PHP, Fatal error: Call to undefined method, why?

Question

I have a simple php structures.

class Ingredient and class Ingredients, i have this code:

class Ingredient
{   
   public function objectIsValid()
   {
      return $validate[0];
   }
}



class Ingredients
{
   public $ingObject;
   function __construct(){   $ingObject = new Ingredient();   }

   public function validateData()
   {
      if($this->ingObject->objectIsValid()      /*** THE ERROR  ***/)
    {   echo "OK";}
      else
    {   echo "NOT";}
   } 
}


$Ingridients = new Ingredients();


$Ingridients->validateData();

I just can't understand why do i get the error..

any help will be appreciated.

thanks!

Artefacto · Accepted Answer · 2010-09-21 13:29:43Z

15

function __construct(){   $ingObject = new Ingredient();   }

ought to be

function __construct(){   $this->ingObject = new Ingredient();   }

In the first case you're setting a local variable, not a field, so it remains null. Then on the validateData you invoke a method on a null variable.

I'm assuming you snipped some code, because your Ingredient class doesn't make sense (there's a $validate variable there that isn't defined).

answered Sep 21, 2010 at 13:29

Artefacto

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1 Comment

smentek Over a year ago

And it would probably be better to transfer $ingObject as an argument in constructor.

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