How to check if class exists within a namespace?

php oop namespaces
Taruo Gene picture Taruo Gene · Mar 14, 2014 · Viewed 79.2k times · Source

I've got this:

    use XXX\Driver\Driver;

...

var_dump(class_exists('Driver')); // false
        $driver = new Driver(); // prints 123123123 since I put an echo in the constructor of this class
        exit;

Well... this behaviour is quite irrational (creating objects of classes that according to PHP do not exist). Is there any way to check if a class exist under given namespace?

Answer

Alma Do picture Alma Do · Mar 14, 2014

In order to check class you must specify it with namespace, full path:

namespace Foo;
class Bar
{
}

and

var_dump(class_exists('Bar'), class_exists('\Foo\Bar')); //false, true

-i.e. you must specify full path to class. You defined it in your namespace and not in global context.

However, if you do import the class within the namespace like you do in your sample, you can reference it via imported name and without namespace, but that does not allow you to do that within dynamic constructions and in particular, in-line strings that forms class name. For example, all following will fail:

namespace Foo;
class Bar {
    public static function baz() {} 
}

use Foo\Bar;

var_dump(class_exists('Bar')); //false
var_dump(method_exists('Bar', 'baz')); //false

$ref = "Bar";
$obj = new $ref(); //fatal

and so on. The issue lies within the mechanics of working for imported aliases. So when working with such constructions, you have to specify full path:

var_dump(class_exists('\Foo\Bar')); //true
var_dump(method_exists('\Foo\Bar', 'baz')); //true

$ref = 'Foo\Bar';
$obj = new $ref(); //ok

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