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

Question

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");

Answer 1

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

Answer 2

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);
}

<button onclick="rep();">click</button>

Answer 3

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);

Answer 4

str = str.split('');
str[3] = 'h';
str = str.join('');

Answer 5

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.

Answer 6

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);

Answer 7

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'

Answer 8

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();

<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" />
<br>
<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"

Answer 9

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
})

Answer 10

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.

Answer 11

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

Answer 12

This works similar to Array.splice:

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

Answer 13

You could try

var strArr = str.split("");

strArr[0] = 'h';

str = strArr.join("");

Answer 14

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"

Answer 15

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);
}

Answer 16

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'));

Answer 17

@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!"

Answer 18

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:

Answer 19

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!"

Answer 20

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);

Answer 21

My safe approach with negative indexes:

/**
 * @param {string} str
 * @param {number} index
 * @param {string} replacement
 * @returns {string}
 */
function 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 this:

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"

Answer 22

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;
   }

Answer 23

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;
}

Answer 24

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()"/>

Answer 25

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, "$1A$`");

Answer 26

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.**

Answer 27

I se this to make a string proper case, that is, the first letter is Upper Case and all the rest are lower case:

function toProperCase(someString){
    
    return someString.charAt(0).toUpperCase().concat(someString.toLowerCase().substring(1,someString.length));
    
};

This first thing done is to ensure ALL the string is lower case - someString.toLowerCase()

then it converts the very first character to upper case -someString.charAt(0).toUpperCase()

then it takes a substring of the remaining string less the first character -someString.toLowerCase().substring(1,someString.length))

then it concatenates the two and returns the new string -someString.charAt(0).toUpperCase().concat(someString.toLowerCase().substring(1,someString.length))

New parameters could be added for the replacement character index and the replacement character, then two substrings formed and the indexed character replaced then concatenated in much the same way.

Answer 28

The solution does not work for negative index so I add a patch to it.

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);
    
}

Answer 29

"hello world".replace(/(.{3})./, "$1h")
// 'helho world'

Answer 30

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.