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PHP Object-Oriented PHP Basics Building the Recipe Static Methods

Posted by ammarkhan
.a{fill-rule:evenodd;}techdegree
ammarkhan
Front End Web Development Techdegree Student 21,661 Points

Why did $recipe replaced $this?

Sorry but i am totally lost here, In the video i am lost on what is going on. Alena passed $recipe as a argument and then replaced $this with $recipe, why? I don't understand, what is the use of this?

3 Answers

Szabolcs Szána
Szabolcs Szána
6,961 Points
Szabolcs Szána
Szabolcs Szána
6,961 Points

Hi,

Beacuse it's a new another (static) class! :)

Peter Furler
Peter Furler
31,647 Points
Peter Furler
Peter Furler
31,647 Points

Elaborating on Szabolcs answer, $recipe is the parameter we used to pass an argument to within our method, which in this case our argument is $recipe1.

The pseudo variable $this is only used to refer to any future instantiated (single occurrence) object from inside the class that the object is created from. Which in our case is $recipe1 which is an object belonging to the Recipe class.

jlampstack
jlampstack
23,932 Points
jlampstack
jlampstack
23,932 Points

I was having problems with this too, until a lightbulb finally went on in my head. Will do my best to explain my understanding below.

The entire method / function is....

public static function displayRecipe($recipe)
{
    return $recipe->title . " by " . $recipe->source;
}

The section of the code we're modifying each time is....

return $recipe->title . " by " . $recipe->source;

Which is equivalent to....

return $recipe1->title . " by " . $recipe1->source;
return $recipe2->title . " by " . $recipe2->source;
return $recipe3->title . " by " . $recipe3->source;
etc.

To sum it up, the argument we pass in when we're calling the method will influence the result.