PHP and variable variables ($$) syntax

Question

Before upgrading to PHP 7, I had this code and it returned true

var_dump(isset($$_SESSION['payment']) );
var_dump(is_object($$_SESSION['payment'])); 
var_dump($_SESSION['payment']); // string 'moneyorder'

After upgrading to PHP 7, I rewrote the same code inside a class, but now it returns false

var_dump(isset(${$_SESSION['payment']})); 
var_dump(is_object(${$_SESSION['payment']}));
var_dump($_SESSION['payment']); // string 'moneyorder'

Do you have an idea why ?

Thank you

Answer 1

Note the PHP documentation for superglobals contains this warning:

Note: Variable variables

Superglobals cannot be used as variable variables inside functions or class methods.

Save it to a local variable instead:

$payment = $_SESSION['payment'];
var_dump(isset(${$payment})); 
var_dump(is_object(${$payment}));