Why cannot I get parent variable in PHP?

Question

Code:

class A {
    public $arrayToReturn = array();
    function __construct() {
        $arrayToReturn['country'] = 'US';
    }
}
class B extends A {
    public function foo() {
        print_r($arrayToReturn);
   }
}
$b = new B();
$b->foo();
print_r($b->arrayToReturn);

Result:

PHP Notice:  Undefined variable: arrayToReturn in xxxx.php on line 11
Array
(
)

I cannot get array in functions of class B, and get an empty array which should has one element by $b->arrayToReturn call. Why?

flagged as "why this code is not working". please give more details. — Alexandre Martin, Aug 19 '16 at 18:26
and then of course, also dupe of http://stackoverflow.com/questions/16959576/reference-what-is-variable-scope-which-variables-are-accessible-from-where-and — Marc B, Aug 19 '16 at 18:29

score 3 · Answer 1 · answered Aug 19 '16 at 18:26

3

You have to use $this->arrayToReturn:

class B extends A {
    public function foo() {
        print_r($this->arrayToReturn);
   }
}

Same in class A:

class A {
    public $arrayToReturn = array();
    function __construct() {
        $this->arrayToReturn['country'] = 'US';
    }
}

answered Aug 19 '16 at 18:26

pavlovich

1,935
15
20

Thanks, it solved my first problem. But the array was added one element in the parent constructor, why did I get an empty array? – sxlllslgh Aug 19 '16 at 18:32
Look at the duplicate. _Why aren't parent constructors being called?_ – AbraCadaver Aug 19 '16 at 18:33
@sxlllslgh you have to create constructor in your class B also and call `parent::__construct();` from it, then you will have it ;) Good luck – pavlovich Aug 19 '16 at 18:37
@pavlovich Yeah now I understand it.Thanks a lot.:) – sxlllslgh Aug 19 '16 at 18:41

Why cannot I get parent variable in PHP?

1 Answers1