PHP检查实际的类是否正在实现接口

Question

On an existing project I'm working on, we have the following situation:

interface A { }

class B implements A { }

class C extends B { }

class D extends B implements A { }

$B = new B();
$C = new C();
$D = new D();

What is the proper way to figure out if the actual class implements the interface A and not only the parent class? The check should return true for $B and $D and false for $C. Normally you would do

if( $C instanceof A ) { //do the work }

but in our situation this will return true which shouldn't.

An approach can be to parse the file and test if the class really implements A with the token_get_all function. But before doing that, I want to ask if there are more elegant solutions.

I know it sounds weird, but the situation is as it is and the class hierarchy can't be changed. Any insights would be helpful.

Answer 1

This function returns true only if interface A is implemented not extended via parent class.

echo checkimplements($B, "A"); //Returns True

function checkimplements($class, $interfacename)
{
    $ownInterfaces = class_implements($class);
    $parent = get_parent_class($class);
    if($parent) {
        $parentInterfaces = class_implements($parent);
    } else {
        $parentInterfaces = array();
    }
    $diff = array_diff($ownInterfaces, $parentInterfaces);
    $found = in_array($interfacename, $diff);
    return $found;
}