TalkPHP - Is referencing a class bad practice?

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MartynMJ

11-30-2007 01:38 PM

Is referencing a class bad practice?

Just wondering is referencing a class in another class bad? Like so

PHP Code:


		
			
<?php



class test

{

   var $testing = "";



   function blah()

   {

      return $this->testing;

   }

}



?>

And then in another file

PHP Code:


		
			
<?php



//include the file with the class above in



class two

{

   var $test = "";



   function two ( &$test )

   {

      $this->test = $test;

   }



   function again ()

   {

      $data = $this->test->blah();

      return $data;

   }



}



?>

Hope that makes sense? :p

Wildhoney

11-30-2007 01:48 PM

No, not at all. Passing in objects is perfectly fine and allows you to place varying logic into separate classes. What you have there is just a simplified example of what was being spoke about in the interfaces tutorial, err, tutorials. Geez! How many do we have? :-)

MartynMJ

11-30-2007 02:41 PM

Heh, Thats good to know.

Just wondering though. Could the way the server's configured make it act differently?

I mean i've developed a script locally and it works fine then i've put it on a web server and its not working properly.

how i mean is i'm referencing a class and then in a function passing some data to an array in the referenced class.

Then back on the main page it called to a functio on the class that was passed data to and the data doesn't seem to be there.

Any idea why it would do this?

Karl	11-30-2007 02:43 PM

The statement "Then back on the main page", sounds like you're trying to keep the class data alive during page reloads, is this the case?

MartynMJ

11-30-2007 07:04 PM

No, Here's an example.

index.php

PHP Code:


		
			
<?php

require_once ( "class1.php" );

require_once ( "class2.php" );



$class1 = new class1;

$class2 = new class2 ( $class1 );



$class2->blah();



echo $class1->execute();



?>

class1.php

PHP Code:


		
			
<?php



class class1

{

   var $data = "";

   

   function execute()

   {

      return $this->data;

   }



}



?>

class2.php

PHP Code:


		
			
<?php



class class2

{

   var $class1 = "";

   

   function class2 ( &$class1 )

   {

       $this->class1 = $class1;

   }



   function blah ()

   {

      $this->class1->data = "TESTING";

   }



}



?>

Now on index.php it should echo TESTING right? But this doesn't appear to be working on a server i'm hosting the site on

Karl	11-30-2007 09:26 PM

That is very strange, what version of PHP is the server using?

sketchMedia

11-30-2007 11:25 PM

Tis indeed strange, it works fine for me

Wildhoney

11-30-2007 11:35 PM

This thread doesn't need another "me too", but it's going to get one anyway! It works for me as well. I guess it all comes to the PHP version, although even still it's still very bizarre.

bluesaga

12-01-2007 12:14 AM

Pre-PHP 5 when you refer to a class in a function call, without using the ampersand operator then PHP would clone the class and not simply pass a reference as you are hoping for.

your code:

PHP Code:


		
			
<?php
require_once ( "class1.php" );
require_once ( "class2.php" );

$class1 = new class1;
$class2 = new class2 ( $class1 );

$class2->blah();

echo $class1->execute();

?>

Should be:

PHP Code:


		
			
$class2 = new class2 ( &$class1 );

MartynMJ

12-01-2007 01:55 AM

Quote:

Originally Posted by bluesaga (Post 4934)

Pre-PHP 5 when you refer to a class in a function call, without using the ampersand operator then PHP would clone the class and not simply pass a reference as you are hoping for.

your code:

PHP Code:


		
			
<?php

require_once ( "class1.php" );

require_once ( "class2.php" );



$class1 = new class1;

$class2 = new class2 ( $class1 );



$class2->blah();



echo $class1->execute();



?>

Should be:

PHP Code:


		
			
$class2 = new class2 ( &$class1 );

First of all PHP Version 4.4.7.

And i tried adding the ampersand but still got the same outcome. As this needs to be done soon i have worked around it by returning the data and then calling to the template functions but this was quite a pain :p

Thanks all ^^

Salathe

12-01-2007 02:15 AM

There's a very simple solution.

Change: (in the class2 constructor)

PHP Code:


		
			
$this->class1 = $class1;

To:

PHP Code:


		
			
$this->class1 =& $class1;

Why does this work? It's all about the way that PHP4 handles passing around references to objects. The &$class1 argument for the constructor ensured that the argument was brought in by reference, but when it was being assigned to the class1 property the assignment was by value -- a copy of the object was made and assigned to the property.

When we changed the data property, it was only changing it in the copy and not the original. Using =& the variable is assigned by reference as we intended. Hopefully you can see now why this behaviour was changed in PHP5 to assign by reference by default. :)

MartynMJ

12-01-2007 02:26 AM

Quote:

Originally Posted by Salathe (Post 4947)

There's a very simple solution.

Change: (in the class2 constructor)

PHP Code:


		
			
$this->class1 = $class1;

To:

PHP Code:


		
			
$this->class1 =& $class1;

^^^^^^^^^^^^^^^^^^^^^^

Works perfectly. Greatly appriciated :D.

Thanks very much :)

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