How to replace all occurrences of a string in JavaScript

Question

I have this string in my JavaScript code:

"Test abc test test abc test test test abc test test abc" 

Doing:

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

Seems to only remove the first occurrence of abc in the string above.

How can 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.

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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); }