Class as non-string argument in PHP - php

Question

Example in file 1:
namespace A;
class Foo{
}
file 2:
use A\Foo;
do_stuff('A\Foo'); // <- need namespace here :(
Foo::someStaticMethod(); // <- namespace not required :D
Is there any way I can pass class names in function arguments like constants or something, so I don't need to prepend the namespace?

Answer 1

Update :)
When I know, that I need to pass the classnames of some classes around as string I'm used to create special class constant
namespace Foo\Bar;
class A {
const __NAMESPACE = __NAMESPACE__;
const __CLASS = __CLASS__;
}
Now you can reference the classname like
use Foo\Bar\A as Baz;
echo Baz::__CLASS;
With PHP5.5 this will be builtin
echo Baz::class;
Full-Qualified-Names (FQN) for namespaces always starts with a namespace separator
do_stuff('\A\Foo');
except (and thats the only exception) in use-statements, because there can only appear complete namespace identifiers, so for convenience you can omit it there.
However, a string is a string and where you use it as a class name is out of scope of the interpreter, so it lost the reference to the former use A\Foo-aliasing. With PHP5.5 you can write Foo::class, but I think thats not an option right now ;)

Answer 2

You could instantiate a new object, then call get_class() to get the fully qualified name for the class.
use A\Foo;
$foo = new Foo();
do_stuff(get_class($foo)); // get_class($foo) = '\A\Foo'
This means that the namespace of Foo is only defined by the use statement (ie. less code maintenance).

Answer 3

Or you can pass class reflection.
ReflectionClass

Answer 4

No, not without tracing the caller, as far as I know. The function you are calling must exists within the same namespace as the object you are trying to pass.
You might want to have a look at the debug_backtrace function if you require the namespace resolution. But this requires the file-paths to be translated into namespace resolutions or similar.
This is however possible: (I see Andrew has answered with the same type of solution.)
function doStuff ($obj)
{
$name = (is_object($obj))
? (new ReflectionClass(get_class($obj)))->getName()
: $obj;
// $name will now contain the fully qualified name
}
namespace Common;
class Test
{}
$testObj = new Test();
// This will work, but requires argument to be
// fully quialified or an instance of the object.
\doStuff($testObj);
\doStuff("\Common\Test");