How to replace all dots in a string using JavaScript

I want to replace all the occurrences of a dot(.) in a JavaScript string

For example, I have:

var mystring = 'okay.this.is.a.string';

I want to get: okay this is a string.

So far I tried:

mystring.replace(/./g,' ')

but this ends up with all the string replaced to spaces.

You need to escape the . because it has the meaning of "an arbitrary character" in a regular expression.

mystring = mystring.replace(/\./g,' ')

One more solution which is easy to understand :)

var newstring = mystring.split('.').join(' ');

/**
 * ReplaceAll by Fagner Brack (MIT Licensed)
 * Replaces all occurrences of a substring in a string
 */
String.prototype.replaceAll = function( token, newToken, ignoreCase ) {
    var _token;
    var str = this + "";
    var i = -1;
    
    if ( typeof token === "string" ) {
        
        if ( ignoreCase ) {
            
            _token = token.toLowerCase();
            
            while( (
                i = str.toLowerCase().indexOf(
                    _token, i >= 0 ? i + newToken.length : 0
                ) ) !== -1
            ) {
                str = str.substring( 0, i ) +
                    newToken +
                    str.substring( i + token.length );
            }
            
        } else {
            return this.split( token ).join( newToken );
        }
        
    }
return str;
};

alert('okay.this.is.a.string'.replaceAll('.', ' '));

Faster than using regex...

EDIT:
Maybe at the time I did this code I did not used jsperf. But in the end such discussion is totally pointless, the performance difference is not worth the legibility of the code in the real world, so my answer is still valid, even if the performance differs from the regex approach.

EDIT2:
I have created a lib that allows you to do this using a fluent interface:

replace('.').from('okay.this.is.a.string').with(' ');

See https://github.com/FagnerMartinsBrack/str-replace.

str.replace(new RegExp(".","gm")," ")

For this simple scenario, i would also recommend to use the methods that comes build-in in javascript.

You could try this :

"okay.this.is.a.string".split(".").join("")

Greetings

I add double backslash to the dot to make it work. Cheer.

var st = "okay.this.is.a.string";
var Re = new RegExp("\\.","g");
st = st.replace(Re," ");
alert(st);

This is more concise/readable and should perform better than the one posted by Fagner Brack (toLowerCase not performed in loop):

String.prototype.replaceAll = function(search, replace, ignoreCase) {
  if (ignoreCase) {
    var result = [];
    var _string = this.toLowerCase();
    var _search = search.toLowerCase();
    var start = 0, match, length = _search.length;
    while ((match = _string.indexOf(_search, start)) >= 0) {
      result.push(this.slice(start, match));
      start = match + length;
    }
    result.push(this.slice(start));
  } else {
    result = this.split(search);
  }
  return result.join(replace);
}

Usage:

alert('Bananas And Bran'.replaceAll('An', '(an)'));

replaceAll(search, replaceWith) [MDN]

 ".a.b.c.".replaceAll('.', ' ')
 // result: " a b c "

 // Using RegEx. You MUST use a global RegEx.
 ".a.b.c.".replaceAll(/\./g, ' ')
 // result: " a b c "
    

replaceAll() replaces ALL occurrences of search with replaceWith.

It's actually the same as using replace() [MDN] with a global regex(*), merely replaceAll() is a bit more readable in my view.

(*) Meaning it'll match all occurrences.

---

Important(!) if you choose regex: > when using a regexp you have to set the global ("g") flag; > otherwise, it will throw a TypeError: "replaceAll must be called with > a global RegExp".

String.prototype.replaceAll = function(character,replaceChar){
	var word = this.valueOf();
	
	while(word.indexOf(character) != -1)
		word = word.replace(character,replaceChar);
	
	return word;
}

Here's another implementation of replaceAll. Hope it helps someone.

    String.prototype.replaceAll = function (stringToFind, stringToReplace) {
        if (stringToFind === stringToReplace) return this;
        var temp = this;
        var index = temp.indexOf(stringToFind);
        while (index != -1) {
            temp = temp.replace(stringToFind, stringToReplace);
            index = temp.indexOf(stringToFind);
        }
        return temp;
    };

Then you can use it:

var myText = "My Name is George";
var newText = myText.replaceAll("George", "Michael");

Example: I want to replace all double Quote (") into single Quote (') Then the code will be like this

var str= "\"Hello\""
var regex = new RegExp('"', 'g');
str = str.replace(regex, '\'');
console.log(str); // 'Hello'

@scripto's made a bit more concise and without prototype:

function strReplaceAll(s, stringToFind, stringToReplace) {
    if (stringToFind === stringToReplace) return s;
    for (let index = s.indexOf(stringToFind); index != -1; index = s.indexOf(stringToFind))
        s = s.replace(stringToFind, stringToReplace);
    return s;
}

Here's how it stacks up: http://jsperf.com/replace-vs-split-join-vs-replaceall/68

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

mystring.replace(new RegExp('.', "g"), ' ');

Simplest way

> "Mr.".split('.').join("");

..............

Console

you can replace all occurrence of any string/character using RegExp javasscript object.

Here is the code,

var mystring = 'okay.this.is.a.string';

var patt = new RegExp("\\.");

while(patt.test(mystring)){

  mystring  = mystring .replace(".","");

}

let a = "once there was a king. spread opeator. let. ver. const.";

let data = a.replaceAll(".","");

Answer : data = "once there was a king spread opeator let ver const";

You need to use replaceAll() method on that string.

var mystring = 'okay.this.is.a.string';
var myNewString = escapeHtml(mystring);

function escapeHtml(text) {
if('' !== text) {
    return text.replace(/&/g, "&")
               .replace(/&lt;/g, "<")
               .replace(/&gt;/g, ">")
               .replace(/\./g,' ')
               .replace(/&quot;/g, '"')
               .replace(/&#39/g, "'");
}