variable in variable in php

Question

$var="profile['Gamer']['last_name']";
echo ${$var};

Gives

Undefined variable: profile['Gamer']['last_name']

. But if i try to echo $profile['Gamer']['last_name'] value exist

I have tried echo $$var that too didn't work

Specifying an array index using curly braces isn't possible. — Unicron
– Unicron, Commented Mar 21, 2011 at 13:42
Why are you trying to do this? What's wrong with echo $profile['Gamer']['last_name']; — Joe
– Joe, Commented Mar 21, 2011 at 13:43

Marcel Jackwerth · Accepted Answer · 2011-03-21 13:43:08Z

4

There is no variable profile['Gamer']['last_name']. There is only a variable named profile.

$var = "profile";
echo ${$var}['Gamer']['last_name'];

answered Mar 21, 2011 at 13:43

Marcel Jackwerth

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troelskn · Accepted Answer · 2011-03-21 13:44:49Z

0

$var="profile['Gamer']['last_name']";
eval('$result = $' . $var . ';');
echo $result;

Keep in mind that there are serious security issues to something like this, and that your code will probably be very hard to understand for anybody but your self.

answered Mar 21, 2011 at 13:44

troelskn

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1 Comment

aWebDeveloper Over a year ago

my aim is to check if user has filled in data for a database field. If filled i assign him some points. To do so in another table i store the fild name and the points associated with it. so what's the best way to implement. and what r the security issues . None one but i have acess to thsi table

Michael S. · Accepted Answer · 2011-03-21 13:53:34Z

0

try this:

$var="profile['Gamer']['last_name']";
eval("echo $".$var.";");

edited Mar 21, 2011 at 13:53

answered Mar 21, 2011 at 13:44

Michael S.

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