How to check if class exists within a namespace? - php

Question

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 1

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

Answer 2

The issue (as mentioned in the class_exists() manual page user notes) is that aliases aren't taken into account whenever a class name is given as a string. This also affects other functions that take a class name, such as is_a(). Consequently, if you give the class name in a string, you must include the full namespace (e.g. '\XXX\Driver\Driver', 'XXX\\Driver\\Driver').
PHP 5.5 introduced the class constant for just this purpose:
use XXX\Driver\Driver;
...
if (class_exists(Driver::class)) {
...
}