How do I replace all occurrences of a string in JavaScript?

Question

Given a string:

s = "Test abc test test abc test test test abc test test abc";

This seems to only remove the first occurrence of abc in the string above:

s = s.replace('abc', '');

How do I replace all occurrences of it?

Answer 1

As of August 2020: Modern browsers have support for the String.replaceAll() method defined by the ECMAScript 2021 language specification.

---

For older/legacy browsers:

function escapeRegExp(string) {
  return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}

function replaceAll(str, find, replace) {
  return str.replace(new RegExp(escapeRegExp(find), 'g'), replace);
}

Here is how this answer evolved:

str = str.replace(/abc/g, '');

In response to comment "what's if 'abc' is passed as a variable?":

var find = 'abc';
var re = new RegExp(find, 'g');

str = str.replace(re, '');

In response to Click Upvote's comment, you could simplify it even more:

function replaceAll(str, find, replace) {
  return str.replace(new RegExp(find, 'g'), replace);
}

Note: Regular expressions contain special (meta) characters, and as such it is dangerous to blindly pass an argument in the find function above without pre-processing it to escape those characters. This is covered in the Mozilla Developer Network's JavaScript Guide on Regular Expressions, where they present the following utility function (which has changed at least twice since this answer was originally written, so make sure to check the MDN site for potential updates):

function escapeRegExp(string) {
  return string.replace(/[.*+?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
}

So in order to make the replaceAll() function above safer, it could be modified to the following if you also include escapeRegExp:

function replaceAll(str, find, replace) {
  return str.replace(new RegExp(escapeRegExp(find), 'g'), replace);
}

Answer 2

For the sake of completeness, I got to thinking about which method I should use to do this. There are basically two ways to do this as suggested by the other answers on this page.

Note: In general, extending the built-in prototypes in JavaScript is generally not recommended. I am providing as extensions on the String prototype simply for purposes of illustration, showing different implementations of a hypothetical standard method on the String built-in prototype.

---

Regular Expression Based Implementation

String.prototype.replaceAll = function(search, replacement) {
    var target = this;
    return target.replace(new RegExp(search, 'g'), replacement);
};

Split and Join (Functional) Implementation

String.prototype.replaceAll = function(search, replacement) {
    var target = this;
    return target.split(search).join(replacement);
};

---

Not knowing too much about how regular expressions work behind the scenes in terms of efficiency, I tended to lean toward the split and join implementation in the past without thinking about performance. When I did wonder which was more efficient, and by what margin, I used it as an excuse to find out.

On my Chrome Windows 8 machine, the regular expression based implementation is the fastest, with the split and join implementation being 53% slower. Meaning the regular expressions are twice as fast for the lorem ipsum input I used.

Check out this benchmark running these two implementations against each other.

---

As noted in the comment below by @ThomasLeduc and others, there could be an issue with the regular expression-based implementation if search contains certain characters which are reserved as special characters in regular expressions. The implementation assumes that the caller will escape the string beforehand or will only pass strings that are without the characters in the table in Regular Expressions (MDN).

MDN also provides an implementation to escape our strings. It would be nice if this was also standardized as RegExp.escape(str), but alas, it does not exist:

function escapeRegExp(str) {
  return str.replace(/[.*+?^${}()|[\]\\]/g, "\\$&"); // $& means the whole matched string
}

We could call escapeRegExp within our String.prototype.replaceAll implementation, however, I'm not sure how much this will affect the performance (potentially even for strings for which the escape is not needed, like all alphanumeric strings).

Answer 3

In the latest versions of most popular browsers, you can use replaceAll as shown here:

let result = "1 abc 2 abc 3".replaceAll("abc", "xyz");
// `result` is "1 xyz 2 xyz 3"

But check Can I use or another compatibility table first to make sure the browsers you're targeting have added support for it first.

---

For Node.js and compatibility with older/non-current browsers:

Note: Don't use the following solution in performance critical code.

As an alternative to regular expressions for a simple literal string, you could use

str = "Test abc test test abc test...".split("abc").join("");

The general pattern is

str.split(search).join(replacement)

This used to be faster in some cases than using replaceAll and a regular expression, but that doesn't seem to be the case anymore in modern browsers.

Benchmark: https://jsben.ch/TZYzj

Conclusion:

If you have a performance-critical use case (e.g., processing hundreds of strings), use the regular expression method. But for most typical use cases, this is well worth not having to worry about special characters.

Answer 4

Here's a string prototype function based on the accepted answer:

String.prototype.replaceAll = function(find, replace) {
    var str = this;
    return str.replace(new RegExp(find, 'g'), replace);
};

If your find contains special characters then you need to escape them:

String.prototype.replaceAll = function(find, replace) {
    var str = this;
    return str.replace(new RegExp(find.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), 'g'), replace);
};

Fiddle: http://jsfiddle.net/cdbzL/

Answer 5

These are the most common and readable methods.

var str = "Test abc test test abc test test test abc test test abc"

Method 1:

str = str.replace(/abc/g, "replaced text");

Method 2:

str = str.split("abc").join("replaced text");

Method 3:

str = str.replace(new RegExp("abc", "g"), "replaced text");

Method 4:

while(str.includes("abc")){
   str = str.replace("abc", "replaced text");
}

Output:

console.log(str);
// Test replaced text test test replaced text test test test replaced text test test replaced text

Answer 6

Use word boundaries (\b)

'a cat is not a caterpillar'.replace(/\bcat\b/gi,'dog');
//"a dog is not a caterpillar"

This is a simple regex that avoids replacing parts of words in most cases. However, a dash - is still considered a word boundary. So conditionals can be used in this case to avoid replacing strings like cool-cat:

'a cat is not a cool-cat'.replace(/\bcat\b/gi,'dog');//wrong
//"a dog is not a cool-dog" -- nips
'a cat is not a cool-cat'.replace(/(?:\b([^-]))cat(?:\b([^-]))/gi,'$1dog$2');
//"a dog is not a cool-cat"

---

Basically, this question is the same as the question here: Replace " ' " with " '' " in JavaScript

Regexp isn't the only way to replace multiple occurrences of a substring, far from it. Think flexible, think split!

var newText = "the cat looks like a cat".split('cat').join('dog');

Alternatively, to prevent replacing word parts—which the approved answer will do, too! You can get around this issue using regular expressions that are, I admit, somewhat more complex and as an upshot of that, a tad slower, too:

var regText = "the cat looks like a cat".replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");

The output is the same as the accepted answer, however, using the /cat/g expression on this string:

var oops = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/cat/g,'dog');
//returns "the dog looks like a dog, not a dogerpillar or cooldog" ??

Oops indeed, this probably isn't what you want. What is, then? IMHO, a regex that only replaces 'cat' conditionally (i.e., not part of a word), like so:

var caterpillar = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
//return "the dog looks like a dog, not a caterpillar or coolcat"

My guess is, this meets your needs. It's not foolproof, of course, but it should be enough to get you started. I'd recommend reading some more on these pages. This'll prove useful in perfecting this expression to meet your specific needs.

• RegExp (regular expression) object

• Regular-Expressions.info

---

Here is an example of .replace used with a callback function. In this case, it dramatically simplifies the expression and provides even more flexibility, like replacing with correct capitalisation or replacing both cat and cats in one go:

'Two cats are not 1 Cat! They\'re just cool-cats, you caterpillar'
   .replace(/(^|.\b)(cat)(s?\b.|$)/gi,function(all,char1,cat,char2)
    {
       // Check 1st, capitalize if required
       var replacement = (cat.charAt(0) === 'C' ? 'D' : 'd') + 'og';
       if (char1 === ' ' && char2 === 's')
       { // Replace plurals, too
           cat = replacement + 's';
       }
       else
       { // Do not replace if dashes are matched
           cat = char1 === '-' || char2 === '-' ? cat : replacement;
       }
       return char1 + cat + char2;//return replacement string
    });
//returns:
//Two dogs are not 1 Dog! They're just cool-cats, you caterpillar

Answer 7

Match against a global regular expression:

anotherString = someString.replace(/cat/g, 'dog');

Answer 8

For replacing a single time, use:

var res = str.replace('abc', "");

For replacing multiple times, use:

var res = str.replace(/abc/g, "");

Answer 9

str = str.replace(/abc/g, '');

Or try the replaceAll method, as recommended in this answer:

str = str.replaceAll('abc', '');

or:

var search = 'abc';
str = str.replaceAll(search, '');

EDIT: Clarification about replaceAll availability

The replaceAll method is added to String's prototype. This means it will be available for all string objects/literals.

Example:

var output = "test this".replaceAll('this', 'that'); // output is 'test that'.
output = output.replaceAll('that', 'this'); // output is 'test this'

Answer 10

Using RegExp in JavaScript could do the job for you. Just simply do something like below code, and don't forget the /g after which standout for global:

var str ="Test abc test test abc test test test abc test test abc";
str = str.replace(/abc/g, '');

If you think of reuse, create a function to do that for you, but it's not recommended as it's only one line function. But again, if you heavily use this, you can write something like this:

String.prototype.replaceAll = String.prototype.replaceAll || function(string, replaced) {
  return this.replace(new RegExp(string, 'g'), replaced);
};

And simply use it in your code over and over like below:

var str ="Test abc test test abc test test test abc test test abc";
str = str.replaceAll('abc', '');

But as I mention earlier, it won't make a huge difference in terms of lines to be written or performance. Only caching the function may affect some faster performance on long strings and is also a good practice of DRY code if you want to reuse.

Answer 11

Say you want to replace all the 'abc' with 'x':

let some_str = 'abc def def lom abc abc def'.split('abc').join('x')
console.log(some_str) //x def def lom x x def

I was trying to think about something more simple than modifying the string prototype.

Answer 12

Use a regular expression:

str.replace(/abc/g, '');

Answer 13

Performance

Today 27.12.2019 I perform tests on macOS v10.13.6 (High Sierra) for the chosen solutions.

Conclusions

• The str.replace(/abc/g, ''); (C) is a good cross-browser fast solution for all strings.
• Solutions based on split-join (A,B) or replace (C,D) are fast
• Solutions based on while (E,F,G,H) are slow - usually ~4 times slower for small strings and about ~3000 times (!) slower for long strings
• The recurrence solutions (RA,RB) are slow and do not work for long strings

I also create my own solution. It looks like currently it is the shortest one which does the question job:

str.split`abc`.join``

str = "Test abc test test abc test test test abc test test abc";
str = str.split`abc`.join``

console.log(str);

Details

The tests were performed on Chrome 79.0, Safari 13.0.4 and Firefox 71.0 (64 bit). The tests RA and RB use recursion. Results

Short string - 55 characters

You can run tests on your machine HERE. Results for Chrome:

Long string: 275 000 characters

The recursive solutions RA and RB gives

RangeError: Maximum call stack size exceeded

For 1M characters they even break Chrome

I try to perform tests for 1M characters for other solutions, but E,F,G,H takes so much time that browser ask me to break script so I shrink test string to 275K characters. You can run tests on your machine HERE. Results for Chrome

Code used in tests

var t="Test abc test test abc test test test abc test test abc"; // .repeat(5000)
var log = (version,result) => console.log(`${version}: ${result}`);


function A(str) {
  return str.split('abc').join('');
}

function B(str) {
  return str.split`abc`.join``; // my proposition
}


function C(str) {
  return str.replace(/abc/g, '');
}

function D(str) {
  return str.replace(new RegExp("abc", "g"), '');
}

function E(str) {
  while (str.indexOf('abc') !== -1) { str = str.replace('abc', ''); }
  return str;
}

function F(str) {
  while (str.indexOf('abc') !== -1) { str = str.replace(/abc/, ''); }
  return str;
}

function G(str) {
  while(str.includes("abc")) { str = str.replace('abc', ''); }
  return str;
}

// src: https://stackoverflow.com/a/56989553/860099
function H(str)
{
    let i = -1
    let find = 'abc';
    let newToken = '';

    if (!str)
    {
        if ((str == null) && (find == null)) return newToken;
        return str;
    }

    while ((
        i = str.indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    }
    return str;
}

// src: https://stackoverflow.com/a/22870785/860099
function RA(string, prevstring) {
  var omit = 'abc';
  var place = '';
  if (prevstring && string === prevstring)
    return string;
  prevstring = string.replace(omit, place);
  return RA(prevstring, string)
}

// src: https://stackoverflow.com/a/26107132/860099
function RB(str) {
  var find = 'abc';
  var replace = '';
  var i = str.indexOf(find);
  if (i > -1){
    str = str.replace(find, replace);
    i = i + replace.length;
    var st2 = str.substring(i);
    if(st2.indexOf(find) > -1){
      str = str.substring(0,i) + RB(st2, find, replace);
    }
  }
  return str;
}




log('A ', A(t));
log('B ', B(t));
log('C ', C(t));
log('D ', D(t));
log('E ', E(t));
log('F ', F(t));
log('G ', G(t));
log('H ', H(t));
log('RA', RA(t)); // use reccurence
log('RB', RB(t)); // use reccurence

<p style="color:red">This snippet only presents codes used in tests. It not perform test itself!<p>

Answer 14

Replacing single quotes:

function JavaScriptEncode(text){
    text = text.replace(/'/g,''')
    // More encode here if required

    return text;
}

Answer 15

Using

str = str.replace(new RegExp("abc", 'g'), "");

worked better for me than the previous answers. So new RegExp("abc", 'g') creates a regular expression what matches all occurrences ('g' flag) of the text ("abc"). The second part is what gets replaced to, in your case empty string (""). str is the string, and we have to override it, as replace(...) just returns result, but not overrides. In some cases you might want to use that.

Answer 16

Loop it until number occurrences comes to 0, like this:

function replaceAll(find, replace, str) {
    while (str.indexOf(find) > -1) {
        str = str.replace(find, replace);
    }
    return str;
}

Answer 17

This is the fastest version that doesn't use regular expressions.

Revised jsperf

replaceAll = function(string, omit, place, prevstring) {
  if (prevstring && string === prevstring)
    return string;
  prevstring = string.replace(omit, place);
  return replaceAll(prevstring, omit, place, string)
}

It is almost twice as fast as the split and join method.

As pointed out in a comment here, this will not work if your omit variable contains place, as in: replaceAll("string", "s", "ss"), because it will always be able to replace another occurrence of the word.

There is another jsperf with variants on my recursive replace that go even faster (http://jsperf.com/replace-all-vs-split-join/12)!

• Update July 27th 2017: It looks like RegExp now has the fastest performance in the recently released Chrome 59.

Answer 18

If what you want to find is already in a string, and you don't have a regex escaper handy, you can use join/split:

    function replaceMulti(haystack, needle, replacement)
    {
        return haystack.split(needle).join(replacement);
    }

    someString = 'the cat looks like a cat';
    console.log(replaceMulti(someString, 'cat', 'dog'));

Answer 19

String.prototype.replaceAll - ECMAScript 2021

The new String.prototype.replaceAll() method returns a new string with all matches of a pattern replaced by a replacement. The pattern can be either a string or a RegExp, and the replacement can be either a string or a function to be called for each match.

const message = 'dog barks meow meow';
const messageFormatted = message.replaceAll('meow', 'woof')

console.log(messageFormatted);

Answer 20

function replaceAll(str, find, replace) {
  var i = str.indexOf(find);
  if (i > -1){
    str = str.replace(find, replace); 
    i = i + replace.length;
    var st2 = str.substring(i);
    if(st2.indexOf(find) > -1){
      str = str.substring(0,i) + replaceAll(st2, find, replace);
    }       
  }
  return str;
}

Answer 21

I like this method (it looks a little cleaner):

text = text.replace(new RegExp("cat","g"), "dog"); 

Answer 22

Of course in 2021 the right answer is:

String.prototype.replaceAll()

console.log(
  'Change this and this for me'.replaceAll('this','that') // Normal case
);
console.log(
  'aaaaaa'.replaceAll('aa','a') // Challenged case
);

If you don't want to deal with replace() + RegExp.

But what if the browser is from before 2020?

In this case we need polyfill (forcing older browsers to support new features) (I think for a few years will be necessary). I could not find a completely right method in answers. So I suggest this function that will be defined as a polyfill.

My suggested options for replaceAll polyfill:

replaceAll polyfill (with global-flag error) (more principled version)

if (!String.prototype.replaceAll) { // Check if the native function not exist
    Object.defineProperty(String.prototype, 'replaceAll', { // Define replaceAll as a prototype for (Mother/Any) String
        configurable: true, writable: true, enumerable: false, // Editable & non-enumerable property (As it should be)
        value: function(search, replace) { // Set the function by closest input names (For good info in consoles)
            return this.replace( // Using native String.prototype.replace()
                Object.prototype.toString.call(search) === '[object RegExp]' // IsRegExp?
                    ? search.global // Is the RegEx global?
                        ? search // So pass it
                        : function(){throw new TypeError('replaceAll called with a non-global RegExp argument')}() // If not throw an error
                    : RegExp(String(search).replace(/[.^$*+?()[{|\\]/g, "\\$&"), "g"), // Replace all reserved characters with '\' then make a global 'g' RegExp
                replace); // passing second argument
        }
    });
}

replaceAll polyfill (With handling global-flag missing by itself) (my first preference) - Why?

if (!String.prototype.replaceAll) { // Check if the native function not exist
    Object.defineProperty(String.prototype, 'replaceAll', { // Define replaceAll as a prototype for (Mother/Any) String
        configurable: true, writable: true, enumerable: false, // Editable & non-enumerable property (As it should be)
        value: function(search, replace) { // Set the function by closest input names (For good info in consoles)
            return this.replace( // Using native String.prototype.replace()
                Object.prototype.toString.call(search) === '[object RegExp]' // IsRegExp?
                    ? search.global // Is the RegEx global?
                        ? search // So pass it
                        : RegExp(search.source, /\/([a-z]*)$/.exec(search.toString())[1] + 'g') // If not, make a global clone from the RegEx
                    : RegExp(String(search).replace(/[.^$*+?()[{|\\]/g, "\\$&"), "g"), // Replace all reserved characters with '\' then make a global 'g' RegExp
                replace); // passing second argument
        }
    });
}

Minified (my first preference):

if(!String.prototype.replaceAll){Object.defineProperty(String.prototype,'replaceAll',{configurable:!0,writable:!0,enumerable:!1,value:function(search,replace){return this.replace(Object.prototype.toString.call(search)==='[object RegExp]'?search.global?search:RegExp(search.source,/\/([a-z]*)$/.exec(search.toString())[1]+'g'):RegExp(String(search).replace(/[.^$*+?()[{|\\]/g,"\\$&"),"g"),replace)}})}

Try it:

if(!String.prototype.replaceAll){Object.defineProperty(String.prototype,'replaceAll',{configurable:!0,writable:!0,enumerable:!1,value:function(search,replace){return this.replace(Object.prototype.toString.call(search)==='[object RegExp]'?search.global?search:RegExp(search.source,/\/([a-z]*)$/.exec(search.toString())[1]+'g'):RegExp(String(search).replace(/[.^$*+?()[{|\\]/g,"\\$&"),"g"),replace)}})}

console.log(
  'Change this and this for me'.replaceAll('this','that')
); // Change that and that for me

console.log(
  'aaaaaa'.replaceAll('aa','a')
); // aaa

console.log(
  '{} (*) (*) (RegEx) (*) (\*) (\\*) [reserved characters]'.replaceAll('(*)','X')
); // {} X X (RegEx) X X (\*) [reserved characters]

console.log(
  'How (replace) (XX) with $1?'.replaceAll(/(xx)/gi,'$$1')
); // How (replace) ($1) with $1?

console.log(
  'Here is some numbers 1234567890 1000000 123123.'.replaceAll(/\d+/g,'***')
); // Here is some numbers *** *** *** and need to be replaced.

console.log(
  'Remove numbers under 233: 236   229  711   200   5'.replaceAll(/\d+/g, function(m) {
    return parseFloat(m) < 233 ? '' : m
  })
); // Remove numbers under 233: 236     711

console.log(
  'null'.replaceAll(null,'x')
); // x


// The difference between My first preference and the original:
// Now in 2022 with browsers > 2020 it should throw an error (But possible it be changed in future)

//   console.log(
//      'xyz ABC abc ABC abc xyz'.replaceAll(/abc/i,'')
//   );

// Browsers < 2020:
// xyz     xyz
// Browsers > 2020
// TypeError: String.prototype.replaceAll called with a non-global RegExp

Browser support:

• Internet Explorer 9 and later (rested on Internet Explorer 11).
• All other browsers (after 2012).

The result is the same as the native replaceAll in case of the first argument input is: null, undefined, Object, Function, Date, ... , RegExp, Number, String, ...

Ref: 22.1.3.19 String.prototype.replaceAll ( searchValue, replaceValue) + RegExp Syntax

Important note: As some professionals mention it, many of recursive functions that suggested in answers, will return the wrong result. (Try them with the challenged case of the above snippet.) Maybe some tricky methods like .split('searchValue').join('replaceValue') or some well managed functions give same result, but definitely with much lower performance than native replaceAll() / polyfill replaceAll() / replace() + RegExp

---

Other methods of polyfill assignment

Naive, but supports even older browsers (be better to avoid)

For example, we can support IE7+ too, by not using Object.defineProperty() and using my old naive assignment method:

if (!String.prototype.replaceAll) {
    String.prototype.replaceAll = function(search, replace) { // <-- Naive method for assignment
        // ... (Polyfill code Here)
    }
}

And it should work well for basic uses on IE7+.
But as here @sebastian-simon explained about, that can make secondary problems in case of more advanced uses. E.g.:

for (var k in 'hi') console.log(k);
// 0
// 1
// replaceAll  <-- ?

Fully trustable, but heavy

In fact, my suggested option is a little optimistic. Like we trusted the environment (browser and Node.js), it is definitely for around 2012-2021. Also it is a standard/famous one, so it does not require any special consideration.

But there can be even older browsers or some unexpected problems, and polyfills still can support and solve more possible environment problems. So in case we need the maximum support that is possible, we can use polyfill libraries like:

https://polyfill.io/

Specially for replaceAll:

<script src="https://polyfill.io/v3/polyfill.min.js?features=String.prototype.replaceAll"></script>

Answer 23

while (str.indexOf('abc') !== -1)
{
    str = str.replace('abc', '');
}

Answer 24

The simplest way to do this without using any regular expression is split and join, like the code here:

var str = "Test abc test test abc test test test abc test test abc";
console.log(str.split('abc').join(''));

Answer 25

var str = "ff ff f f a de def";
str = str.replace(/f/g,'');
alert(str);

http://jsfiddle.net/ANHR9/

Answer 26

If the string contains a similar pattern like abccc, you can use this:

str.replace(/abc(\s|$)/g, "")

Answer 27

As of August 2020 there is a Stage 4 proposal to ECMAScript that adds the replaceAll method to String.

It's now supported in Chrome 85+, Edge 85+, Firefox 77+, Safari 13.1+.

The usage is the same as the replace method:

String.prototype.replaceAll(searchValue, replaceValue)

Here's an example usage:

'Test abc test test abc test.'.replaceAll('abc', 'foo'); // -> 'Test foo test test foo test.'

---

It's supported in most modern browsers, but there exist polyfills:

• core-js
• es-shims

It is supported in the V8 engine behind an experimental flag --harmony-string-replaceall. Read more on the V8 website.

Answer 28

The previous answers are way too complicated. Just use the replace function like this:

str.replace(/your_regex_pattern/g, replacement_string);

Example:

var str = "Test abc test test abc test test test abc test test abc";

var res = str.replace(/[abc]+/g, "");

console.log(res);

Answer 29

After several trials and a lot of fails, I found that the below function seems to be the best all-rounder when it comes to browser compatibility and ease of use. This is the only working solution for older browsers that I found. (Yes, even though old browser are discouraged and outdated, some legacy applications still make heavy use of OLE browsers (such as old Visual Basic 6 applications or Excel .xlsm macros with forms.)

Anyway, here's the simple function.

function replaceAll(str, match, replacement){
   return str.split(match).join(replacement);
}

Answer 30

If you are trying to ensure that the string you are looking for won't exist even after the replacement, you need to use a loop.

For example:

var str = 'test aabcbc';
str = str.replace(/abc/g, '');

When complete, you will still have 'test abc'!

The simplest loop to solve this would be:

var str = 'test aabcbc';
while (str != str.replace(/abc/g, '')){
   str.replace(/abc/g, '');
}

But that runs the replacement twice for each cycle. Perhaps (at risk of being voted down) that can be combined for a slightly more efficient but less readable form:

var str = 'test aabcbc';
while (str != (str = str.replace(/abc/g, ''))){}
// alert(str); alerts 'test '!

This can be particularly useful when looking for duplicate strings.
For example, if we have 'a,,,b' and we wish to remove all duplicate commas.
[In that case, one could do .replace(/,+/g,','), but at some point the regex gets complex and slow enough to loop instead.]

Answer 31

Although people have mentioned the use of regex, there's a better approach if you want to replace the text irrespective of the case of the text. Like uppercase or lowercase. Use the below syntax:

// Consider the below example
originalString.replace(/stringToBeReplaced/gi, '');

// The output will be all the occurrences removed irrespective of casing.

You can refer to the detailed example here.

Answer 32

With the regular expression i flag for case insensitive

console.log('App started'.replace(/a/g, '')) // Result: "App strted"
console.log('App started'.replace(/a/gi, '')) // Result: "pp strted"

Answer 33

Just add /g

document.body.innerHTML = document.body.innerHTML.replace('hello', 'hi');

to

// Replace 'hello' string with /hello/g regular expression.
document.body.innerHTML = document.body.innerHTML.replace(/hello/g, 'hi');

/g means global

Answer 34

The following function works for me:

String.prototype.replaceAllOccurence = function(str1, str2, ignore)
{
    return this.replace(new RegExp(str1.replace(/([\/\,\!\\\^\$\{\}\[\]\(\)\.\*\+\?\|\<\>\-\&])/g,"\\$&"),(ignore?"gi":"g")),(typeof(str2)=="string")?str2.replace(/\$/g,"$$$$"):str2);
} ;

Now call the functions like this:

"you could be a Project Manager someday, if you work like this.".replaceAllOccurence ("you", "I");

Simply copy and paste this code in your browser console to TEST.

Answer 35

In terms of performance related to the main answers, these are some online tests.

While the following are some performance tests using console.time() (they work best in your own console because the time is very short to be seen in the following snippet).

console.time('split and join');
"javascript-test-find-and-replace-all".split('-').join(' ');
console.timeEnd('split and join')

console.time('regular expression');
"javascript-test-find-and-replace-all".replace(new RegExp('-', 'g'), ' ');
console.timeEnd('regular expression');

console.time('while');
let str1 = "javascript-test-find-and-replace-all";
while (str1.indexOf('-') !== -1) {
    str1 = str1.replace('-', ' ');
}
console.timeEnd('while');

The interesting thing to notice is that if you run them multiple times, the results are always different even though the regular expression solution seems the fastest on average and the while loop solution the slowest.

Answer 36

This can be achieved using regular expressions. A few combinations that might help someone:

var word = "this,\\ .is*a*test,    '.and? / only /     'a \ test?";
var stri = "This      is    a test         and only a        test";

To replace all non alpha characters,

console.log(word.replace(/([^a-z])/g,' ').replace(/ +/g, ' '));
Result: [this is a test and only a test]

To replace multiple continuous spaces with one space,

console.log(stri.replace(/  +/g,' '));
Result: [This is a test and only a test]

To replace all * characters,

console.log(word.replace(/\*/g,''));
Result: [this,\ .isatest,    '.and? / only /     'a  test?]

To replace question marks (?)

console.log(word.replace(/\?/g,'#'));
Result: [this,\ .is*a*test,    '.and# / only /     'a  test#]

To replace quotation marks,

console.log(word.replace(/'/g,'#'));
Result: [this,\ .is*a*test,    #.and? / only /     #a  test?]

To replace all ' characters,

console.log(word.replace(/,/g,''));
Result: [this\ .is*a*test    '.and? / only /     'a  test?]

To replace a specific word,

console.log(word.replace(/test/g,''));
Result: [this,\ .is*a*,    '.and? / only /     'a  ?]

To replace backslash,

console.log(word.replace(/\\/g,''));
Result: [this, .is*a*test,    '.and? / only /     'a  test?]

To replace forward slash,

console.log(word.replace(/\//g,''));
Result: [this,\ .is*a*test,    '.and?  only      'a  test?]

To replace all spaces,

console.log(word.replace(/ /g,'#'));
Result: [this,\#.is*a*test,####'.and?#/#only#/#####'a##test?]

To replace dots,

console.log(word.replace(/\./g,'#'));
Result: [this,\ #is*a*test,    '#and? / only /     'a  test?]

Answer 37

I use split and join or this function:

function replaceAll(text, busca, reemplaza) {
  while (text.toString().indexOf(busca) != -1)
    text = text.toString().replace(busca, reemplaza);
  return text;
}

Answer 38

Check this answer. Maybe it will help, and I used it in my project.

function replaceAll(searchString, replaceString, str) {
    return str.split(searchString).join(replaceString);
}

replaceAll('abc', '',"Test abc test test abc test test test abc test test abc" ); // "Test  test test  test test test  test test "

Answer 39

For replacing all kind of characters, try this code:

Suppose we have a need to send " and \ in my string. Then we will convert it " to " and \ to \.

So this method will solve this issue.

String.prototype.replaceAll = function (find, replace) {
     var str = this;
     return str.replace(new RegExp(find.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), 'g'), replace);
 };

var message = $('#message').val();
message = message.replaceAll('\\', '\\\\'); /*it will replace \ to \\ */
message = message.replaceAll('"', '\\"');   /*it will replace " to \\"*/

I was using Ajax, and I had the need to send parameters in JSON format. Then my method is looking like this:

 function sendMessage(source, messageID, toProfileID, userProfileID) {

     if (validateTextBox()) {
         var message = $('#message').val();
         message = message.replaceAll('\\', '\\\\');
         message = message.replaceAll('"', '\\"');
         $.ajax({
             type: "POST",
             async: "false",
             contentType: "application/json; charset=utf-8",
             url: "services/WebService1.asmx/SendMessage",
             data: '{"source":"' + source + '","messageID":"' + messageID + '","toProfileID":"' + toProfileID + '","userProfileID":"' + userProfileID + '","message":"' + message + '"}',
             dataType: "json",
             success: function (data) {
                 loadMessageAfterSend(toProfileID, userProfileID);
                 $("#<%=PanelMessageDelete.ClientID%>").hide();
                 $("#message").val("");
                 $("#delMessageContainer").show();
                 $("#msgPanel").show();
             },
             error: function (result) {
                 alert("message sending failed");
             }
         });
     }
     else {
         alert("Please type message in message box.");
         $("#message").focus();

     }
 }

 String.prototype.replaceAll = function (find, replace) {
     var str = this;
     return str.replace(new RegExp(find.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), 'g'), replace);
 };

Answer 40

I use p to store the result from the previous recursion replacement:

function replaceAll(s, m, r, p) {
    return s === p || r.contains(m) ? s : replaceAll(s.replace(m, r), m, r, s);
}

It will replace all occurrences in the string s until it is possible:

replaceAll('abbbbb', 'ab', 'a') → 'abbbb' → 'abbb' → 'abb' → 'ab' → 'a'

To avoid infinite loop I check if the replacement r contains a match m:

replaceAll('abbbbb', 'a', 'ab') → 'abbbbb'

Answer 41

You can simply use below method

/**
 * Replace all the occerencess of $find by $replace in $originalString
 * @param  {originalString} input - Raw string.
 * @param  {find} input - Target key word or regex that need to be replaced.
 * @param  {replace} input - Replacement key word
 * @return {String}       Output string
 */
function replaceAll(originalString, find, replace) {
  return originalString.replace(new RegExp(find, 'g'), replace);
};

Answer 42

Method 1

Try to implement a regular expression:

"Test abc test test abc test test test abc test test abc".replace(/\abc/g, ' ');

Method 2

Split and join. Split with abc and join with empty space.

"Test abc test test abc test test test abc test test abc".split("abc").join(" ")

Answer 43

My implementation, very self explanatory

function replaceAll(string, token, newtoken) {
    if(token!=newtoken)
    while(string.indexOf(token) > -1) {
        string = string.replace(token, newtoken);
    }
    return string;
}

Answer 44

Most people are likely doing this to encode a URL. To encode a URL, you shouldn't only consider spaces, but convert the entire string properly with encodeURI.

encodeURI("http://www.google.com/a file with spaces.html")

to get:

http://www.google.com/a%20file%20with%20spaces.html

Answer 45

In my applications, I use a custom function that is the most powerful for this purpose, and even wrapping the split/join solution in the simpler case, it is a little bit faster in Chrome 60 and Firefox 54 (JSBEN.CH) than other solutions. My computer runs Windows 7 64 bits.

The advantage is that this custom function can handle many substitutions at the same time using strings or characters, which can be a shortcut for some applications.

Like the above split/join solution, the solution below doesn't have any problems with escape characters, differently than the regular expression approach.

function replaceAll(s, find, repl, caseOff, byChar) {
    if (arguments.length<2)
        return false;
    var destDel = ! repl;       // If destDel delete all keys from target
    var isString = !! byChar;   // If byChar, replace set of characters
    if (typeof find !== typeof repl && ! destDel)
        return false;
    if (isString && (typeof find !== "string"))
        return false;

    if (! isString && (typeof find === "string")) {
        return s.split(find).join(destDel ? "" : repl);
    }

    if ((! isString) && (! Array.isArray(find) ||
        (! Array.isArray(repl) && ! destDel)))
        return false;

    // If destOne replace all strings/characters by just one element
    var destOne = destDel ? false : (repl.length === 1);

    // Generally source and destination should have the same size
    if (! destOne && ! destDel && find.length !== repl.length)
        return false

    var prox, sUp, findUp, i, done;
    if (caseOff)  { // Case insensitive

    // Working with uppercase keys and target
    sUp = s.toUpperCase();
    if (isString)
       findUp = find.toUpperCase()
    else
       findUp = find.map(function(el) {
                    return el.toUpperCase();
                });
    }
    else { // Case sensitive
        sUp = s;
        findUp = find.slice(); // Clone array/string
    }

    done = new Array(find.length); // Size: number of keys
    done.fill(null);

    var pos = 0;  // Initial position in target s
    var r = "";   // Initial result
    var aux, winner;
    while (pos < s.length) {       // Scanning the target
        prox  = Number.MAX_SAFE_INTEGER;
        winner = -1;  // No winner at the start
        for (i=0; i<findUp.length; i++) // Find next occurence for each string
            if (done[i]!==-1) { // Key still alive

                // Never search for the word/char or is over?
                if (done[i] === null || done[i] < pos) {
                    aux = sUp.indexOf(findUp[i], pos);
                    done[i] = aux;  // Save the next occurrence
                }
                else
                    aux = done[i]   // Restore the position of last search

                if (aux < prox && aux !== -1) { // If next occurrence is minimum
                    winner = i; // Save it
                    prox = aux;
                }
        } // Not done

        if (winner === -1) { // No matches forward
            r += s.slice(pos);
            break;
        } // No winner

        // Found the character or string key in the target

        i = winner;  // Restore the winner
        r += s.slice(pos, prox); // Update piece before the match

        // Append the replacement in target
        if (! destDel)
            r += repl[destOne ? 0 : i];
        pos = prox + (isString ? 1 : findUp[i].length); // Go after match
    }  // Loop

    return r; // Return the resulting string
}

The documentation is below:

 replaceAll

 Syntax
 ======

      replaceAll(s, find, [repl, caseOff, byChar)

 Parameters
 ==========

   "s" is a string target of replacement.
   "find" can be a string or array of strings.
   "repl" should be the same type than "find" or empty

  If "find" is a string, it is a simple replacement for
    all "find" occurrences in "s" by string "repl"

  If "find" is an array, it will replaced each string in "find"
    that occurs in "s" for corresponding string in "repl" array.
  The replace specs are independent: A replacement part cannot
  be replaced again.


  If "repl" is empty all "find" occurrences in "s" will be deleted.
  If "repl" has only one character or element,
      all occurrences in "s" will be replaced for that one.

  "caseOff" is true if replacement is case insensitive
       (default is FALSE)

  "byChar" is true when replacement is based on set of characters.
  Default is false

  If "byChar", it will be replaced in "s" all characters in "find"
  set of characters for corresponding character in "repl"
  set of characters

 Return
 ======

  The function returns the new string after the replacement.

To be fair, I ran the benchmark with no parameter test.

Here is my test set, using Node.js:

function l() {
    return console.log.apply(null, arguments);
}

var k = 0;
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do", "fa"]));  // 1
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do"])); // 2
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"])); // 3
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
     "aeiou", "", "", true)); // 4
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "a", "", true)); // 5
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "uoiea", "", true)); // 6
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "uoi", "", true)); // 7
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do", "fa", "leg"])); // 8
l(++k, replaceAll("BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER",
      ["ri", "nea"], ["do", "fa"])); // 9
l(++k, replaceAll("BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER",
      ["ri", "nea"], ["do", "fa"], true)); // 10
return;

And the results:

1 'banana is a dope fruit harvested far the dover'
2 'banana is a dope fruit harvested dor the dover'
3 'banana is a pe fruit harvested r the ver'
4 'bnn s rp frt hrvstd nr th rvr'
5 'banana as a rapa fraat harvastad naar tha ravar'
6 'bununu is u ripo frait hurvostod nour tho rivor'
7 false
8 false
9 'BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER'
10 'BANANA IS A doPE FRUIT HARVESTED faR THE doVER'

Answer 46

If using a library is an option for you then you will get the benefits of the testing and community support that goes with a library function. For example, the string.js library has a replaceAll() function that does what you're looking for:

// Include a reference to the string.js library and call it (for example) S.
str = S(str).replaceAll('abc', '').s;

Answer 47

There is now a finished proposal for integrating String.prototype.replaceAll into the official specification. Eventually, developers will not have to come up with their own implementations for replaceAll - instead, modern JavaScript engines will support it natively.

The proposal is at stage 4, which means that everything is complete, and all that's left is for browsers to start implementing it.

It has shipped in the latest versions of Chrome, Firefox, and Safari.

Here are the implementation details:

Per the current TC39 consensus, String.prototype.replaceAll behaves identically to String.prototype.replace in all cases, except for the following two cases:

1. If searchValue is a string, String.prototype.replace only replaces a single occurrence of the searchValue, whereas String.prototype.replaceAll replaces all occurrences of the searchValue (as if .split(searchValue).join(replaceValue) or a global & properly-escaped regular expression had been used).
2. If searchValue is a non-global regular expression, String.prototype.replace replaces a single match, whereas String.prototype.replaceAll throws an exception. This is done to avoid the inherent confusion between the lack of a global flag (which implies "do NOT replace all") and the name of the method being called (which strongly suggests "replace all").

Notably, String.prototype.replaceAll behaves just like String.prototype.replace if searchValue is a global regular expression.

You can see a specification-compliant polyfill here.

In supported environments, the following snippet will log foo-bar-baz, without throwing an error:

const str = 'foo bar baz';
console.log(
  str.replaceAll(' ', '-')
);

Answer 48

Here's a very simple solution. You can assign a new method to a String object

String.prototype.replaceAll = function(search, replace){
   return this.replace(new RegExp(search, 'g'), replace)
}

var str = "Test abc test test abc test test test abc test test abc";
str = str.replaceAll('abc', '');

console.log(str) // -> Test  test test  test test test  test test

Answer 49

Here is the working code with prototype:

String.prototype.replaceAll = function(find, replace) {
    var str = this;
    return str.replace(new RegExp(find.replace(/([.*+?^=!:${}()|\[\]\/\\])/g, "\\$1"), 'g'), replace);
};

Answer 50

function replaceAll(str, find, replace) {
    var $r="";
    while($r!=str){ 
        $r = str;
        str = str.replace(find, replace);
    }
    return str;
}

Answer 51

In string first element search and replace

var str = '[{"id":1,"name":"karthikeyan.a","type":"developer"}]'
var i = str.replace('"[','[').replace(']"',']');
console.log(i,'//first element search and replace')

In string global search and replace

var str = '[{"id":1,"name":"karthikeyan.a","type":"developer"}]'
var j = str.replace(/\"\[/g,'[').replace(/\]\"/g,']');
console.log(j,'//global search and replace')

Answer 52

For unique replaceable values

String.prototype.replaceAll = function(search_array, replacement_array) {
  //
  var target = this;
  //
  search_array.forEach(function(substr, index) {
    if (typeof replacement_array[index] != "undefined") {
      target = target.replace(new RegExp(substr, 'g'), replacement_array[index])
    }
  });
  //
  return target;
};

//  Use:
var replacedString = "This topic commented on :year. Talking :question.".replaceAll([':year', ':question'], ['2018', 'How to replace all occurrences of a string in JavaScript']);
//
console.log(replacedString);

Answer 53

In November 2019, a new feature is added to the JavaScript, string.prototype.replaceAll().

Currently it's only supported with Babel, but maybe in the future it can be implemented in all the browsers. For more information, read here.

Answer 54

Use split and join:

var str = "Test abc test test abc test test test abc test test abc";
var replaced_str = str.split('abc').join('');
console.log(replaced_str);

Answer 55

This can be solved using regular expressions and the flag g, which means to not stop after finding the first match. Really, regular expressions are life savers!

function replaceAll(string, pattern, replacement) {
    return string.replace(new RegExp(pattern, "g"), replacement);
}

// or if you want myString.replaceAll("abc", "");

String.prototype.replaceAll = function(pattern, replacement) {
    return this.replace(new RegExp(pattern, "g"), replacement);
};

Answer 56

I just want to share my solution, based on some of the functional features of last versions of JavaScript:

   var str = "Test abc test test abc test test test abc test test abc";

   var result = str.split(' ').reduce((a, b) => {
      return b == 'abc' ? a : a + ' ' + b;   })

  console.warn(result)

Answer 57

The best solution, in order to replace any character we use the indexOf(), includes(), and substring() functions to replace the matched string with the provided string in the current string.

• The String.indexOf() function is to find the n^th match index position.
• The String.includes() method determines whether one string may be found within another string, returning true or false as appropriate.
• String.substring() function is to get the parts of String(preceding,exceding). Add the replace String in-between these parts to generate final return String.

The following function allows to use any character.
where as RegExp will not allow some special character like ** and some characters need to be escaped, like $.

String.prototype.replaceAllMatches = function(obj) { // Obj format: { 'matchkey' : 'replaceStr' }
    var retStr = this;
    for (var x in obj) {
        //var matchArray = retStr.match(new RegExp(x, 'ig'));
        //for (var i = 0; i < matchArray.length; i++) {
        var prevIndex = retStr.indexOf(x); // matchkey = '*', replaceStr = '$*' While loop never ends.
        while (retStr.includes(x)) {
            retStr = retStr.replaceMatch(x, obj[x], 0);
            var replaceIndex = retStr.indexOf(x);
            if( replaceIndex <  prevIndex + (obj[x]).length) {
                break;
            } else {
                prevIndex = replaceIndex;
            }
        }
    }
    return retStr;
};
String.prototype.replaceMatch = function(matchkey, replaceStr, matchIndex) {
    var retStr = this, repeatedIndex = 0;
    //var matchArray = retStr.match(new RegExp(matchkey, 'ig'));
    //for (var x = 0; x < matchArray.length; x++) {
    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;
};

---

We can also use the regular expression object for matching text with a pattern. The following are functions which will use the regular expression object.

You will get SyntaxError when you are using an invalid regular expression pattern like '**'.

• The String.replace() function is used to replace the specified String with the given String.
• The String.match() function is to find how many time the string is repeated.
• The RegExp.prototype.test method executes a search for a match between a regular expression and a specified string. Returns true or false.

String.prototype.replaceAllRegexMatches = function(obj) { // Obj format: { 'matchkey' : 'replaceStr' }
    var retStr = this;
    for (var x in obj) {
        retStr = retStr.replace(new RegExp(x, 'ig'), obj[x]);
    }
    return retStr;
};

Note that regular expressions are written without quotes.

---

Examples to use the above functions:

var str = "yash yas $dfdas.**";
console.log('String: ', str);

// No need to escape any special character
console.log('Index matched replace: ', str.replaceMatch('as', '*', 2));
console.log('Index Matched replace: ', str.replaceMatch('y', '~', 1));
console.log('All Matched replace: ', str.replaceAllMatches({'as': '**', 'y':'Y', '$':'-'}));
console.log('All Matched replace : ', str.replaceAllMatches({'**': '~~', '$':'&$&', '&':'%', '~':'>'}));

// You need to escape some special Characters
console.log('REGEX all matched replace: ', str.replaceAllRegexMatches({'as' : '**', 'y':'Y', '\\$':'-'}));

Result:

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

All Matched replace:  Y**h Y** -dfd**.**
All Matched replace:  yash yas %$%dfdas.>>

REGEX All Matched replace:  Y**h Y** -dfd**.**

---

Answer 58

var myName = 'r//i//n//o//l////d';
var myValidName = myName.replace(new RegExp('\//', 'g'), ''); > // rinold
console.log(myValidName);

var myPetName = 'manidog';
var renameManiToJack = myPetName.replace(new RegExp('mani', 'g'), 'jack'); > // jackdog

Answer 59

This should work.

String.prototype.replaceAll = function (search, replacement) {
    var str1 = this.replace(search, replacement);
    var str2 = this;
    while(str1 != str2) {
        str2 = str1;
        str1 = str1.replace(search, replacement);
    }
    return str1;
}

Example:

Console.log("Steve is the best character in Minecraft".replaceAll("Steve", "Alex"));

Answer 60

str = "Test abc test test abc test test test abc test test abc"

str.split(' ').join().replace(/abc/g,'').replace(/,/g, ' ')

Answer 61

This solution combines some previous answers and conforms somewhat better to the proposed August 2020 standard solution. This solution is still viable for me in September 2020, as String.replaceAll is not available in the Node.js binary I am using.

---

RegExp.escape is a separate issue to deal with, but it is important here, because the official proposed solution will automatically escape string-based find input. This String.replaceAll polyfill would not without the RegExp.escape logic.

I have added an answer which doesn't polyfill RegExp.Escape, in the case that you don't want that.

---

If you pass a regular expression to find, you must include g as a flag. This polyfill won't provide a nice TypeError for you and will cause you a major bad time.

If you need exact standards conformance, for an application which is rigorously relying on the standard implementation, then I suggest using Babel or some other tool to get you the 'right answer' every time instead of Stack Overflow. That way you won't have any surprises.

---

Code:

if (!Object.prototype.hasOwnProperty.call(RegExp, 'escape')) {
  RegExp.escape = function(string) {
    // https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions#Escaping
    // https://github.com/benjamingr/RegExp.escape/issues/37
    return string.replace(/[.*+\-?^${}()|[\]\\]/g, '\\$&'); // $& means the whole matched string
  };
}

if (!Object.prototype.hasOwnProperty.call(String, 'replaceAll')) {
  String.prototype.replaceAll = function(find, replace) {
    // https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replaceAll
    // If you pass a RegExp to 'find', you _MUST_ include 'g' as a flag.
    // TypeError: "replaceAll must be called with a global RegExp" not included, will silently cause significant errors. _MUST_ include 'g' as a flag for RegExp.
    // String parameters to 'find' do not require special handling.
    // Does not conform to "special replacement patterns" when "Specifying a string as a parameter" for replace
    // Does not conform to "Specifying a function as a parameter" for replace
    return this.replace(
          Object.prototype.toString.call(find) == '[object RegExp]' ?
            find :
            new RegExp(RegExp.escape(find), 'g'),
          replace
        );
  }
}

Code, Minified:

Object.prototype.hasOwnProperty.call(RegExp,"escape")||(RegExp.escape=function(e){return e.replace(/[.*+\-?^${}()|[\]\\]/g,"\\$&")}),Object.prototype.hasOwnProperty.call(String,"replaceAll")||(String.prototype.replaceAll=function(e,t){return this.replace("[object RegExp]"==Object.prototype.toString.call(e)?e:new RegExp(RegExp.escape(e),"g"),t)});

---

Example:

console.log(
  't*.STVAL'
    .replaceAll(
      new RegExp(RegExp.escape('T*.ST'), 'ig'),
      'TEST'
    )
);

console.log(
  't*.STVAL'
    .replaceAll(
      't*.ST',
      'TEST'
    );
);

---

Code without RegExp.Escape:

if (!Object.prototype.hasOwnProperty.call(String, 'replaceAll')) {
  String.prototype.replaceAll = function(find, replace) {
    // https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replaceAll
    // If you pass a RegExp to 'find', you _MUST_ include 'g' as a flag.
    // TypeError: "replaceAll must be called with a global RegExp" not included, will silently cause significant errors. _MUST_ include 'g' as a flag for RegExp.
    // String parameters to 'find' do not require special handling.
    // Does not conform to "special replacement patterns" when "Specifying a string as a parameter" for replace
    // Does not conform to "Specifying a function as a parameter" for replace
    return this.replace(
          Object.prototype.toString.call(find) == '[object RegExp]' ?
            find :
            new RegExp(find.replace(/[.*+\-?^${}()|[\]\\]/g, '\\$&'), 'g'),
          replace
        );
  }
}

Code without RegExp.Escape, Minified:

Object.prototype.hasOwnProperty.call(String,"replaceAll")||(String.prototype.replaceAll=function(e,t){return this.replace("[object RegExp]"==Object.prototype.toString.call(e)?e:new RegExp(e.replace(/[.*+\-?^${}()|[\]\\]/g,"\\$&"),"g"),t)});

Answer 62

JavaScript provides a direct way to replace a part of a string with another string and there are some tricks also by which you can do this.

To replace all the occurrences you can use replace() or replaceAll method in JavaScript.

1. replace() method - To replace all elements using this method use a regular expression as a pattern to find the matching string and then replace it with a new string. Please consider using the /g flag with it.

const str = "To do or not to do";
const pattern = /do/g;
const replaceBy = "Code";
console.log(str.replace(pattern, replaceBy));

2. replaceAll() method - To replace all elements using this method, use either a string or regular expression as a pattern to find the matching string and then replace it with a new string. We must use the /g flag with a regular expression in the replaceAll method.

const str = "To do or not to do";
const pattern = "do";
const replaceBy = "Code";
console.log(str.replaceAll(pattern, replaceBy));

const pattern2 = /do/g;
console.log(str.replaceAll(pattern2, replaceBy));

Alternate method: By using split and join method

Split the string at what you want to replace and join by using the new string as a separator. See the example.

const str = "To do or not to do";
const newStr = str.split("do").join("Code");
console.log(newStr);

Answer 63

Repeat until you have replaced them all:

const regex = /^>.*/im;
while (regex.test(cadena)) {
    cadena = cadena.replace(regex, '*');
}

Answer 64

Try this:

String.prototype.replaceAll = function (sfind, sreplace) {
    var str = this;

    while (str.indexOf(sfind) > -1) {
        str = str.replace(sfind, sreplace);
    }

    return str;
};

Answer 65

The simplest solution -

let str = "Test abc test test abc test test test abc test test abc";

str = str.split(" ");
str = str.filter((ele, key)=> ele!=="abc")
str = str.join(" ")

Or simply -

str = str.split(" ").filter((ele, key) => ele != "abc").join(" ")

Answer 66

You can do it without Regex, but you need to be careful if the replacement text contains the search text.

e.g.

replaceAll("nihIaohi", "hI", "hIcIaO", true)

So here is a proper variant of replaceAll, including string-prototype:

function replaceAll(str, find, newToken, ignoreCase)
{
    let i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
}

Or, if you want to have a string-prototype function:

String.prototype.replaceAll = function (find, replace) {
    let str = this;

    let i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
};

Answer 67

In August 2020

No more regular expression stuff

const str = "Test abc test test abc test test test abc test test abc";
const modifiedStr = str.replaceAll('abc', '');
console.log(modifiedStr);

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replaceAll

Answer 68

You can try like this:

Example data:

var text = "heloo,hai,hei"

text = text.replace(/[,]+/g, '')

or

text.forEach((value) => {
  hasil = hasil.replace(',', '')
})

Answer 69

All the answers are accepted, and you can do this by many ways. One of the trick to do this is this:

const str = "Test abc test test abc test test test abc test test abc";

const compare = "abc";
arrayStr = str.split(" ");
arrayStr.forEach((element, index) => {
  if (element == compare) {
    arrayStr.splice(index, 1);
  }
});
const newString = arrayStr.join(" ");
console.log(newString);

Answer 70

We can use the replace method in JavaScript:

var result = yourString.replace('regexPattern', "replaceString");

var str = "Test abc test test abc test test test abc test test abc";

var expectedString = str.replace(/abc(\s|$)/g, "");

console.log(expectedString);

Answer 71

I know this isn’t the best way to do this, but you can try this:

var annoyingString = "Test abc test test abc test test test abc test test abc";

while (annoyingString.includes("abc")) {
    annoyingString = annoyingString.replace("abc", "")
}

Answer 72

I added the function below to this performance test page in the "library" section:

https://jsben.ch/LFfWA

function _replace(t, s, r){
    var i = t.indexOf(s);
    if (i == -1) return t;
    return t.slice(0, i) + r + _replace(t.slice(i + s.length, t.length), s,r);
}

..and put this in as the test:

var replaced = _replace(testString, 'abc', '123');

.. and that function performs about 34% faster for me than split or regex. The idea / hope was to end up pasting smaller and smaller pieces of the string onto the stack and then building the entire result by unrolling the stack, thereby minimizing extra string copies and extra searches through the same string data and hopefully optimizing use of the CPU cache.

Part of the thought was that if the string isn't too big, it may end up in the CPU cache; passing it and pasting pieces of it puts those bits into the cache, and then the searching can operate entirely using CPU cached data. Now whether or not that's actually what ends up happening I'm sure is entirely JavaScript implementation-dependent.

This isn't as fast as possible, but it's as fast as I could manage without mutable strings. Arrays in JavaScript probably have a pointer for each element, so, a solution involving a lot of array elements is not likely to be as CPU cache friendly as this.

Answer 73

Starting from v85, Chrome now supports String.prototype.replaceAll natively. Note this outperform all other proposed solutions and should be used once majorly supported.

Feature status: https://chromestatus.com/feature/6040389083463680

var s = "hello hello world";
s = s.replaceAll("hello", ""); // s is now "world"
console.log(s)

Answer 74

There is a way to use the new replaceAll() method.

But you need to use a cutting-edge browser or a JavaScript run time environment.

You can check the browser compatibility in here.

Answer 75

// Try this way

const str = "Test abc test test abc test test test abc test test abc";
const result = str.split('abc').join('');
console.log(result);

Answer 76

I have read this question and answers, but I didn't find an appropriate solution for me. Although the answers are quite useful, I decided to create my own solution from scratch. The problems about this kind of replacing are:

• Often it's not enough to simply find a string which matches any upper- or lower case. E.g., for search results I need to replace it with the same case.
• If I deal with innerHTML, I can easily damage HTML tags (e.g., an occurrence of hr in the href attribute).
• The replaceAll() method is too fresh for many use cases, and it doesn't solve all the issues.

So, I was writing functions to highlight search results in a table, where table data cells may have links inside, as well as other HTML tags. And these links were intended to be kept, so innerText was not enough.

The solution I decided to provide for everyone with the same issues. Surely, you can use it not only for tables, but for any elements. Here is the code:

/* Iterate over table data cells to insert a highlight tag */
function highlightSearchResults(textFilter) {
  textFilter = textFilter.toLowerCase().replace('<', '&lt;').replace('>', '&gt;');
  let tds;
  tb = document.getElementById('sometable'); //root element where to search
  if (tb) {
    tds = tb.getElementsByTagName("td"); //sub-elements where to make replacements
  }
  if (textFilter && tds) {
    for (td of tds) {
      //specify your span class or whatever you need before and after
      td.innerHTML = insertCaseInsensitive(td.innerHTML, textFilter, '<span class="highlight">', '</span>');
    }
  }
}

/* Insert a highlight tag */
function insertCaseInsensitive(srcStr, lowerCaseFilter, before, after) {
  let lowStr = srcStr.toLowerCase();
  let flen = lowerCaseFilter.length;
  let i = -1;
  while ((i = lowStr.indexOf(lowerCaseFilter, i + 1)) != -1) {
    if (insideTag(i, srcStr)) continue;
    srcStr = srcStr.slice(0, i) + before + srcStr.slice(i, i+flen) + after + srcStr.slice(i+flen);
    lowStr = srcStr.toLowerCase();
    i += before.length + after.length;
  }
  return srcStr;
}

/* Check if an ocurrence is inside any tag by index */
function insideTag(si, s) {
  let ahead = false;
  let back = false;
  for (let i = si; i < s.length; i++) {
    if (s[i] == "<") {
      break;
    }
    if (s[i] == ">") {
      ahead = true;
      break;
    }
  }
  for (let i = si; i >= 0; i--) {
    if (s[i] == ">") {
      break;
    }
    if (s[i] == "<") {
      back = true;
      break;
    }
  }
  return (ahead && back);
}

Answer 77

I would suggest adding a global method for the string class by appending it to the prototype chain.

String.prototype.replaceAll = function(fromReplace, toReplace, {ignoreCasing} = {}) { return this.replace(new RegExp(fromReplace, ignoreCasing ? 'ig': 'g'), toReplace);}

And it can be used like:

'stringwithpattern'.replaceAll('pattern', 'new-pattern')

Answer 78

Check this. I'm sure it will help you:

<!DOCTYPE html>
<html>
<body>
<p>Click the button to do a global search and replace for "is" in a string.</p>
<button onclick="myFunction()">Try it</button>
<p id="demo"></p>
<script>
function myFunction() {
  var str = 'Is this "3" dris "3"?';
  var allvar= '"3"';
  var patt1 = new RegExp( allvar, 'g' );
  document.getElementById("demo").innerHTML = str.replace(patt1,'"5"');
}
</script>
</body>
</html>

Here is the JSFiddle link.