database connection into class

Question

Using this:

$db = mysqli_connect
(
    $db_host,
    $db_user,
    $db_pwd
) or die ("FATAL ! : The server ".$db_host." is not responding to ".$db_user."!");

how to get the $db value into this class (from include "class.php") - without errors ;-)

class User
{
    public $db;
    public $id;

    public function getUser()
    {
        if($this->id)
        {
            $sql = "
            SELECT
                users.*
            FROM
                users
            WHERE
                users.u_id='".$this->id."'
            ";
            $res = mysqli_query($db, $sql) or die(mysqli_error($db));
            $user_row = mysqli_fetch_object($db, $res);
            return $user_row;
        }
        else
        {
            return false;
        }
    }
    } // end class User

All I get is - well, nothing. Exept my main page loads empty... If I stop using the class, all is well (exept no data for the user is shown ;-))

Honestly, just create a database class and extend these models off it. No need for these extra steps. OR if you have multiple databases, setup a singleton or factory pattern and call it statically. — wesside
– wesside, Commented Oct 31, 2012 at 18:15

JvdBerg · Accepted Answer · 2012-10-31 18:19:24Z

1

You must instantiate your class, and you must use the $this operator to access class members. Futhermore, it is better to inject parameters on construction and usage:

class User
{
    private $db;

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

    public function getUser($id)
    {
        if($id)
        {
            $sql = "
            SELECT
                users.*
            FROM
                users
            WHERE
                users.u_id=" . (int)$id ;

            $res = mysqli_query($this->db, $sql) or die(mysqli_error($this->db));
            $user_row = mysqli_fetch_object($this->db, $res);
            return $user_row;
        }
        else
        {
            return false;
        }
    }
 }

Usage:

$user = new User($db);
$userdata = $user->getUser(2);

edited Oct 31, 2012 at 18:19

answered Oct 31, 2012 at 18:05

JvdBerg

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1 Comment

GBD Over a year ago

mysqli_fetch_object($db, $res); this also need $this

NappingRabbit · Accepted Answer · 2012-10-31 18:03:37Z

0

when you instantiate the class (User), if you have already assigned the variable ($db)... you can do

$user = new User;
$user->db = $db;
$user->getUser();

yes AND inside your class when you are referencing properties of the class from its methods, use the $this notation... i.e. change $db to $this->db

that should work. you mentioned something about suppressing errors, that is a different issue. I am unsure what your goal is, but normal exception handling prior to the class's instantiation would be adequate.

answered Oct 31, 2012 at 18:03

NappingRabbit

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2 Comments

osomanden Over a year ago

Worked like a charm. I am of course aware of a need for a more Classy (get it ;-)) correct method. Will be in a next release ..

NappingRabbit Over a year ago

you should note some of the classes written in other answers, I stuck only to what I thought the question to be. they are right though in trying to move the connect to inside the class. It is a more complete solution. glad you got what you were looking for tho.

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