How to retrieve Name from Email Address

Question

With javascript, how do we remove the @gmail.com or @aol.com from a string so that what only remains is the name?

var string = "johndoe@yahoo.com";

Will be just "johdoe"? I tried with split but it did not end well. thanks.

Answer 1

var email = "john.doe@example.com";
var name   = email.substring(0, email.lastIndexOf("@"));
var domain = email.substring(email.lastIndexOf("@") +1);

console.log( name );   // john.doe
console.log( domain ); // example.com

The above will also work for valid names containing @ (tools.ietf.org/html/rfc3696^{Page 5}):

john@doe
"john@@".doe
"j@hn".d@e

• String.prototype.substring()
• String.prototype.lastIndexOf()

Using RegExp:

Given the email value is already validated, String.prototype.match() can be than used to retrieve the desired name, domain:

String match:

const name   = email.match(/^.+(?=@)/)[0];    
const domain = email.match(/(?<=.+@)[^@]+$/)[0]; 

Capturing Group:

const name   = email.match(/(.+)@/)[1];    
const domain = email.match(/.+@(.+)/)[1];

To get both fragments in an Array, use String.prototype.split() to split the string at the last @ character:

const [name, domain] = email.split(/(?<=^.+)@(?=[^@]+$)/);
console.log(name, domain);

or simply with /@(?=[^@]*$)/.
Here's an example that uses a reusable function getEmailFragments( String )

const getEmailFragments = (email) => email.split(/@(?=[^@]*$)/);

[ // LIST OF VALID EMAILS:
  `info@example.com`,
  `john@doe@example.com`,
  `"john@@".doe@example.com`,
  `"j@hn".d@e@example.com`,
]
.forEach(email => {
  const [name, domain] = getEmailFragments(email);
  console.log("DOMAIN: %s NAME: %s ", domain, name);
});

Answer 2

You should take note that a valid email address is an incredibly sophisticated object and may contain multiple @ signs (ref. http://cr.yp.to/im/address.html).

"The domain part of an address is everything after the final @."

Thus, you should do something equivalent to:

var email = "johndoe@yahoo.com";
var name = email.substring(0, email.lastIndexOf("@"));

or even shorter,

var name = email.replace(/@[^@]+$/, '');

If you want both the name and the domain/hostname, then this will work:

var email = "johndoe@yahoo.com";
var lasta = email.lastIndexOf('@');
var name, host;
if (lasta != -1) {
    name = email.substring(0, lasta);
    host = email.substring(lasta+1);
    /* automatically extends to end of string when 2nd arg omitted */
} else {
    /* respond to invalid email in some way */
}

Answer 3

And another alternative using split:

var email = "john.doe@email.com";
var sp = email.split('@');

console.log(sp[0]); // john.doe
console.log(sp[1]); // email.com

Answer 4

Try it using substring() and indexOf()

var name = email.substring(0, email.indexOf("@"));

Answer 5

var email = "johndoe@yahoo.com";
email=email.replace(/@.*/,""); //returns string (the characters before @)

Answer 6

You can try with the replace() and regular expression. You can read more about replace() using regex here

var myEmail = 'johndoe@yahoo.com';
var name= myEmail.replace(/@.*/, "");
console.log(name);

This returns the string before @

Answer 7

As shown in How to remove or get domain name from an e-mail address in Javascript?, you can do it using the following code:

const getDomainFromEmail = email => {
  let emailDomain = null;
  const pos = email.search('@'); // get position of domain
  if (pos > 0) {
    emailDomain = email.slice(pos+1); // use the slice method to get domain name, "+1" mean domain does not include "@"
  }
  return emailDomain;
 };

const yourEmail = "jonth.nt49@4codev.com"
 console.log(getDomainFromEmail(yourEmail));
 // result : 4codev.com

Answer 8

This simple regex will do the needful.
/^.*(?=@)/g.

Example:

"johndoe@yahoo.com".match(/^.*(?=@)/g); // returns Array [ "johndoe" ]

Answer 9

This is only extract your name and remove the nos and special characters present in your name this will work only if your mail id has your name in it

var email = user.email;
var name = email.replace(/@.*/, "");
var noNumbername=name.replace(/[0-9]/g, '');
function capitalizeFirstLetter(string) {
return string.charAt(0).toUpperCase() + string.slice(1);
}
var makefirstNameCaps=capitalizeFirstLetter(noNumbername);
console.log(makefirstNameCaps);