Arrays as objects

Question

I there a method to allow me to use an array as an object?

I have the following code in my main class

function  __construct()
{
        require 'config.php';
        $this->config = new PHPSO_Config();
}

where the PHPSO_Config class looks something like this

class PHPSO_Config
{
    /**
     * The administrators email address
     */
    var $adminEmail = 'user@domain.tld'; //I'm using this email to test
}

so can I accomplish the same thing and be able to access the config vars as an object without creating a second class?

Answer 1

You can cast an array to an object using

$myObject = (object) $myarray;

thus the following should work (untested):

function  __construct()
{
        require 'config.php';
        $this->config = (object) array("username" => "root");
}

then $class->config->username should work.

Answer 2

You could use PHP5 overloading to make it seem that your class has the properties provided by the config class, e.g. something like this:

class MyClass
{
    private $config;

    function  __construct()
    {
        require 'config.php';
        $this->config = new PHPSO_Config();
    }


    public function __get($name) {
        if (property_exists($this->config, $name)) {
            return $this->config->$name;
        }
        return null;
}

Then doing something like this:

$config=new MyClass;
echo $config->adminEmail;

Would result in seeing the adminEmail member variable of the internal config class.

This is almost a Strategy Pattern, where the runtime implementation of a class can be selected at runtime.

Answer 3

I'm not sure exactly what you want, but you can cast an array to an object. Make sure you don't use numeric indices though as they don't work quite right;

$config = array('adminEmail' => 'user@domain.tld');
$this->config = (object)$config;

Answer 4

Maybe I am missing something here but are you trying to replace your config class with an array, then cast it to an object? Sounds a little crazy to me. I would just have a config class, its way better design.

Its best not to go around making arrays where you should have classes.

Answer 5

Well yeah

there are a technique that personally I would use it:

you create a file call whatever you want (i.e. config.php) then write this in it forexample:

Whenever you wanted use these information you have to just put "require_once "config.php";" above your code like in index.php and all definition will be in that page. For example you can read database server name by typing $DB_SERVER.

If I were you I would stick to it because you can use it everywhere in your project and maintain it just from a file which makes life easier.