How to use a variable in a namespace?

Question

Question by Chris

I’m trying to dig up my long-lost PHP skills and have therefore stumbled on a problem. I have a function that initializes (not sure if it’s a correct word for it) a class in the following fashion:

$foo = new MasterSlave$bar();

$bar is a defined class, obviously. But the entire thing doesn’t work. This only seems to work when I do the following:

$foo = new $bar();

But with my first example, it outputs the following error:

unexpected T_VARIABLE, expecting T_STRING

Which means that I have to manually enter the class name, correct? But, what if I’m a stubborn nerd who doesn’t want to and doesn’t see the efficiency in it? Hence my question; how to pull this off without getting a bloody error?

UPDATE: Got the thing working with wrapping the MasterSlave$bar in a $variable. Not sure if it’s the correct way of doing this, but it works and props go to Visual Idiot

Answer 1

Answer by Peter Kiss

Peter Kiss

Variables never bound to any namespace they will be always in the global scope.

Answer 2

Answer by Starx

Starx

AFAIK, the namespacing does not work the way you are trying to.

This is how you should do it

namespace MasterSlave;
$bar = "theclassname";
class $bar() {
}

Docs

How to use a variable in a namespace?

Question by Chris

Answer by Peter Kiss

Answer by Starx

Author: Nabin Nepal (Starx)

Please fill the form - I will response as fast as I can!