what does "$var1->$var2" mean?

[Derleek]

So i run into this code every once and a while when i'm trying to find some example code or something:

 

$var1->$var2

 

Now what exactly does this represent?

[Drezard]

Look into OOP. $var1 is a class, and $var2 is the variable within the class.

 

So like:

 

$class1->$var1 = 10;

 

Daniel

[Derleek]

Ok, i'm actually familiar with some OOP (i have some history with C++).

 

So

$class1->$var1 = 10;

basically means; set $var1 to 10 in $class1?

[cooldude832]

yes however there are scope things you need to be observant of in OOP

 

in the class $this->$var1 = 10 means set the instant of the class that triggered this  object var value is 10

outside you use its globalscope name

[Drezard]

That last post was confusing for me too :S sorry cooldude832.

 

But what he was trying to say was:

 

when declaring the variable INSIDE the class (such as):

 

class class1 {

    public $var1 = 1;

    function test () {
        $this->var1 = 10;
        // NOT $var1 = 10;
    }

}

 

But in OUTSIDE use (such as):

 

$newclass = new class1;

$newclass->var1 = 10;
// NOT $var1 = 10

 

Hope that helps

 

Daniel

[cooldude832]

yup way to articulate something u said you couldn't understand

[Derleek]

yup that is what i meant drezard.  I missed the little syntax thing "$class1->var1" not "$class1->$var1"

 

I'll have to look into the specifics of OOP in php i guess....

[Drezard]

yup way to articulate something u said you couldn't understand

 

Hehehe... Yea i know

[Andy-H]

Or it can be a result of a mysql_fetch_object() call...

 

$query = mysql_query("SELECT * FROM table WHERE field = '$value' LIMIT 1")or die(mysql_error());

$fetch = mysql_fetch_object($query);

 

$someVar = $fetch->someField;

[cooldude832]

which is an object thus you are working on an object when you do so