Replace multiple strings with multiple other strings

Question

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."

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?

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".

Answer 1

function strCleaner(realArr,fakeArr,str){

   var i = 0;

   for (; i < fakeArr.length; i++) {
     str = str.replace(fakeArr[i],realArr[i]);
     if(str.includes(fakeArr[i])) i--;
   }
   console.log(str);
   return str;
}
var realArr = ["rezzan","fatih","busra","emir"];
var fakeArr = ["1f","1m","2f","2m"];
var str = "mama is 1f 1f 1f, daddy 1m, daughter is 2f, son is 2m";

strCleaner(realArr,fakeArr,str);

It could be a map or an object but it's very clean and clear.

Answer 2

using Array.prototype.reduce():

UPDATED (much better) answer (using object): This function will replace all occurrences and is case insensitive

/**
 * 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

ORIGINAL answer (using array of objects):

    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

Answer 3

You can find and replace string using delimiters.

var obj = {
  'firstname': 'John',
  'lastname': 'Doe'
}

var text = "Hello {firstname}, Your firstname is {firstname} and 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;
}

Answer 4

const str = 'Thanks for contributing an answer to Stack Overflow!'
    const substr = ['for', 'to']

    function boldString(str, substr) {
        let boldStr
        boldStr = str
        substr.map(e => {
                const strRegExp = new RegExp(e, 'g');
                boldStr= boldStr.replace(strRegExp, `<strong>${e}</strong>`);
            }
        )
        return boldStr
}

Answer 5

Try my solution. feel free to improve

function multiReplace(strings, regex, replaces) {
  return str.replace(regex, function(x) {
    // check with replaces key to prevent error, if false it will return original value
    return Object.keys(replaces).includes(x) ? replaces[x] : x;
  });
}
var str = "I have a Cat, a dog, and a goat.";
//(json) use value to replace the key
var replaces = {
  'Cat': 'dog',
  'dog': 'goat',
  'goat': 'cat',
}
console.log(multiReplace(str, /Cat|dog|goat/g, replaces))

Answer 6

you can try this. buy not smart.

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

Out put: I have a dog, a goat, and a cat.

Answer 7

one possible solution could be by using the mapper expression function.

const regex = /(?:cat|dog|goat)/gmi;
const str = `I have a cat, a dog, and a goat.`;

let mapper = (key) => {
  switch (key) {
    case "cat":
      return "dog"
    case "dog":
      return "goat";
    case "goat":
      return "cat"
  }
}
let result = str.replace(regex, mapper);

console.log('Substitution result: ', result);
//Substitution result1:  I have a dog, a goat, and a cat.

Answer 8

We can also use split() and join() methods:

var str = "I have a cat, a dog, and a goat.";

str=str.split("cat").map(x => {return x.split("dog").map(y => {return y.split("goat").join("cat");}).join("goat");}).join("dog");

console.log(str);

Answer 9

This solution can be adapted to only replace whole words - so for example, "catch", "ducat" or "locator" wouldn't be found when searching for "cat". This can be done by using negative lookbehind (?<!\w) and negative lookahead (?!\w) on word characters before and after each word in the regular expression:

(?<!\w)(cathy|cat|ducat|locator|catch)(?!\w)

JSFiddle demo: http://jsfiddle.net/mfkv9r8g/1/

Answer 10

As an answer to:

looking for an up-to-date answer

If you are using "words" as in your current example, you might extend the answer of Ben McCormick using a non capture group and add word boundaries \b at the left and at the right to prevent partial matches.

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

• \b A word boundary to prevent a partial match
• (?: Non capture group
  • cathy|cat|catch match one of the alternatives
• ) Close non capture group
• \b A word boundary to prevent a partial match

Example for the original question:

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);

Example for the example in the comments that not seems to be working well:

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);

Answer 11

• I have modified Ben McCormick's answer to work with your new test case.
• I simply added word boundaries to the regular expression:

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

"Run code snippet" to see the results below:

var str = "I have a cat, a catch, and a cathy.";
var mapObj = {
   cathy:"cat",
   cat:"catch",
   catch:"cathy"
};
str = str.replace(/\b(cathy|cat|catch)\b/gi, function(matched){
  return mapObj[matched];
});

console.log(str);

Answer 12

NOTE!

If you are using dynamically provided mapping, NONE of the solutions here are sufficient enough!

In this case, there are two ways to solve this problem, (1) using split-join technique, (2) using Regex with special character escaping technique.

1. This one is a split-join technique, which is much more faster than the other one (at least 50% faster):

var str = "I have {abc} a c|at, a d(og, and a g[oat] {1} {7} {11."
var mapObj = {
   'c|at': "d(og",
   'd(og': "g[oat",
   'g[oat]': "c|at",
};
var entries = Object.entries(mapObj);
console.log(
  entries
    .reduce(
      // Replace all the occurrences of the keys in the text into an index placholder using split-join
      (_str, [key], i) => _str.split(key).join(`{${i}}`), 
      // Manipulate all exisitng index placeholder -like formats, in order to prevent confusion
      str.replace(/\{(?=\d+\})/g, '{-')
    )
    // Replace all index placeholders to the desired replacement values
    .replace(/\{(\d+)\}/g, (_,i) => entries[i][1])
    // Undo the manipulation of index placeholder -like formats
    .replace(/\{-(?=\d+\})/g, '{')
);

2. This one, is the Regex special character escaping technique, which also works, but much slower:

var str = "I have a c|at, a d(og, and a g[oat]."
var mapObj = {
   'c|at': "d(og",
   'd(og': "g[oat",
   'g[oat]': "c|at",
};
console.log(
  str.replace(
    new RegExp(
      // Convert the object to array of keys
      Object.keys(mapObj)
        // Escape any special characters in the search key
        .map(key => key.replace(/[-[\]{}()*+?.,\\^$|#\s]/g, '\\$&'))
        // Create the Regex pattern
        .join('|'), 
      // Additional flags can be used. Like `i` - case-insensitive search
      'g'
    ), 
    // For each key found, replace with the appropriate value
    match => mapObj[match]
  )
);

The advantage of the latter, is that it can also work with case-insensitive search.

Answer 13

All solutions work great, except when applied in programming languages that closures (e.g. Coda, Excel, Spreadsheet's REGEXREPLACE).

Two original solutions of mine below use only 1 concatenation and 1 regex.

Method #1: Lookup for replacement values

The idea is to append replacement values if they are not already in the string. Then, using a single regex, we perform all needed replacements:

var str = "I have a cat, a dog, and a goat.";
str = (str+"||||cat,dog,goat").replace(
   /cat(?=[\s\S]*(dog))|dog(?=[\s\S]*(goat))|goat(?=[\s\S]*(cat))|\|\|\|\|.*$/gi, "$1$2$3");
document.body.innerHTML = str;

Explanations:

• cat(?=[\s\S]*(dog)) means that we look for "cat". If it matches, then a forward lookup will capture "dog" as group 1, and "" otherwise.
• Same for "dog" that would capture "goat" as group 2, and "goat" that would capture "cat" as group 3.
• We replace with "$1$2$3" (the concatenation of all three groups), which will always be either "dog", "cat" or "goat" for one of the above cases
• If we manually appended replacements to the string like str+"||||cat,dog,goat", we remove them by also matching \|\|\|\|.*$, in which case the replacement "$1$2$3" will evaluate to "", the empty string.

Method #2: Lookup for replacement pairs

One problem with Method #1 is that it cannot exceed 9 replacements at a time, which is the maximum number of back-propagation groups. Method #2 states not to append just replacement values, but replacements directly:

var str = "I have a cat, a dog, and a goat.";
str = (str+"||||,cat=>dog,dog=>goat,goat=>cat").replace(
   /(\b\w+\b)(?=[\s\S]*,\1=>([^,]*))|\|\|\|\|.*$/gi, "$2");
document.body.innerHTML = str;

Explanations:

• (str+"||||,cat=>dog,dog=>goat,goat=>cat") is how we append a replacement map to the end of the string.
• (\b\w+\b) states to "capture any word", that could be replaced by "(cat|dog|goat) or anything else.
• (?=[\s\S]*...) is a forward lookup that will typically go to the end of the document until after the replacement map.
  • ,\1=> means "you should find the matched word between a comma and a right arrow"
  • ([^,]*) means "match anything after this arrow until the next comma or the end of the doc"
• |\|\|\|\|.*$ is how we remove the replacement map.

Answer 14

by using prototype function we can replace easily by passing object with keys and values and replacable text

String.prototype.replaceAll=function(obj,keydata='key'){
 const keys=keydata.split('key');
 return Object.entries(obj).reduce((a,[key,val])=> a.replace(`${keys[0]}${key}${keys[1]}`,val),this)
}

const data='hids dv sdc sd ${yathin} ${ok}'
console.log(data.replaceAll({yathin:12,ok:'hi'},'${key}'))

Answer 15

You can use https://www.npmjs.com/package/union-replacer for this purpose. It is basically a string.replace(regexp, ...) counterpart, which allows multiple replaces to happen in one pass while preserving full power of string.replace(...).

Disclosure: I am the author. The library was developed to support more complex user-configurable replacements and it addresses all the problematic things like capture groups, backreferences and callback function replacements.

The solutions above are good enough for exact string replacements though.

Answer 16

Solution with Jquery (first include this file): Replace multiple strings with multiple other strings:

var replacetext = {
    "abc": "123",
    "def": "456"
    "ghi": "789"
};

$.each(replacetext, function(txtorig, txtnew) {
    $(".eng-to-urd").each(function() {
        $(this).text($(this).text().replace(txtorig, txtnew));
    });
});

Answer 17

    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."

Answer 18

Use numbered items to prevent replacing again. eg

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])

How it works:- %\d+ finds the numbers which come after a %. The brackets capture the number.

This number (as a string) is the 2nd parameter, n, to the lambda function.

The +n-1 converts the string to the number then 1 is subtracted to index the pets array.

The %number is then replaced with the string at the array index.

The /g causes the lambda function to be called repeatedly with each number which is then replaced with a string from the array.

In modern JavaScript:-

replace_n=(str,...ns)=>str.replace(/%(\d+)/g,(_,n)=>ns[n-1])

Answer 19

With my replace-once package, you could do the following:

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.'

Answer 20

I expanded on @BenMcCormicks a bit. His worked for regular strings but not if I had escaped characters or wildcards. Here's what I did

str = "[curl] 6: blah blah 234433 blah blah";
mapObj = {'\\[curl] *': '', '\\d: *': ''};


function replaceAll (str, mapObj) {

    var arr = Object.keys(mapObj),
        re;

    $.each(arr, function (key, value) {
        re = new RegExp(value, "g");
        str = str.replace(re, function (matched) {
            return mapObj[value];
        });
    });

    return str;

}
replaceAll(str, mapObj)

returns "blah blah 234433 blah blah"

This way it will match the key in the mapObj and not the matched word'

Answer 21

user regular function to define the pattern to replace and then use replace function to work on input string,

var i = new RegExp('"{','g'),
    j = new RegExp('}"','g'),
    k = data.replace(i,'{').replace(j,'}');

Answer 22

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:

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:

var person = '{title} {first} {last}'.fmt({ title: 'Agent', first: 'Jack', last: 'Bauer' });
// person = 'Agent Jack Bauer'

Answer 23

I wrote this npm package stringinject https://www.npmjs.com/package/stringinject which allows you to do the following

var string = stringInject("this is a {0} string for {1}", ["test", "stringInject"]);

which will replace the {0} and {1} with the array items and return the following string

"this is a test string for stringInject"

or you could replace placeholders with object keys and values like so:

var str = stringInject("My username is {username} on {platform}", { username: "tjcafferkey", platform: "GitHub" });

"My username is tjcafferkey on Github" 

Answer 24

Just in case someone is wondering why the original poster's solution is not working:

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."

Answer 25

<!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='&nbsp';}
if(x=='car'){return x='BMW'}
else{return x;}//must need



});

    document.getElementById("demo").innerHTML = res;
}
</script>

</body>
</html>

Answer 26

This worked for me:

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"

Answer 27

Specific Solution

You can use a function to replace each one.

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 example

Generalizing it

If you want to dynamically maintain the regex and just add future exchanges to the map, you can do this

new RegExp(Object.keys(mapObj).join("|"),"gi"); 

to generate the regex. So then it would look like this

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. 

fiddle with dynamic regex

Making it Reusable

If you want this to be a general pattern you could pull this out to a function like this

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.

fiddle with function

To ensure Object.keys works in older browsers, add a polyfill eg from MDN or Es5.

Answer 28

String.prototype.replaceSome = function() {
    var replaceWith = Array.prototype.pop.apply(arguments),
        i = 0,
        r = this,
        l = arguments.length;
    for (;i<l;i++) {
        r = r.replace(arguments[i],replaceWith);
    }
    return r;
}

/* replaceSome method for strings it takes as ,much arguments as we want and replaces all of them with the last argument we specified 2013 CopyRights saved for: Max Ahmed this is an example:

var string = "[hello i want to 'replace x' with eat]";
var replaced = string.replaceSome("]","[","'replace x' with","");
document.write(string + "<br>" + replaced); // returns hello i want to eat (without brackets)

*/

jsFiddle: http://jsfiddle.net/CPj89/