5 replies to this topic

[Mr. Jay]

I was wondering how to get $this->test->main() to work

<?php
	class test
	{
		function main()
		{
			return rand(1,10000);
		}
	}
	echo $this->test->main();
?>

Thanks

[Hayden]

$this->main();

that is only used within the class as such

class test {
   function main() {
	  $hello = "Hello World";
	  return $hello;
   }
   echo $this->main();
}

if you want to use it outside of the class, for instance in your index.php where you have included the file where the class is located.

then you would do this...

<?php
include "file_with_class.php";

$temp = new test();

echo $temp->main();
?>

[Mr. Jay]

Thank you SpatialVisionary

[rc69]

Slight correct on the "used within a class" statement.

class test {
   function bar(){
	  return 'The bar() method.';
   }

   function main() {
	  echo $this->bar();
   }
}

$obj = new test();
$obj->main();

Last i checked, all that a class definition could contain was function definitions, not calls to functions.

More info can be found at php.net, nothing specifically about $this though (which is sort of depressing).
http://php.net/manua...anguage.oop.php
http://php.net/manua...ef.classobj.php

Edited by rc69, 13 May 2007 - 02:01 AM.

[Mr. Jay]

One more thing is it possible to do $temp->username->main('lalala')

username = $_GET['name'] or w/e

In return it'll show (username): lalala

<?php
	class test
	{
		function main($text)
		{
			return (username?).": ".$text;
		}
	}
?>

<?php
include "test.php";

$temp = new test();

echo $temp->username->main('lalala');
?>

Thanks

Edited by Mr. Jay, 14 May 2007 - 02:33 AM.

[rc69]

For that it would probably be best to store the username as a variable, not as an object.

<?php
class test{
	var $username;
	function main($text){
		return $this->username.": ".$text;
	}
}
?>

But yes, you can do what you want, but i've never seen the need for objects within objects, so you'll have to figure out the class set up.

<?php
class test{
	function main($text){
		return $this->name.": ".$text;
	}
}

class temp{
	var $username;
	var $name;
	function temp($name){
		$this->name = $name;
		$username = new test();
	}
}

$temp = new temp('blah');
?>

Again, go to php.net if you have more questions. As it stands, i think you need a better understanding of OOP before messing around with it to much. Not that it is dangerous or anything, it is just that you don't always need to use OOP to accomplish what you want to do.