PHP OOP declaring class into class - php

Question

I have 2 classes and I would like to have 1 more for comfort :)
So I have class for database and system. I would like to have 3rd classs for functions. This class MUST use functions from database since there are SQL Queries. This is What I ended with:
class database {
public function __construct() {
$this->mysqli = new mysqli('localhost', 'root', '', 'roids');
$this->func = new functions();
}
}
class functions {
function __construct(){
$db = new database();
}
public function banned() {
$q = $db->select($this->prefix."banned", "*", "banned_ip", $this->getIP());
if (0 == 0) {
header('Location: banned.php'); // For testing
}
}
}
And I this kind of version ends in cycle. Any solutions ? Thanks a lot

Answer 1

Some of your classes must be a source, so other classes may use instances of it.
Let me show you one example:
class DataBase { //THIS class is a source.
public $mysqli ;
public function __construct(){
$this->mysqli = new mysqli('localhost', 'root', '', 'roids');
$this->mysqli->set_charset("utf-8") ;
}
public function ready(){
return ($this->mysqli instanceof MySQLi && $this->mysqli->ping()) ;
}
}
class User {
protected $db;
public function __construct(DataBase $db){
$this->db = $db ; //Now you can use your source
}
public function isBanned(){
if ($this->db->ready()){
$result = $this->db->mysqli->query("SELECT id FROM banned WHERE name='Hacker' ; ") ;
return ($result && $result->num_rows >= 1) ;
}
return false ;
}
}
Then instantiate the source and give it to instance of User:
$database = new DataBase() ;
$user = new User($database) ;
Now you can use a more complexed function:
if ($user->isBanned()) exit("You are banned") ;
The same way you can make class Functions and use it (other classes may use it) as a source.