php - check if class name stored in a string is implementing an interface

Question

I understand that my question is somehow wrong, but I'm still trying to solve this problem.

I have an interface Programmer:

interface Programmer {
    public function writeCode();
}

and a couple of namespaced classes:

Students\BjarneProgrammer (implements Programmer)
Students\CharlieActor (implements Actor)

I have this class names stored in array $students = array("BjarneProgrammer", "CharlieActor");

I want to write a function, that will return an instance of class if it's implementing Programmer interface.

Examples:

getStudentObject($students[0]); - It should return an instance of BjarneProgrammer because it's implementing Programmer.

getStudentObject($students[1]); - It should return false because Charlie is not a Programmer.

I tried it using instanceof operator, but the main problem is that I do not want to instantiate an object if it's not implementing Programmer.

I checked How to load php code dynamically and check if classes implement interface, but there is no appropriate answer as I don't want to create object unless it's returned by function.

@how: The OP says they don't want to instantiate the class.

Ilmari Karonen
– Ilmari Karonen

2015-11-18 23:06:23 +00:00
Commented Nov 18, 2015 at 23:06 — Ilmari Karonen
– Ilmari Karonen, Commented Nov 18, 2015 at 23:06

The Alpha · Accepted Answer · 2024-06-24 10:23:48Z

37

You can use class_implements (requires PHP 5.1.0)

interface MyInterface { }
class MyClass implements MyInterface { }

$interfaces = class_implements('MyClass');
if($interfaces && in_array('MyInterface', $interfaces)) {
    // Class MyClass implements interface MyInterface
}

You can pass the class name as a string as function's argument. Also, you may use Reflection

$class = new ReflectionClass('MyClass');
if ( $class->implementsInterface('MyInterface') ) {
    // Class MyClass implements interface MyInterface
}

Update : (You may try something like this)

interface Programmer {
    public function writeCode();
}

interface Actor {
    // ...
}

class BjarneProgrammer implements Programmer {
    public function writeCode()
    {
        echo 'Implemented writeCode method from Programmer Interface!';
    }
}

Function that checks and returns instanse/false

function getStudentObject($cls)
{
    $class = new ReflectionClass($cls);
    if ( $class->implementsInterface('Programmer') ) {
        return new $cls;
    }
    return false;
}

Get an instance or false

$students = array("BjarneProgrammer", "CharlieActor");
$c = getStudentObject($students[0]);
if($c) {
    $c->writeCode();
}

Update: Since PHP - 5.3.9+ you may use is_a, example taken from PHP Manual:

interface Test {
    public function a();
}

class TestImplementor implements Test {
    public function a() {
        print "A";
    }
}

$testImpl = new TestImplementor();

var_dump(is_a($testImpl, 'Test')); // true

Working example here.

edited Jun 24, 2024 at 10:23

answered Nov 24, 2013 at 1:10

The Alpha

147k30 gold badges294 silver badges313 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Stichoza Over a year ago

Thanks, your answer helped me. I used just ReflectionClass - codepad.org/EvZRbaWZ

Stichoza Over a year ago

I'm using method from ReflectionClass return $class->newInstance(); instead of return new $cls;. Thanks again :>

The Alpha Over a year ago

That's fine I think :-)

ircmaxell · Accepted Answer · 2013-11-24 13:00:49Z

27

If you're using a modern version of PHP (5.3.9+), then the easiest (and best) way would be to use is_a() with the third parameter true:

$a = "Stdclass";

var_dump(is_a($a, "stdclass", true));
var_dump(is_a($a, $a, true));

Both of those will return true.

answered Nov 24, 2013 at 13:00

ircmaxell

166k36 gold badges269 silver badges316 bronze badges

2 Comments

Ivan Yarych Over a year ago

PHP documentation is kind of misleading (if you don't care to read every sentence :) ) - it says first parameter of is_a is an object. But if you read on - you find out it can actually be a string...

Alberto Suárez Over a year ago

For me, this was the more elegant solution

Michael Käfer · Accepted Answer · 2021-06-17 15:36:52Z

10

Use PHP's function is_subclass_of():

use App\MyClass;
use App\MyInterface;

if (is_subclass_of(MyClass::class, MyInterface::class)) {
    // ...
}

answered Jun 17, 2021 at 15:36

Michael Käfer

1,8062 gold badges24 silver badges41 bronze badges

Collectives™ on Stack Overflow

php - check if class name stored in a string is implementing an interface

3 Answers 3

3 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related