {
'firstname': 'John',
'lastname': 'Doe'
}
var text = "My firstname is {firstname} and my lastname is {lastname}"
console.log(mutliStringReplace(obj,text))
function mutliStringReplace(object, string) {
var val = string
var entries = Object.entries(object);
entries.forEach((para)=> {
var find = '{' + para[0] + '}'
var regExp = new RegExp(find,'g')
val = val.replace(regExp, para[1])
})
return val;
}回答by Anand Kumar Singh
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."
回答by Kodie Grantham
With my replace-oncepackage, you could do the following:
使用我的替换一次包，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.'
回答by KARTHIKEYAN.A
user regular function to define the pattern to replace and then use replace function to work on input string,
用户常规函数定义要替换的模式，然后使用替换函数处理输入字符串，
var i = new RegExp('"{','g'),
j = new RegExp('}"','g'),
k = data.replace(i,'{').replace(j,'}');
