Var inside a class not being assigned error

BigBadKev

Hi all,

I have this class called browser class and it outputs the browser and other details using $_SERVER["HTTP_USER_AGENT"] and I have a Var called $platform that i am wishing to assign the platform returned within the class but it never gets populated? can anyone help? here is the class

//class called
$b = new Browser;

//The Class
class Browser {

  var $platform;

  function Browser() {
     $this->_get_browser_details(); 
  }

  function platform() {
    return $this->platform;
  }

  function _get_browser_details() {
    $useragent = $_SERVER["HTTP_USER_AGENT"];
    //echo $useragent = Mozilla/4.0 (compatible; MSIE 5.5; Windows 98)

//I know $useragent is a string because I used is_string($useragent)
if (strstr($useragent, 'W')) {
  //Here is the start of the error
  $this->platform = 'Win'; //This fails.....
  $platform = "Win";       //This works.....
}
  }
}

Can someone explain to me why this is failing? is it because the "$this" is pointing to the function platform rather than the variable?

Thanks for any help

BBK
😃

jimson

i copied ur code exactly as u wrote

paste in my textpad.

$this->platform = 'Win';
//$platform = "Win";

i print_r($b) and i can see platform=Win

//$this->platform = 'Win';
$platform = "Win";

and i get platform with no values...

well... basically, u need more coffee. 🙂

BigBadKev

Hi thanks for replying,

I am totally confused?? I also get what you do so we at least kno w the Var $platform has the value = "Win"

With your print_r($b); function I get this : browser Object ( [platform] => Win )

But when I run it through my debugger it is empty, my function is this : $d->s("platform = " . $platform);

$d is the Debug Class
$s is a function in the Debug Class that checks if what is passed through is either is_string or is_bool or is_integer or is_object or is_array, I am assuming that the $platform is something else? like a resource type var?

My Debug basically grabs the type and prints it out so I am thinking i havent covered what it is?

Thanks for any help

BBK
😃

BigBadKev

Hi,

You got me thinking while i was drinking my coffee 🙂 about print_r and I finaly clicked that it is a property of the object so when I was doing this : $d->s("platform = " . $platform); I was trying to append the var which was empty

So i did this : $d->s("platform = " . $this->platform); and guess what...it printed out "Win".....which is what it is pointing to

Can you answer another question for me? why does the var $platform only point to "Win" rather than hold the value "Win"?

Thank you for showing me the print_r function as it made me realise what type of object I ws looking at.

BBK
😃

jimson

first, i am sorry for late reply... coz i got some difficulty to reach phpbuilder.com

Can you answer another question for me? why does the var $platform only point to "Win" rather than hold the value "Win"?

i really hope you can explain a bit your question... :?
i don't seem to understand.

but basically,

when you say a variable is pointing to a value, it means,

$a = 'apple';
$b = &$a;
$b = 'bag';
echo $a;
// you will find $a = bag
// this mean $b is a reference to $a
// and usually they use reference in function parameter to prevent system making another copy of variable.
// of course, to save computer's memory!
// eg.

function test(&$plus) {
     $plus++;
}

$c = 5;
test($c);
echo $c // will output $c = 6

holding a value means

$a = 44;
$b = 55;
// so variable $a holds 44 (integer) and $b holds 55 (integer).
// thats all.

you can read more on this in "references explained" inside php manual.