PHP OOP declaring a variable to an upper class

Question

First of all, I want to write two class.

<?php

class A
{
    function load($v)
    {
        $this->$v = "stack";
    }
}

class Base
{
    var $lib = null;

    function __construct()
    {
        $this->lib = new A();
    }
}


$bs = new Base();
$bs->lib->load("libx");

// Current calling method
$bs->lib->libx;




// But I want this
$bs->libx;

?>

I write a MVC fremawork. CodeIgniter can do this but I could not. My english is poor. Because of this, don't talk me complicated please.

No I need to not use extend for this. Because there are a lot of class like this. — CodeMax
– CodeMax, Commented Mar 3, 2015 at 17:12
I think then you are better off using a shortcut method as mentioned in an answer here (which has been deleted it seems) — Maximus2012
– Maximus2012, Commented Mar 3, 2015 at 17:13
unfortunately i must call like this because; when I write html in View method, I can't call "$bs->lib->xx". because there is no "lib" variable in view. So I don't want to break my code and I want to do this like this. — CodeMax
– CodeMax, Commented Mar 3, 2015 at 17:21

mhall · Accepted Answer · 2015-03-03 17:20:40Z

1

You can add a magic __get method to Base to look for the property in lib, and return it if it exists:

class A
{
    function load($v)
    {
        $this->$v = "stack";
    }
}

class Base
{
    var $lib = null;

    function __construct()
    {
        $this->lib = new A();
    }

    public function __get($v)
    {
        if (property_exists($this->lib, $v)) {
            return $this->lib->$v;
        }
    }
}

Usage:

$bs = new Base();
$bs->lib->load("libx");

echo $bs->lib->libx, PHP_EOL;
echo $bs->libx, PHP_EOL;

Output:

stack
stack

answered Mar 3, 2015 at 17:20

mhall

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1 Comment

CodeMax Over a year ago

Thank you very much. I wanted to this :)

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