CODEWITHSUNDEEP
php
user1270172
user1270172

Reputation: 73

PHP: __call called instead of __callStatic

I'm developing a custom ORM and here's a simplified example to demo the issue:

class Foo {
    public static function __callStatic($name, $args){
        if($name == "getBar"){
            return "Bar";
        }
        return NULL;
    }

    public function __call($name, $args){
        if($name == "getFoo"){
            $model = "Foo";
            $method = "getBar";            
            return $model::$method();            
        }
        return NULL;
    }
}

class Foo2 {
    public function __call($name, $args){
        if($name == "getFoo"){
            $model = "Foo";
            $method = "getBar";            
            return $model::$method();            
        }
        return NULL;
    }
}

echo Foo::getBar();//Bar

$foo = new Foo;
var_dump($foo->getFoo()); //Null though I'm expecting Bar

$foo = new Foo2;
var_dump($foo->getFoo()); //Bar

So why same method when called inside Foo triggers __call and inside Foo2 triggers __callStatic?

Upvotes: 2

Views: 268

Answers (1)

xate
xate

Reputation: 6379

Thats because in the first call where you get Null returned, your in an object context. So it is calling __call() instead of __callStatic().

EDIT: In the first case, Foo::getBar() is called in the scope of an instance of class Foo, so Foo::getBar() is actually the same as (instance)->getBar() which of course isn't a static call.

EDIT2: Just found a question + answer regarding your question: Why does PHP prefer __call() to __callStatic()?

Upvotes: 2

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