How to set public variable dynamically when a class instance is created

Question

When I initialize a new class instance I want to set a public variable based on the current url.

How do I go about setting the public variable dynamically when the class instance is created so it is available without having to call a function.

 class CONFIGURATOR{
    static public $ACTIVE = true;
    public $CURRENT_URL="<current_url here>"
  }

You are unclear with your question, but you can either populate that variable via constructor when you're creating the object or after you created the object via $obj->CURRENT_URL = 'your url'; — N.B.
– N.B., Commented May 15, 2013 at 8:56

Alarid · Accepted Answer · 2013-05-15 08:58:28Z

2

Use the constructor :

class CONFIGURATOR{
   static public $ACTIVE = true;
   public $CURRENT_URL="<current_url here>"

   public function __construct($url)
   {
       $this->CURRENT_URL = $url;
   }
}

You can call your object like this :

$configurator = new CONFIGURATOR($your_url);

answered May 15, 2013 at 8:58

Alarid

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Comments

Subhra Sekhar Mukhopadhyay · Accepted Answer · 2013-05-15 08:59:36Z

0

Use constructor , it may be solve your problem

class CONFIGURATOR{
    static public $ACTIVE = true;
    public $CURRENT_URL="";

    function __construct()
    {
        $this->$CURRENT_URL="<current_url here>";
    }
  }

Thank You

answered May 15, 2013 at 8:59

Subhra Sekhar Mukhopadhyay

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