TalkPHP - View Single Post - The "$this->" variable...

xenon · 05-13-2009, 05:38 PM

You've got it somewhat right, but not fully. $this only refers to the current instance of an object created from a class. So, given the following simple class:

Code:

class SimpleClass
{
    public $var1 = 'my first variable';
}

$object1 = new SimpleClass();
echo $object1->var1; // outputs 'my first variable'

$object2 = new SimpleClass();
$object2->var1 = 'the second one';
echo $object2->var1; // outputs 'the second one'

echo $object1->var1; // still outputs 'my first variable'

Note that the code above is not optimized, it's only used for demonstration purposes.

Now, you can see that we transform the $var1 public member (that's the name of class variables). In each case, we have a 'separate' $this variable at the class level (inside it), that is tied to each object you create from that class. $this is a special variable referring to "the current object". And that's really what it is...so you don't have to create instances of the class you're writing inside the very same class. Instead, you're only programming a prototype of an object, customizing each instance as you need. $this, however, is not available outside a class (as you've probably already seen), because then it would loose its meaning.

I hope I've been clear enough. If not, feel free to ask questions, and you will be answered

Keep up the good work

05-13-2009, 05:38 PM	#2 (permalink)
xenon The Frequenter Join Date: Dec 2007 Location: Bucharest, Romania Posts: 438 Thanks: 3	You've got it somewhat right, but not fully. $this only refers to the current instance of an object created from a class. So, given the following simple class: Code: class SimpleClass { public $var1 = 'my first variable'; } $object1 = new SimpleClass(); echo $object1->var1; // outputs 'my first variable' $object2 = new SimpleClass(); $object2->var1 = 'the second one'; echo $object2->var1; // outputs 'the second one' echo $object1->var1; // still outputs 'my first variable' Note that the code above is not optimized, it's only used for demonstration purposes. Now, you can see that we transform the $var1 public member (that's the name of class variables). In each case, we have a 'separate' $this variable at the class level (inside it), that is tied to each object you create from that class. $this is a special variable referring to "the current object". And that's really what it is...so you don't have to create instances of the class you're writing inside the very same class. Instead, you're only programming a prototype of an object, customizing each instance as you need. $this, however, is not available outside a class (as you've probably already seen), because then it would loose its meaning. I hope I've been clear enough. If not, feel free to ask questions, and you will be answered Keep up the good work __________________ I have optimistic thoughts, even though sometimes (if not always) life's a bitch.