Extend prepare() to Database class - php

Question

I'm using:
> $db = new Database(); // my own Database class
> $result = $db -> query(sql);
> $row = $result -> fetch_array(MYSQLI_BOTH);
... and it does so fine and without any issues. But often times the sql is changed a bit based on user input (or can be altered by user input), which I've learned is extremely susceptible to sql injection. Yikes.
So I've been reading up on prepared statements, and I'm ready to switch to something more like this:
> $db = new Database();
> $stmt = $db -> prepare(sql);
> if ($stmt->execute(array($_GET['name'])) {
> > while ($row = $stmt->fetch()) {
> > > print_r($row);
> > }
> }
... which I've grabbed from example #3. I've altered my code to fit what I want but I'm getting Call to undefined method Database::prepare(), which I realize means I don't have the prepare() method available in the class. How can I extend this functionality? And if you don't mind, a little explanation might be helpful. :)
Edit: Here are the contents of my current Database class.
class Database {
private $link;
private $host = "#####";
private $username = "#####";
private $password = "####";
private $db = "####";
public function __construct(){
$this->link = new mysqli($this->host, $this->username, $this->password, $this->db)
OR die("There was a problem connecting to the database.");
return true;
}
public function query($query) {
$result = mysqli_query($this->link, $query);
if (!$result) die('Invalid query: ' . mysql_error());
return $result;
}
public function __destruct() {
mysqli_close($this->link)
OR die("There was a problem disconnecting from the database.");
}
}

Answer 1

Make Database extend mysqli
class Database extends mysqli
{
private $link;
private $host = "#####";
private $username = "#####";
private $password = "####";
private $db = "####";
public function __construct()
{
parent::__construct($this->host, $this->username, $this->password, $this->db)
OR die("There was a problem connecting to the database.");
}
public function __destruct()
{
mysqli_close($this->link)
OR die("There was a problem disconnecting from the database.");
}
}
Then you can call $db as if it's a mysqli object
$db = new Database();
$db->query($sql);
BUT you should just use the mysqli object directly if you're not really adding any functionality to the class...

Answer 2

Return the mysqli link from your construct or you could write a get method for returning the link.
public function __construct(){
$this->link = new mysqli($this->host, $this->username, $this->password, $this->db)
OR die("There was a problem connecting to the database.");
return $this->link;
}
And try this:
$db = new Database();
$name = $_GET['name']; // try validating the user input here
if( $stmt = $db->prepare($sql) ){
$stmt->bind_param("s", $name);
$stmt->execute();
while ($row = $stmt->fetch()) {
print_r($row);
}
}