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

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

str.replaceAt(0,"h");
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.

C
Cem Kalyoncu

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.substring(0, index) + replacement + this.substring(index + replacement.length);
}

And use it like this:

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

Note that it's generally not a good idea to extend base JavaScript classes. Use a plain utility function instead.
I must ask why? Prototype support is there for this purpose.
@CemKalyoncu: It can make code play bad with other libraries. But I haven't ever been bitten by that myself.
@alex: you are right about that, you really need to check if that function exists before doing it. But still even if it does, it should do exactly the same operation.
I disagree, even if you detect whether the function exists a later library may still use/create a similar function -- 2 bad ideas making a worse one. In general, just never play around with other peoples code (include the core system). Create a new function.
Z
Zachary Christopoulos

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'); alert(str); } function setCharAt(str,index,chr) { if(index > str.length-1) return str; return str.substring(0,index) + chr + str.substring(index+1); }


I prefere this solution because you can replace a single character with it. The solution that has the most upvotes doesn't work if you want to replace only one character. It only works when replacing two characters! Also, substr should be replaced with substring as mentioned in many other comments.
In my tests this method was quickest by far... and even faster if you don't need to verify that [index] isn't bigger than the string length. Modified: function setCharAt(str,idx,newChr){ return str.substring(0,idx)+newChr+str.substring(idx+1);}
@ScottTesler - This answer uses substring, not substr... and where in your link does it say that either function is "legacy" or should be avoided? ...substring and substr serve different purposes.
@ashleedawg It looks like it's no longer deprecated.
z
zendka

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);

Actually, yes, you can, but it would be an overkill. You could set up a regular expression to skip the first index - 1 characters and match the character at the index. You can use String.replace with that regular expression to do a direct, in-place replacement. But it's an overkill. So, in practice you can't. But in theory, you can.
@Ates: The replace function doesn't do an in-place replacement, it creates a new string and returns it.
MDN suggests to use String.prototype.substring over String.prototype.substr.
c
colllin
str = str.split('');
str[3] = 'h';
str = str.join('');

Clear and simple, but currently won't work with emoji (such as 😀) when using split and join
With es6 I think that Array.from(str) might be a better option now, but I came here with that knowledge and learned something just as concise from you so thank you!
@DavidKamer I'm late to the party, but it seems Array.from is significantly slower than str.split. blog.shovonhasan.com/…
@Green if your string contains emojis it's probably a user input, hence I don't think there are obvious reasons to change a character in the string using this method, if you want to change a part in an input transformation scenario you should use replace method with regexp instead.
@KadenSkinner It's an array, so you could str.splice(3, h.length, ...h.split('')) or str.splice.apply(str, [3, h.length].concat(h.split(''))) but this seems crazy compared to str.substring(0,3) + h + str.substring(5).
O
OzrenTkalcecKrznaric

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).

Results:

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.


Moved your test to jsperf, with vastly different results: jsperf.com/string-replace-character. It's all about choosing the right tool for the job ;)
@colllin: I absolutely agree. So I almost literally cloned the tests from jsfiddle to jsperf. However, aside from using the right tool, you have to do it in the right way. Pay attention to the problem, we're talking about what will happen if we have to make quite a few replacements in an algorithm. The concrete example includes 10000 replacements. This means one test should include 10000 replacements. So, using jsperf, I got quite consistent results, see: jsperf.com/….
this is why i don't like javascript. such frequent, intuitive, obvious use cases aren't provisioned for. You'd think it would occur to the writers of javascript that people may want to substitute characters within strings, and that they'd decide to put in convenience methods into the language.
@colllin did something different to OzrenTkalcecKrznaric. OzrenTkalcecKrznaric made the assumption that the string is static and and the split can be optimized before the runs. Which is not necessarily true. if you have to split at every run then the substring approach is about twice-4x faster on small strings, and only faster the bigger the string gets. In my opinion this makes the substring the better implementation for most use cases, barring large numbers of changes on a large static string.
The problem is that on test2 the split and join are outside the for loop, where you compare single char replacement with a multiple one (unfair), the first method is faster for single replacements and the second one is better for chained replacements one after another
F
Fabio Phms

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, "#"); document.write(str);


this is bad for large strings, no need to create an Array with crazy amount of cells if you can just use substr to cut it...
Useful if you need to change a lot of characters, and it is simple.
I created a test on jsperf: jsperf.com/replace-character-by-index -- Substr is way faster, and consistent from 32bytes up to 100kb. I upvoted this and as much as it hurts me to say, even though this feels nicer, substr is the better choice for strings of any size.
This is quite unoptimal.
This is a good quick solution if you know you are going to use small strings. If you are okay with extending the prototypes, then you could achieve similar behavior in 2 lines: Array.prototype.replace = function(index, val) { this[index] = val; return this; }; 'word'.split('').replace(1, '0').join(''); // returns 'w0rd'
c
colllin

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'


V
Vikramaditya

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; console.log(result); } dothis();

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"

Up voted because I found the extra description to be useful and the implementation simple and powerful. Thank you! :-D
Array.from might also be a good idea for string to array conversion
Yes Array.from can also be used here. but Array.from was added in ES2015 and has no support in older browsers. Also, i personally prefer split function because it gives more freedom to play with the string patterns.
If you want to replace just 1 character you can use arr[6]='F'; instead of .splice().
What you have suggested was already suggested by @fabio-phms in the answer above. i was just looking through another point of view where we have flexibility to replace one or more characters.
A
Afanasii Kurakin

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"

Explained:

//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
})

beautifully simplistic - great option that doesn't force extending the prototype or creating new functions. Thank you!
S
Stephen Quan

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; else 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.


A
Avi Cohen

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


A
Aram Kocharyan

This works similar to Array.splice:

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

M
Mehulkumar

You could try

var strArr = str.split("");

strArr[0] = 'h';

str = strArr.join("");

Not much explanation here. Could you explain how it works? Is "0" supposed to be the index that I want to change?
@SethEden Yes, "0" is supposed to be the index that you want to change.
S
Scrimothy

@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!"

Not sure why this was voted down. It's a legit answer. It works as indicated. And it's consistent with Array.splice type functionality.
M
Magne

You could try

var strArr = str.split("");

strArr[0] = 'h';

str = strArr.join("");

Check out this function for printing steps

steps(3)
//       '#  '
//       '## '
//       '###'

function steps(n, i = 0, arr = Array(n).fill(' ').join('')) {
  if (i === n) {
    return;
  }

  str = arr.split('');
  str[i] = '#';
  str = str.join('');
  console.log(str);

  steps(n, (i = i + 1), str);
}


J
Jan Turoň

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!"

M
Madmadi

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"

As far as contemporary JavaScript goes, in 2022, this is by far the best answer.
m
mbadeveloper

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:


G
Gyum Fox

Using the spread syntax, you may convert the string to an array, assign the character at the given position, and convert back to a string:

const str = "hello world"; function replaceAt(s, i, c) { const arr = [...s]; // Convert string to array arr[i] = c; // Set char c at pos i return arr.join(''); // Back to string } // prints "hallo world" console.log(replaceAt(str, 1, 'a'));


I would be careful in putting this into a function like this. Because inevitably someone will run the function inside a loop. Which in this case will be terribly suboptimal compared to if they had to simply write the "convert to array" and "join" themselves, when they needed it. Then they'd naturally put it outside the loop.
M
MD SHAYON

You can concatenate using sub-string function at first select text before targeted index and after targeted index then concatenate with your potential char or string. This one is better

const myString = "Hello world";
const index = 3;
const stringBeforeIndex = myString.substring(0, index);
const stringAfterIndex = myString.substring(index + 1);
const replaceChar = "X";
myString = stringBeforeIndex + replaceChar + stringAfterIndex;
console.log("New string - ", myString)

or

const myString = "Hello world";
let index = 3;
myString =  myString.substring(0, index) + "X" + myString.substring(index + 1);

r
revelt

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 = [";","_",">","<","'","%","$","&","/","|",":","=","*"];
        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;
   }

M
Matt Wright

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: https://codesandbox.io/s/ov7zxp9mjq

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

k
karlo1zg

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}`); }


j
jonas kjellerup

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);

}

k
ksp

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, "$1A$`");

Any idea why is this tagged not useful?
I haven't checked if this regex works, but maybe it's downvoted because it's probably a lot faster to create a new string, than compiling a state machine (regular expression), which is very expensive.
Y
Yash

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) && (retStr.indexOf(matchkey) > -1); x++) {
        if (repeatedIndex == 0 && 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;
};

Test:

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

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

Output:

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

A
Alexandre Daubricourt

My safe approach with negative indexes

/**
 * @param {string} str 
 * @param {number} index 
 * @param {string} replacement 
 * @returns {string}
 */
static replaceAt (str, index, replacement)
{
    if (index < 0) index = str.length + index
    if (index < 0 || index >= str.length) throw new Error(`Index (${index}) out of bounds "${str}"`)
    return str.substring(0, index) + replacement + str.substring(index + 1)
}

Use it like that:

replaceAt('my string', -1, 'G') // 'my strinG'
replaceAt('my string', 2, 'yy') // 'myyystring'
replaceAt('my string', 22, 'yy') // Uncaught Error: Index (22) out of bounds "my string"

m
mikey

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.


Not actually. In fact it will replace the first occurrence of the character. Still wrong.