如何将mysqli类扩展到其他类方法中

Question

Hey sorry for the stupid question. I have read lots about oop in php so i decided to try. i have a blog running procedural php and all works fine but i have mysql not mysqli on the code and i decided to upgrade to the new mysqli. the problem here is really query the database to get results. i have been on the problem for two days now. here's a code example of what i want to implement.

class myDB extends mysqli
{
  //perform mysqli connection to host here and return  $connection_handle
}

class siteConfig
{
   private function getConfig($id)
  {
   $res = $connection_handle->query("SELECT * from config where id='1'");
   $row = $res->fetch_assoc();
   return $row['option'];
  }

}

on the main index file i would do this

$c = new siteConfig();
echo $c->getConfig(1);
//and this will return the result of the query.

also please pardon my programming logic and still a very terrible newbie to the oop world

any ideas will be helpful thank you.

PS

this is working example of the code am using

$db_link = mysqli_connect(DB,HOST,DB_USER,DB,PASS,DB_NAME) or die(mysql_error());

class myDB extends mysqli
{
  public function getConfig($id, $db_link) // thanks cbuckley
{
 $db_link = $this->db_link;
     $res = mysqli_query($this->db_link,"SELECT * from config where id='1'");
     $row = mysqli_fetch_array($res);
     return $row['option'];
}    
}

i have defined the constants already and the connection is successful but the select is not working. on the index page this goes [code] include('inc/dbc.php');

 $db = new MyDB();
 echo $db->getConfig(1, $db_link);  

please let me know where am missing it or if possible refactor the code for me where neccessary

Answer 1

Why not just extend the mysqli class and add your functions to myDB?

class myDB extends mysqli
{
   public function getConfig($id) // thanks cbuckley
  {
   $res = $this->query("SELECT * from config where id='1'");
   $row = $res->fetch_assoc();
   return $row['option'];
  }    
}

$db = new myDB;
$db->getConfig($id);

You can call any mysqli function inside myDb by using the $this keyword (which refers to itself and since "it self" extends mysqli, it will have all the functions mysqli has + the ones you add)

Answer 2

It's good to keep the separation between your database connection and any business logic, so your example is good. There are just a couple of extra things that need sorting out, namely how the application logic has access to the database handler. Here's an example:

class MyDB extends mysqli {
    // any custom DB stuff here
}

class SiteConfig {
    protected $db;

    public function __construct(MyDB $db) {
        $this->db = $db;
    }

    public function getConfig($id) {
        $statement = $this->db->prepare('SELECT * FROM config WHERE id = ?');
        $statement->bind_param('i', $id);

        if ($statement->execute()) {
            $row = $statement->get_result()->fetch_assoc();
            return $row['option'];
        }
    }
}

$db = new MyDB('host', 'user', 'password', 'db');
$config = new SiteConfig($db);
var_dump($config->getConfig(1));

A couple of points of interest here:

• Dependency injection : passing $db into your config class
• Separation of concerns : the SiteConfig class is not responsible for setting up the database connection, thus keeping its role clean and well-defined
• Method and property visibility : $config->db is declared protected, so it can't be accessed outside of the scope of the class
• Type hinting : forcing the first parameter of SiteConfig's constructor to be an instance of MyDB
• Class properties : setting the $db instance as a property of the SiteConfig class, so it can be used in getConfig()
• Method and property visibility : $config->db is declared protected, so it can't be accessed outside of the scope of the class
• Prepared statements : more efficient queries that get protection against SQL injection for free