string – How do I replace a character at a particular index in JavaScript?

The Question :

598 people think this question is useful

I have a string, let’s say Hello world and I need to replace the char at index 3. How can I replace a char by specifying a index?

var str = "hello world";

I need something like


The Question Comments :
  • What’s weird is that str[0] = 'x' doesn’t seem to throw any errors, yet doesn’t have the desired effect!
  • @Michael with that you would get the index at 0, set it to ‘x’, that statement at itself would return the new value; ‘x’. but all of it doesnt change the origional, so its perfectly valid, just not what you expected. its not a reference
  • @Michael it does if "use strict" is activated: Uncaught TypeError: Cannot assign to read only property '0' of string 'hello world' (at least in webkit browsers)
  • Javascript strings are immutable, they cannot be modified “in place” so you cannot modify a single character. in fact every occurence of the same string is ONE object.

The Answer 1

669 people think this answer is useful

In JavaScript, strings are immutable, which means the best you can do is to create a new string with the changed content and assign the variable to point to it.

You’ll need to define the replaceAt() function yourself:

String.prototype.replaceAt = function(index, replacement) {
    return this.substr(0, index) + replacement + this.substr(index + replacement.length);

And use it like this:

var hello = "Hello World";
alert(hello.replaceAt(2, "!!")); // Should display He!!o World

The Answer 2

108 people think this answer is useful

There is no replaceAt function in JavaScript. You can use the following code to replace any character in any string at specified position:

function rep() {
    var str = 'Hello World';
    str = setCharAt(str,4,'a');

function setCharAt(str,index,chr) {
    if(index > str.length-1) return str;
    return str.substring(0,index) + chr + str.substring(index+1);
<button onclick="rep();">click</button>

The Answer 3

81 people think this answer is useful

You can’t. Take the characters before and after the position and concat into a new string:

var s = "Hello world";
var index = 3;
s = s.substring(0, index) + 'x' + s.substring(index + 1);

The Answer 4

35 people think this answer is useful

There are lot of answers here, and all of them are based on two methods:

  • METHOD1: split the string using two substrings and stuff the character between them
  • METHOD2: convert the string to character array, replace one array member and join it

Personally, I would use these two methods in different cases. Let me explain.

@FabioPhms: Your method was the one I initially used and I was afraid that it is bad on string with lots of characters. However, question is what’s a lot of characters? I tested it on 10 “lorem ipsum” paragraphs and it took a few milliseconds. Then I tested it on 10 times larger string – there was really no big difference. Hm.

@vsync, @Cory Mawhorter: Your comments are unambiguous; however, again, what is a large string? I agree that for 32…100kb performance should better and one should use substring-variant for this one operation of character replacement.

But what will happen if I have to make quite a few replacements?

I needed to perform my own tests to prove what is faster in that case. Let’s say we have an algorithm that will manipulate a relatively short string that consists of 1000 characters. We expect that in average each character in that string will be replaced ~100 times. So, the code to test something like this is:

var str = "... {A LARGE STRING HERE} ...";

for(var i=0; i<100000; i++)
  var n = '' + Math.floor(Math.random() * 10);
  var p = Math.floor(Math.random() * 1000);
  // replace character *n* on position *p*

I created a fiddle for this, and it’s here. There are two tests, TEST1 (substring) and TEST2 (array conversion).


  • TEST1: 195ms
  • TEST2: 6ms

It seems that array conversion beats substring by 2 orders of magnitude! So – what the hell happened here???

What actually happens is that all operations in TEST2 are done on array itself, using assignment expression like strarr2[p] = n. Assignment is really fast compared to substring on a large string, and its clear that it’s going to win.

So, it’s all about choosing the right tool for the job. Again.

The Answer 5

34 people think this answer is useful

Work with vectors is usually most effective to contact String.

I suggest the following function:

String.prototype.replaceAt=function(index, char) {
    var a = this.split("");
    a[index] = char;
    return a.join("");

Run this snippet:

String.prototype.replaceAt=function(index, char) {
    var a = this.split("");
    a[index] = char;
    return a.join("");

var str = "hello world";
str = str.replaceAt(3, "#");


The Answer 6

32 people think this answer is useful
str = str.split('');
str[3] = 'h';
str = str.join('');

The Answer 7

30 people think this answer is useful

In Javascript strings are immutable so you have to do something like

var x = "Hello world"
x = x.substring(0, i) + 'h' + x.substring(i+1);

To replace the character in x at i with ‘h’

The Answer 8

11 people think this answer is useful

One-liner using String.replace with callback (no emoji support):

// 0 - index to replace, 'f' - replacement string
'dog'.replace(/./g, (c, i) => i == 0? 'f': c)
// "fog"


//String.replace will call the callback on each pattern match
//in this case - each character
'dog'.replace(/./g, function (character, index) {
   if (index == 0) //we want to replace the first character
     return 'f'
   return character //leaving other characters the same

The Answer 9

11 people think this answer is useful

function dothis() {
  var x = document.getElementById("x").value;
  var index = document.getElementById("index").value;
  var text = document.getElementById("text").value;
  var length = document.getElementById("length").value;
  var arr = x.split("");
  arr.splice(index, length, text);
  var result = arr.join("");
  document.getElementById('output').innerHTML = result;
<input id="x" type="text" value="White Dog" placeholder="Enter Text" />
<input id="index" type="number" min="0"value="6" style="width:50px" placeholder="index" />
<input id="length" type="number" min="0"value="1" style="width:50px" placeholder="length" />
<input id="text" type="text" value="F" placeholder="New character" />
<button id="submit" onclick="dothis()">Run</button>
<p id="output"></p>

This method is good for small length strings but may be slow for larger text.

var x = "White Dog";
var arr = x.split(""); // ["W", "h", "i", "t", "e", " ", "D", "o", "g"]
arr.splice(6, 1, 'F');

  Here 6 is starting index and 1 is no. of array elements to remove and 
  final argument 'F' is the new character to be inserted. 
var result = arr.join(""); // "White Fog"

The Answer 10

4 people think this answer is useful

@CemKalyoncu: Thanks for the great answer!

I also adapted it slightly to make it more like the Array.splice method (and took @Ates’ note into consideration):

spliceString=function(string, index, numToDelete, char) {
      return string.substr(0, index) + char + string.substr(index+numToDelete);

var myString="hello world!";
spliceString(myString,myString.lastIndexOf('l'),2,'mhole'); // "hello wormhole!"

The Answer 11

4 people think this answer is useful

This works similar to Array.splice:

String.prototype.splice = function (i, j, str) {
    return this.substr(0, i) + str + this.substr(j, this.length);

The Answer 12

4 people think this answer is useful

You could try

var strArr = str.split("");

strArr[0] = 'h';

str = strArr.join("");

The Answer 13

3 people think this answer is useful

If you want to replace characters in string, you should create mutable strings. These are essentially character arrays. You could create a factory:

  function MutableString(str) {
    var result = str.split("");
    result.toString = function() {
      return this.join("");
    return result;

Then you can access the characters and the whole array converts to string when used as string:

  var x = MutableString("Hello");
  x[0] = "B"; // yes, we can alter the character
  x.push("!"); // good performance: no new string is created
  var y = "Hi, "+x; // converted to string: "Hi, Bello!"

The Answer 14

3 people think this answer is useful

You can extend the string type to include the inset method:

String.prototype.append = function (index,value) {
  return this.slice(0,index) + value + this.slice(index);

var s = "New string";
alert(s.append(4,"complete "));

Then you can call the function:

The Answer 15

3 people think this answer is useful

Generalizing Afanasii Kurakin’s answer, we have:

function replaceAt(str, index, ch) {
    return str.replace(/./g, (c, i) => i == index ? ch : c);

let str = 'Hello World';
str = replaceAt(str, 1, 'u');
console.log(str); // Hullo World

Let’s expand and explain both the regular expression and the replacer function:

function replaceAt(str, index, newChar) {
    function replacer(origChar, strIndex) {
        if (strIndex === index)
            return newChar;
            return origChar;
    return str.replace(/./g, replacer);

let str = 'Hello World';
str = replaceAt(str, 1, 'u');
console.log(str); // Hullo World

The regular expression . matches exactly one character. The g makes it match every character in a for loop. The replacer function is called given both the original character and the index of where that character is in the string. We make a simple if statement to determine if we’re going to return either origChar or newChar.

The Answer 16

2 people think this answer is useful

var str = "hello world";
var arr = [...str];
arr[0] = "H";
str = arr.join("");

The Answer 17

1 people think this answer is useful

I did a function that does something similar to what you ask, it checks if a character in string is in an array of not allowed characters if it is it replaces it with ”

    var validate = function(value){
        var notAllowed = [";","_",">","<","'","%","$","&amp;","/","|",":","=","*"];
        for(var i=0; i<value.length; i++){
            if(notAllowed.indexOf(value.charAt(i)) > -1){
               value = value.replace(value.charAt(i), "");
               value = validate(value);
      return value;

The Answer 18

1 people think this answer is useful

this is easily achievable with RegExp!

const str = 'Hello RegEx!';
const index = 11;
const replaceWith = 'p';

//'Hello RegEx!'.replace(/^(.{11})(.)/, `$1p`);
str.replace(new RegExp(`^(.{${ index }})(.)`), `$1${ replaceWith }`);

//< "Hello RegExp"

The Answer 19

1 people think this answer is useful

Here is a version I came up with if you want to style words or individual characters at their index in react/javascript.

replaceAt( yourArrayOfIndexes, yourString/orArrayOfStrings ) 

Working example:

function replaceAt(indexArray, [...string]) {
    const replaceValue = i => string[i] = <b>{string[i]}</b>;
    return string;

And here is another alternate method

function replaceAt(indexArray, [...string]) {
    const startTag = '<b>';
    const endTag = '</b>';
    const tagLetter = i => string.splice(i, 1, startTag + string[i] + endTag);
    return string.join('');

And another…

function replaceAt(indexArray, [...string]) {
    for (let i = 0; i < indexArray.length; i++) {
        string = Object.assign(string, {
          [indexArray[i]]: <b>{string[indexArray[i]]}</b>
    return string;

The Answer 20

0 people think this answer is useful

I know this is old but the solution does not work for negative index so I add a patch to it. hope it helps someone

String.prototype.replaceAt=function(index, character) {
    if(index>-1) return this.substr(0, index) + character + this.substr(index+character.length);
    else return this.substr(0, this.length+index) + character + this.substr(index+character.length);


The Answer 21

0 people think this answer is useful

Lets say you want to replace Kth index (0-based index) with 'Z'. You could use Regex to do this.

var re = var re = new RegExp("((.){" + K + "})((.){1})")
str.replace(re, "$1A$`");

The Answer 22

0 people think this answer is useful

You can use the following function to replace Character or String at a particular position of a String. To replace all the following match cases use String.prototype.replaceAllMatches() function.

String.prototype.replaceMatch = function(matchkey, replaceStr, matchIndex) {
    var retStr = this, repeatedIndex = 0;
    for (var x = 0; (matchkey != null) &amp;&amp; (retStr.indexOf(matchkey) > -1); x++) {
        if (repeatedIndex == 0 &amp;&amp; x == 0) {
            repeatedIndex = retStr.indexOf(matchkey);
        } else { // matchIndex > 0
            repeatedIndex = retStr.indexOf(matchkey, repeatedIndex + 1);
        if (x == matchIndex) {
            retStr = retStr.substring(0, repeatedIndex) + replaceStr + retStr.substring(repeatedIndex + (matchkey.length));
            matchkey = null; // To break the loop.
    return retStr;


var str = "yash yas $dfdas.**";

console.log('Index Matched replace : ', str.replaceMatch('as', '*', 2) );
console.log('Index Matched replace : ', str.replaceMatch('y', '~', 1) );


Index Matched replace :  yash yas $dfd*.**
Index Matched replace :  yash ~as $dfdas.**

The Answer 23

-1 people think this answer is useful

Here is my solution using the ternary and map operator. More readable, maintainable end easier to understand if you ask me.

It is more into es6 and best practices.

function replaceAt() {
  const replaceAt = document.getElementById('replaceAt').value;

  const str = 'ThisIsATestStringToReplaceCharAtSomePosition';
  const newStr = Array.from(str).map((character, charIndex) => charIndex === (replaceAt - 1) ? '' : character).join('');

  console.log(`New string: ${newStr}`);
<input type="number" id="replaceAt" min="1" max="44" oninput="replaceAt()"/>

The Answer 24

-3 people think this answer is useful

The methods on here are complicated. I would do it this way:

var myString = "this is my string";
myString = myString.replace(myString.charAt(number goes here), "insert replacement here");

This is as simple as it gets.

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