Instantiating class from string wont really work in php

Question

Ive just started learning namespaces. If I do this:

$a = new Devices\Desktop();

this works. But I have dynamic class names, so I have to do it from a variable.

$a = 'Devices\Desktop';
$a = new $a();

this is not working, although its the same. The class is not found. Why?

possible duplicate of http://stackoverflow.com/questions/5072352/instantiating-class-by-string-using-php-5-3-namespaces — underscore, May 20 '13 at 15:53
no, not possible duplicate, I read that, and didnt solve the problem. — user1929946, May 20 '13 at 15:55
judging on the answers to the other question, it looks like the code you posted is correct. the problem appears to lie elsewhere (perhaps the class is declared in a different namespace?) — user428517, May 20 '13 at 15:56

deceze · Accepted Answer · 2013-05-20T16:42:08.917

3

Going out on a limb:

namespace Foo;

class Bar { }

new Bar;

$bar = 'Bar';
new $bar;

This won't work. String class names are always absolute, they're not resolved relative to the current namespace (because you can pass strings from one namespace to another, which should it resolve against then?). To make those two instantiations equivalent, you need to use a fully qualified class name:

$bar = 'Foo\Bar';

The __NAMESPACE__ constant can be useful here:

$bar = __NAMESPACE__ . '\Bar';

edited May 20 '13 at 16:42

answered May 20 '13 at 15:56

deceze

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I see. Is this in the official document? – user1929946 May 20 '13 at 15:58
Probably somewhere, yes. :-3 – deceze May 20 '13 at 15:59

Instantiating class from string wont really work in php

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