Javascript replaceAll function

[Black Bart]

I continue to see this foolish technique presented as javascript replaceAll function.

Can anyone see the problem with this?

function replaceAll(str, str1, str2)
{
   var index = str.indexOf( str1 );
 
   while (index != -1)
   {
      str = str.replace(str1, str2);
 
      index = str.indexOf(str1);
   }
 
   return str;
}

[Ron White]

That replaceAll function will create an endless loop under the following conditions:

var myString = "x";

myString = replaceAll(myString,  "x", "xyz");

Try something like this:

function replaceAll(str, str1, str2)
{
   return str.replace(new RegExp(str1, "g"), str2);
}

It's simple and effective.

[Black Bart]

The problem with using a regular expression is that there are too many special characters that need to be escaped.

Consider the following will produce $100.00-
instead of -100.00 which is what you might have expected.

replaceAll("$100.00",  "$", "-");

Also,

replaceAll("$100.00", ".", "")

produces the empty string, rather than $10000

Lastly,

replaceAll("Hello?", "?", "!")

Will give you a javascript error rather than Hello!

I could go on with this.

There has to be a better way.

[qwerty]

Here is the latest function posted as code:

Code:

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);
}

Here is a modified version that corrects the issues you raise and also supports "ignore case" as an optional boolean.

You just need to escape all the regex special characters.

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

This code adds a replaceAll method to the String object.

You call it like this:

var myString = "Hello???";

myString.replaceAll("?", "!");

And it will return this:

Hello!!!

See if you can break it.

[Bailyn]

Hi qwerty ... that's pretty impressive.  Compact, simple but powerful! ... That's about my contribution to this topic.  I will let you know when I break it.

[qwerty]

Thanks Bailyn,

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

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

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

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

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

[Black Bart]

Is this a valid test?

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

[qwerty]

This is a correct test

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

The built in javascript function returns the following:

"x".replace("", "xyz");
xyzx

but it only replaces the first occurrence.

[Black Bart]

qwerty,

I found a problem with your replaceAll function.

It doesn't handle this case correctly:

"x".replaceAll("x", "$$");

It should return this
$$
But it returns this instead
$

Try again:)

Thanks Bailyn,

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

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

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

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

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

[qwerty]

Good find Bart,

This new version resolves the issue, and it's faster.

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);
};

Looks like I have a new test case
"x".replaceAll("x", "$$");
$$

[qwerty]

I'm going to repost this function as code:

Code:

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);
};