Variable value as part of array key

Question

$pname = "postback_".$data;
$_SESSION["$pname"] = $data;

It is not working. How to use it?

Single quoted strings are literal and do not have variable substitution (you don't even need quotes in your case). — Prix
– Prix, Commented Jul 3, 2010 at 8:05
@Prix: I don't see single quoted strings here ;) (but of course you are right). — Felix Kling
– Felix Kling, Commented Jul 3, 2010 at 8:13
@Felix i am just saying that when using single quotes it does not matter if u have a variable on it as it will not be parsed hence you should just use the session variable name for example $_SESSION['time'] would be time other then this using quotes are not a good pratice in my point of view. My bad making it complicated. — Prix
– Prix, Commented Jul 3, 2010 at 8:15
@monkey_boys - Please click the tick mark sign, beside the correct answer, if you have already found one. — Knowledge Craving
– Knowledge Craving, Commented Jul 8, 2010 at 7:14

vick · Accepted Answer · 2010-07-03 08:00:42Z

5

<?php
session_start();

$pname = "postback_".$data;
$_SESSION[$pname] = $data;

?>

answered Jul 3, 2010 at 8:00

vick

4766 silver badges18 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sarfraz · Accepted Answer · 2010-07-03 07:58:13Z

0

Make sure that you have put session_start(); on top of your script.

answered Jul 3, 2010 at 7:58

Sarfraz

383k81 gold badges561 silver badges613 bronze badges