REGEX in javascript not working for dots (.) in JavaScript

Question

I am using a regex to validate an email address in JavaScript.

The regex is pretty simple. It checks three things: 1)'@' , 2)'.' ('dot' as in [email protected]), and 3) 'a-z' in an email address. If all three return true, email address is valid (according to my validation, atleast)

Here is the code:

function checkemail(){
  var e = document.getElementById("email").value;
  if((e.match(/@/g)==null)||(e.match(/[a-z]/ig)==null)||(e.match(/./g)==null)){
  //display error message
  }
}

My question is:

(e.match(/./g)==null); //returns false even though there are no dots in the string e

returns false even when there are no dots in string.

For example:

("thisIsMyEmail".match(/./ig))==null //returns false

Why does it return false when it should be true?

periods are special in a regex, it means "any character", not just a period, to make it mean a period, it has to be escaped. — adeneo
– adeneo, Commented Dec 6, 2014 at 17:03
Instead of using match for literal strings, you should use String.prototype.indexOf() — Casimir et Hippolyte
– Casimir et Hippolyte, Commented Dec 6, 2014 at 17:08

T.J. Crowder · Accepted Answer · 2014-12-06 17:03:58Z

5

/./g (or /./ig) will match any string that as at least one character in it. . is special in regular expressions, it means "any character here."

For an actual dot, escape it with a backslash: /\./g.

answered Dec 6, 2014 at 17:03

T.J. Crowder

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Comments

Ian Hazzard · Accepted Answer · 2014-12-06 20:32:02Z

2

First off, you don't need to check if the string is null. Simply use this:

var email = "[email protected]";

if (email.match(/^\S+\@\S+\.\S+$/i)){
alert("Email address passed validation.");
}

edited Dec 6, 2014 at 20:32

answered Dec 6, 2014 at 17:06

Ian Hazzard

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2 Comments

raneshu Over a year ago

Thanks for your response. It does pretty much the same as mine. It does not seem to check for letters after the final dot. Eg: bob@gmail. returns true. But it helps. Thanks

Ian Hazzard Over a year ago

@raneshu, I'm sorry, my code was a bit messy. Try it now.

Carlo Gonzales · Accepted Answer · 2014-12-06 17:10:17Z

1

you have to escape the . The unescaped period means matches any character.

Meaning having a string "abc" using your expression would result in an array containing the characters 'a', 'b', and 'c'.

In your snippet the correct answer is

(e.match(/\./g)==null);

This should result to what you're expecting

answered Dec 6, 2014 at 17:10

Carlo Gonzales

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Comments

Oleksandr T. · Accepted Answer · 2015-02-08 08:09:36Z

1

Try this

(e.match(/\./g)==null);

. matches any character so needs escaping /\./g

edited Feb 8, 2015 at 8:09

answered Dec 6, 2014 at 17:03

Oleksandr T.

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Comments

Samiddha · Accepted Answer · 2022-06-16 10:05:42Z

I know you have already got the answer.

But I just want to give an advice. My advice is - don't use the javascript code to validate any email address; because as per your code, @domain., @domain.com these all are also valid email, but everybody knows these are not a valid email address.

So use the below code:

let email = $(this).val();
var positionOfAt = email.indexOf("@");
var positionOfDot = email.lastIndexOf(".");

if(email.search("@") == -1 || //if '@' is not present
email.search(" ") >= 1 || //if blank space is present
email.search(".") == -1 || //if "." is not present
positionOfAt < 1 || //if there is no character before "@", at least one character should be present before "@"
positionOfDot - positionOfAt <= 2 || //between '@' and '.', if there is not at least two character
email.length - positionOfDot <= 2) //if after '.' there is not at least two character)
{
    console.log("Invalid email id")
}
else
{
    console.log("Valid email id")
}

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