Fastest method to replace all instances of a character in a string

Fastest method to replace all instances of a character in a string

What is the fastest way to replace all instances of a string/character in a string in JavaScript? A while, a for-loop, a regular expression?


Solution 1:

The easiest would be to use a regular expression with g flag to replace all instances:

str.replace(/foo/g, "bar")

This will replace all occurrences of foo with bar in the string str. If you just have a string, you can convert it to a RegExp object like this:

var pattern = "foobar",
    re = new RegExp(pattern, "g");

Solution 2:

Try this replaceAll:

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

It is very fast, and it will work for ALL these conditions
that many others fail on:

"x".replaceAll("x", "xyz");
// xyz

"x".replaceAll("", "xyz");
// xyzxxyz

"aA".replaceAll("a", "b", true);
// bb

"Hello???".replaceAll("?", "!");
// Hello!!!

Let me know if you can break it, or you have something better, but make sure it can pass these 4 tests.

Solution 3:

var mystring = 'This is a string';
var newString = mystring.replace(/i/g, "a");

newString now is ‘Thas as a strang’

Solution 4:

Also you can try:


Solution 5:

Just thinking about it from a speed issue I believe the case sensitive example provided in the link above would be by far the fastest solution.

var token = "\r\n";
var newToken = " ";
var oldStr = "This is a test\r\nof the emergency broadcasting\r\nsystem.";
newStr = oldStr.split(token).join(newToken);

newStr would be
“This is a test of the emergency broadcast system.”

Solution 6:

You can use the following:

newStr = str.replace(/[^a-z0-9]/gi, '_');


newStr = str.replace(/[^a-zA-Z0-9]/g, '_');

This is going to replace all the character that are not letter or numbers to (‘_’). Simple change the underscore value for whatever you want to replace it.