go to your backoffice and users, then click the admin account, then uncheck 'show online' then click save.

Thanks, but that is not what I want to do. I have a variable $user (integer) in File called users.php (it's an add-on) that I want to sheck if Admin is logged-in then echo $user should be $user -1;

Ah ok, try;
array_keys($Sessions->_registered_Users); should give you a array of all the user id's that are logged in at the time. You should be able to figure the rest out.

Thanks again, but from the backoffice I know the Admin ID. Can you please tell the variable name that test that ID.

yeh so;
$logged_in_users = array_keys($Sessions->_registered_Users);
in_array($admin_id, $logged_in_users);

should return whether the admin is logged in or not.

Many, Many thanks

kskhater since you're running 0.9.2 I'm not sure balupton's info will work. He's a 1.* guy is the thing. If it doesn't get you where you want to be try this:

if( is_logged_in() ) { // gotta be true since you're in the back office?
if( $current_User->('ID') == 1 ) { // this is The Admin ID
$whatever_you_need = true;
}
}

Actually it's been so long I'm not sure if you use $current_User->('ID') or $current_User->get('ID') but I think it's the first. The second way (with a get) will probably display the ID on the page.

Many Thanks, you are right, the previous metod did not work. yYour method worked.

balupton,

I'm using 1.9.2 now and have tried to use your code like this:

<? $logged_in_users = array_keys($Sessions->_registered_Users);
  if( in_array($admin_id, $logged_in_users) { 
	$user = $user -1;
	} ?>

but I get the following errors:

Warning: array_keys() [function.array-keys]: The first argument should be an array in /home/alkha4/public_html/blog/addOns/visitors/users.php on line 43

Notice: Undefined variable: admin_id in /home/alkha4/public_html/blog/addOns/visitors/users.php on line 44

Warning: in_array() [function.in-array]: Wrong datatype for second argument in /home/alkha4/public_html/blog/addOns/visitors/users.php on line 44