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

Just a precision : Before the data was in a simple files and now the same datas are in class. — kurama
– kurama, Commented Jun 3, 2016 at 20:42
So does your original code run in PHP 7 or not? Because it seems like your question should say "After upgrading to PHP 7, rewriting some lines of code, and putting them in a class, it works differently." — miken32
– miken32, Commented Jun 3, 2016 at 20:49
I suggest you redesign your code so you don't need variable variables. Anything you do with them should probably be done using a associative array. — Barmar
– Barmar, Commented Jun 3, 2016 at 21:46

miken32 · Accepted Answer · 2016-06-03 21:11:39Z

1

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}));

answered Jun 3, 2016 at 21:11

miken32

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