[RESOLVED] Notice: Undefined offset: 1

[jeffshead]

Googled for hours and can't find an answer😕

I get a "Notice: Undefined offset: 1" warning referring to line 2 in the the code below:

/

* add domain name to username if whatnot and soforth */
    if (!list($login_host, $other) = split(':', $_SERVER['HTTP_HOST'])) {
      $login_host = $_SERVER['HTTP_HOST'];
    }
    if (!strstr($login_username, '@')) {
      /* strip any leading 'www.' so that only a username is required */
      $login_host = preg_replace('/^www\./i', '', $login_host);
      $login_host = preg_replace('/^webmail\./i', '', $login_host);
      $login_host = preg_replace('/^mail\./i', '', $login_host);
      $login_username .= '@' . $login_host;
    }

I did not write the code, just trying to stop the warning. This code snippet is from a webmail redirect page which prevents users from reposting their form data after a successful logout.

I'm sure the fix is is to add "isset" somewhere, but I can't figure it out.

Can someone help?

[Jack_McSlay]

that means it's trying to look for the second element of split(':', $SERVER['HTTP_HOST']) while there is only one element, which is no surprised given that $SERVER['HTTP_HOST'] is supposed to give only the domain part of the url (subdomain.domain.com), therefore not having a ":" character for the split function result in an array of more than one element

[jeffshead]

Jack McSlay wrote:

that means it's trying to look for the second element of split(':', $SERVER['HTTP_HOST']) while there is only one element, which is no surprised given that $SERVER['HTTP_HOST'] is supposed to give only the domain part of the url (subdomain.domain.com), therefore not having a ":" character for the split function result in an array of more than one element

Thanks for your reply Jack🙂

What is the best solution? What should the code be changed to?

[NogDog]

Assuming that the $other variable is not used anywhere else, you could replace this:

    if (!list($login_host, $other) = split(':', $_SERVER['HTTP_HOST'])) {
      $login_host = $_SERVER['HTTP_HOST'];
    }

...with this:

$login_host = preg_replace('/:.*$/', '', $_SERVER['HTTP_HOST']);

[jeffshead]

NogDog wrote:

Assuming that the $other variable is not used anywhere else, you could replace this:

    if (!list($login_host, $other) = split(':', $_SERVER['HTTP_HOST'])) {
      $login_host = $_SERVER['HTTP_HOST'];
    }

...with this:

$login_host = preg_replace('/:.*$/', '', $_SERVER['HTTP_HOST']);

Thanks NogDog,

Seems to work just fine.🙂