PHP syntax that will set a variable within a foreach loop using an array of variable names

Question

I am trying to define a set of variables by setting them from SESSION and POST variables of the same name. (I.e. I want to transfer $_SESSION['a'] to $a.

I am trying to use an array of variables names (a, b, c) to define such a set using a foreach loop, but am having when trying to define them I end up with syntax like $$variable within the loop, which does not work. I've tried single and double quotes around the $variable, but no joy

$data = array (
        'a',
        'b',
        'c',
              );

foreach($data as $variable)
 {
   if (isset($_POST['$variable'])) $_SESSION['$variable']=$_POST['$variable'];
   if (isset($_SESSION['$variable'])) {$$variable=$_SESSION['$variable'];} else {$$variable="";}
 }

Any help greatly appreciated.

I'm trying end up with many instances of something like:

 if (isset($_POST['$a'])) $_SESSION['$a']=$_POST['$a'];
 if (isset($_SESSION['$a'])) {$a=$_SESSION['$a'];} else {$a="";}

Oh dang, I accidentally deleted the intro explanation, let me edit — Gideon
– Gideon, Commented Nov 8, 2012 at 1:09
Variables are only interpolated in double quotes, not single quotes. And here was no need for either in the first place. — mario
– mario, Commented Nov 8, 2012 at 1:12
possible duplicate of Why is PHP not replacing the variable in string? — mario
– mario, Commented Nov 8, 2012 at 1:13
possible duplicate of Difference between single quote and double quote string in php — mario
– mario, Commented Nov 8, 2012 at 1:14
Have you read this page? php.net/manual/en/language.variables.variable.php — Nic Wortel
– Nic Wortel, Commented Nov 8, 2012 at 1:14

trapper · Accepted Answer · 2012-11-08 01:12:41Z

1

The single quotes prevent $variable from being evaluated, just take them out.

$data = array (
        'a',
        'b',
        'c',
              );

foreach($data as $variable)
 {
   if (isset($_POST[$variable])) $_SESSION[$variable]=$_POST[$variable];
   if (isset($_SESSION[$variable])) {$$variable=$_SESSION[$variable];} else {$$variable="";}
 }

answered Nov 8, 2012 at 1:12

trapper

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Thanks very much, I knew it would be something simple I was ignoring, got distracted by presuming $$ would break it.

Gideon
– Gideon

11/08/2012 01:17:28
Commented Nov 8, 2012 at 1:17

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