从类引用中获取类名称空间和名称，而不更改我的类

Question

In php you often check if the object you get is of the correct type. If not you throw an exception with a message like this:

use My\Class\MyClass

...

if (!$object instanceof MyClass) {
    throw new Exception(
        sprintf(
            "object must be of type '%s'", 
            'My\Class\MyClass'
        )
    );
}

Right now I pass the full namespace and the name of the class in a string to sprintf.

How can I get this from the class reference so that i can do something like this

sprintf("object must be of type '%s'", MyClass::getName())

EDIT:

I would like to achieve this for all classes without adding new methods. So it should be a solution using some existing method or one of the php __ MAGIC__ methods.

Answer 1

As of php 5.5 there's a magic constant that gives you the FQCN of any class. You can use it like this:

namespace My\Long\Namespace\Foo\Bar;

MyClass::class;
// will return "My\Long\Namespace\Foo\Bar\MyClass"

It's documented on the new features page.