PHP OOP - I am trying to output an array which has been loaded into a different class

Question

class_1.php

class class_1
{
    public function __construct()
    {
        require_once $_SERVER['DOCUMENT_ROOT'].'/array.php';
    }
}

class_2.php

class class_2
{
    static function output_array()
    {
        return $array_NUMBERS;
    }
}

array.php

$array_NUMBERS = array('1', '2', '3');

page.php

require_once $_SERVER['DOCUMENT_ROOT'].'/class_1.php';

require_once $_SERVER['DOCUMENT_ROOT'].'/class_2.php';

$obj = new class_1();

$numbers = class_2::output_array();

echo '<pre>';
print_r($numbers);

What am I doing wrong here? Aren't you supposed to use "require_once" within classes?

Problem: It's not outputting the array values.

Please edit the post to include your error message, if any, or how your program functions differently than your expectations. We are not mind readers. — user229044
– user229044 ♦, Commented Mar 17, 2011 at 22:33

user229044 · Accepted Answer · 2011-03-17 22:43:38Z

3

You're creating a local variable in class_1::__construct(), which falls out of scope immediately and is lost forever.

Even if it weren't, variables declared inside a function are local to that function, and cannot be access from other functions. class_2::output_array() has no concept of the variable $array_NUMBERS. You need to read up on scope.

To make this work, you'd have to make the variable a public member of class_1:

class class_1 {
    public $array_NUMBERS;
    function __construct() {
        require_once('array.php');
    }
}

class class_2 {
    public static function output_array() {
        $class1 = new class_1();
        return $class1->array_NUMBERS;
    }
}

array.php:

$this->array_NUMBERS = array(...);

edited Mar 17, 2011 at 22:43

answered Mar 17, 2011 at 22:35

user229044♦

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Comments

Keith Gaughan · Accepted Answer · 2011-03-17 22:37:43Z

0

Think of it this way: include() and co. are for pulling code in at a certain place, essentially what's happening in class_1.php is that you're ending up with the equivalent of this:

class class_1
{
    public function __construct()
    {
        $array_NUMBERS = array('1', '2', '3');
    }
}

Thus, $array_NUMBERS is a local variable within the constructor, which is why class_2 can't see it.

answered Mar 17, 2011 at 22:37

Keith Gaughan

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Comments

jeroen · Accepted Answer · 2011-03-17 22:37:59Z

0

With your require_once your code becomes:

class class_1
{
    public function __construct()
    {
        $array_NUMBERS = array('1', '2', '3');
    }
}

So what you are doing, is assigning a variable in a function. As a result, the scope of that variable is only inside that function.

There are ugly hacks to get the variable to your other class / function, but the whole approach seems wrong (also see @Mike Lewis's answer).

A solution would be to define a variable in class_1, assign the array to that variable and pass the instance of that class as an argument to your output_array function.

answered Mar 17, 2011 at 22:37

jeroen

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