Constant, isset and empty evaluation - php

Question

Does anyone know how isset and empty is interpreted by the translator, or how the translator treats undefined variables?
To expand on my question, I have the below code in my include file:
define('USER_AUTH', !empty($_SESSION['username']) && !empty($_SESSION['status'])); //user is verified
define('ACC_IS_PENDING', $_SESSION['status'] == '0');//checks to see if user status is pending which has a status of 0
USER_AUTH is used as a quick hand to check if user is authenticated. ACC_IS_PENDING is used as a quick hand for when the account status is pending. However, PHP gives me a notice to advise me that $_SESSION['status'] in my second line of code is undefined. I know that it is undefined, but I haven't used it yet! How dare you tell me what I already know. LoL
However, when I trick the code with the below:
define('USER_AUTH', !isempty($_SESSION['username']) && !isempty($_SESSION['status']));
define('ACC_IS_PENDING', $_SESSION['status'] == '0');
Where isempty() is a custom made function that will always return FALSE. Then no notice!
Alternatively, if I use the below code:
define('USER_AUTH', notempty($_SESSION['username']) && notempty($_SESSION['status']));
define('ACC_IS_PENDING', $_SESSION['status'] == '0');
Where notempty() always return TRUE, then again no notice.
Am I right in saying that the translator checks that the variable has been tested once, and that if the test resulted in true, then the translator sees this as the variable has been defined?
If this was the case, then what about isset and empty? They both seem to give me notices no matter if the evaluation is true or false.
define('USER_AUTH', isset($_SESSION['username']) && isset($_SESSION['status']));
define('ACC_IS_PENDING', $_SESSION['status'] == '0');
and
define('USER_AUTH', empty($_SESSION['username']) && empty($_SESSION['status']));
define('ACC_IS_PENDING', $_SESSION['status'] == '0');
Apologies for the long winded question. This seems trivial, but it would be nice to have a quick defined constant without having to get notices! Any help in explanation or a better solution for such trivial task would be appreciated, thanks!

Answer 1

PHP complains because the index 'status' is not defined in the array. You would need to write
!isset($_SESSION['status']) || empty($_SESSION['status']
When you "trick" the code as described, PHP will never try to access the non-existing array index, which is why you don't get any notice.
In the fourth code example (with isset), you are still accessing the non-existing array index in the second line of code, so I suspect that's why there's still a notice.

Answer 2

a) If you wants to use $_SESSION['status'], you have to check first that the variable is not empty ( see http://www.php.net/manual/fr/function.isset.php for more details) :
if (isset($_SESSION['status'])) {
// here the value exists and can be used
Do something...
} else {
// here the value does not exist and cannot be used
}
b) I believe that
empty($_SESSION['username']) && !empty($_SESSION['status'])
is not constant : it varies from one run to the other. You may want to use
$user_is_logged = empty($_SESSION['username']) && !empty($_SESSION['status']);
and use the variable $user_is_logged instead of a constant. See this section http://www.php.net/manual/fr/language.constants.php for a speek about constants.