[Solved] Notice: Undefined variable: [closed]

The error is here:

if($admin_username && $admin_password && $admin_token == session_id())

It seems quite likely this is line 63. The problem is you are only defining $admin_username if $_SESSION['admin_username'] is set.

Although you don’t show it in your code example, I’d presume that the same thing is true of $signin_username.

Solution #1 – use isset

There are a number of solutions to your problem. The first, and perhaps simplest, is to check that the variables are set before checking their value.

if (isset($admin_username, $admin_password, $admin_token) && session_id() == $admin_token)

Solution #2 – define defaults

If there are default values for the variables before checking to see if the corresponding $_SESSION variables exist, no error can occur:

$admin_username = $admin_password = $admin_token = FALSE;

if(isset($_SESSION['admin_username'])){ $admin_username = $_SESSION ['admin_username']; }
if(isset($_SESSION['admin_password'])){ $admin_password = $_SESSION['admin_password']; }
if(isset($_SESSION['admin_token'])){ $admin_token = $_SESSION['admin_token']; }

Solution #3 – define when missing

If these variables are defined irrespective of whether the relevant $_SESSION variable exists, again these error cannot occur:

$admin_username = isset($_SESSION['admin_username']) ? $_SESSION['admin_username'] : FALSE;

$admin_password = isset($_SESSION['admin_password']) ? $_SESSION['admin_password'] : FALSE;

$admin_token = isset($_SESSION['admin_token']) ? $_SESSION['admin_token'] : FALSE;