Using regex to extract username from email address

Question

My string of text looks like this:

johndoe@domain.com (John Doe)

I need to get just the part before the @ and nothing else. The text is coming from a simple XML object if that matters any.

The code I have looks like this:

$authorpre = $key->{"author"};
$re1 = '((?:[a-z][a-z]+))';

if ($c = preg_match_all ("/".$re1."/is", $authorpre, $matches))
{
    $author = $matches[1][0];
}

Sometimes the username might have numbers or an underscore before the @ symbol, which is where the regex stops it seems.

Answer 1

The regular expression that will match and capture any character until it reaches the @ character:

([^@]+)

That seems like what you need. It'll handle all kinds of freaky variations on e-mail addresses.

---

I'm not sure why Ben James deleted his answer, since I feel it's better than mine. I'm going to post it here (unless he undeletes his answer):

Why use regex instead of string functions?

$parts = explode("@", "johndoe@domain.com");
$username = $parts[0];

You don't need regular expressions in this situation at all. I think using explode is a much better option, personally.

---

As Johannes Rössel points out in the comments, e-mail address parsing is rather complicated. If you want to be 100% sure that you will be able to handle any technically-valid e-mail address, you're going to have to write a routine that will handle quoting properly, because both solutions listed in my answer will choke on addresses like "a@b"@example.com. There may be a library that handles this kind of parsing for you, but I am unaware of it.

Answer 2

@OP, if you only want to get everything before @, just use string/array methods. No need complicated regex. Explode on "@", then remove the last element which is the domain part

$str = '"peter@john@doe"@domain.com (John Doe)';
$s = explode("@",$str);
array_pop($s); #remove last element.
$s = implode("@",$s);
print $s;

output

$ php test.php
"peter@john@doe"

Answer 3

Maybe this variant is a bit slower than explode(), but it takes only one string:

$name = preg_replace('/@.*?$/', '', $email);

Answer 4

<?php
$email  = 'name@example.com';
$domain = strstr($email, '@');
echo $domain; // prints @example.com

$user = strstr($email, '@', true); // As of PHP 5.3.0
echo $user; // prints name
?>

source

Answer 5

I used preg_replace

$email_username = preg_replace('/@.*/', '', $_POST['email']);

Answer 6

My suggestion:

$email = 'johndoe@domain.com';
$username = substr($email, 0, strpos($email, '@'));

// Output (in $username): johndoe

Answer 7

I'd go with $author = str_replace(strrchr($authorpre, '@'), '', $authorpre);

Answer 8

You could start by using mailparse_rfc822_parse_addresses to parse the address and extract just the address specification without any display name. Then, you could extract the part before @ with the regexp (.*)@.

Answer 9

Use something like this:

list($username, $domain) = explode('@', $email . "@"); // ."@" is a trick: look note below

With this solution you'll have already populated two variables with email address parts in one row.

."@": This is made to avoid in short critical errors with the list command and ensure that explode will produce at least two variables as needed.

Answer 10

In case someone is still looking in 2020 ..here's the regex that can pick the text before the '@'

^(\S+)(?=@)

Answer 11

Basic example:

    $email = "linuxUser@IsGrand.com";
    if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
        list($user, $domain) = explode('@', trim($email) . "@");
    } else {
        echo "Unable to get account info ....";
    }

Complex Example: Something like this to populate Firstname and Last name fields:

1) valid email ?  if yes get the two parts  user and domain.
2) else set to something default etc.
3) use the email address if we don't have a decoded value.

Code:

    if (filter_var($email, FILTER_VALIDATE_EMAIL)) {
        list($fname, $lname) = explode('@', trim($email) . "@");
    } else {
        $fname = "Xdefault";
        $lname = "Ydefault";
    }

    $fname = (!empty($decoded['firstname'][0]))  ? $decoded['firstname'][0] : $fname ;
    $lname = (!empty($decoded['lastname'][0]))  ? $decoded['lastname'][0] : $lname ;

Answer 12

I would also like to suggest a non regex solution here as it may be useful in most cases:

strstr('n.shah@xyz.co', '@', true)

output:

n.shah