CODEWITHSUNDEEP
javascript
snowcoder
snowcoder

Reputation: 491

remove email address format from username

var login_id = '[email protected]';
console.log(login_id.substring(0, login_id.lastIndexOf("@")));

Above script works perfectly if I pass input with '@abc.com'. Does not work if string do not have '@domain.com' We have some user name with [email protected] and few are just username. I want extract @domain.com from user name. Expected output is if input is [email protected] return output = username and if input is username, output should be username.

Please help if there is any way.

Upvotes: 0

Views: 315

Answers (4)

HijenHEK
HijenHEK

Reputation: 1296

check first if it contains the '@' charactar first

    login_id.includes('@') ? console.log(login_id.substring(0, 
      login_id.lastIndexOf("@"))) : console.log(login_id) ;

Running example

function username(login_id) {
  return login_id.includes('@') ? login_id.substring(0,login_id.lastIndexOf("@")) :  login_id ;
   }


console.log(username("[email protected]")) ;
console.log(username("sa-testaccount0125")) ;

Upvotes: 0

Arihant Jain
Arihant Jain

Reputation: 847

Use .split() method to split your string in parts

It will still work if you do not have @domain.com

Case-1:

const login_id = '[email protected]';
console.log(login_id.split("@")[0])

Output

"sa-testaccount0125"

Case-2:

const login_id = 'sa-testaccount0125';
console.log(login_id.split("@")[0])

Output

"sa-testaccount0125"

Upvotes: 2

Wyck
Wyck

Reputation: 11740

If you split on @ then you get an array of items. The first item of the array will be the username, whether there was a @ in it or not.

const test = str => str.split('@')[0]

console.log(test('[email protected]'));
console.log(test('sa-testaccount0125'));

Upvotes: 1

Real Quick
Real Quick

Reputation: 2800

You're already using a indexOf. Usee that to check @ if exists as well:

function check(login_id) {
  if (login_id.indexOf('@') >= 0) {

    return login_id.substring(0, login_id.lastIndexOf("@"));
  }
  return login_id;
}

console.log(check('[email protected]'));
console.log(check('sa-asd'));
console.log(check('[email protected]'));

Upvotes: 0

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