Another Class problem [SOLVED]

PHP programming forum. Ask questions or help people concerning PHP code. Don't understand a function? Need help implementing a class? Don't understand a class? Here is where to ask. Remember to do your homework!

Moderator: General Moderators

Post Reply
anthony88guy
Forum Contributor
Posts: 246
Joined: Thu Jan 20, 2005 8:22 pm

Another Class problem [SOLVED]

Post by anthony88guy »

I'm really glad I finally tried classes. But I'm a newbie and here is my problem:

Code: Select all

class Main
{
	function config()
	{
		global $tb;
		
		$config = array();
		$query['config'] = mysql_query("SELECT * FROM `" . $tb['config'] . "") or die(mysql_error());
			
		while($row = mysql_fetch_array($query['config']))
		{
			$config[$row['config_name']] = $row['config_value'];
		}
		
		return $config;
	}
}

Code: Select all

$main = new Main;
$main->config();
print_r($config);
Notice: Undefined variable: config in /home/*/public_html/includes/header.php on line 28
If I do

Code: Select all

print_r($config);

inside the function config() it works.

What I want to do is when I call $main->config(); I have an array with my config settings.
Last edited by anthony88guy on Sat Jul 22, 2006 3:47 pm, edited 1 time in total.
User avatar
John Cartwright
Site Admin
Posts: 11470
Joined: Tue Dec 23, 2003 2:10 am
Location: Toronto
Contact:

Post by John Cartwright »

The actual variable $config will only exist within the scope of the function. What your wanting to do is

Code: Select all

$main = new Main;
print_r($main->config());
anthony88guy
Forum Contributor
Posts: 246
Joined: Thu Jan 20, 2005 8:22 pm

Post by anthony88guy »

Jcart wrote:The actual variable $config will only exist within the scope of the function. What your wanting to do is

Code: Select all

$main = new Main;
print_r($main->config());
Output:
Array ( [test] => 500 )
Great it works,

Now I need to print out the value of 'test' in the array? How would I go about doing that?

Thanks...
User avatar
John Cartwright
Site Admin
Posts: 11470
Joined: Tue Dec 23, 2003 2:10 am
Location: Toronto
Contact:

Post by John Cartwright »

Like you would any other array.

Code: Select all

$main = new Main;
$config = $main->config();

echo $config['test'];
anthony88guy
Forum Contributor
Posts: 246
Joined: Thu Jan 20, 2005 8:22 pm

Post by anthony88guy »

wow... thank you very much.
Post Reply