Reputation: 5
Im gonna edit the question to make it clearer, so you can see what i have got now, and get an easier understanding of the problem.
<?php
$mysqli = new mysqli("localhost", "user", "password", "test");
class building
{
private $mysqli;
public $buildingid;
public $userid;
public $buildinglevel;
public function __construct($buildingid, $userid, \mysqli $mysqli)
{
$this->buildinglevel;
$this->mysqli = $mysqli;
}
public function getLevel()
{
return $this->mysqli->query("SELECT ".$this->buildingid." FROM worlds WHERE city_userid=".$this->userid."");
}
}
}
?>
Then I use this to create and use the function:
$cityHall = new building("cityHall",$user['id'],$mysqli);
echo $cityHall->getLevel();
This turns out blank, and nothing happens.
Upvotes: 0
Views: 1687
Reputation: 191729
Objects are units that encapsulate behavior which is exposed to other objects via methods. A wrapper around public properties retrieved from the DB does not object-oriented programming make. In fact, mysqli
can do this for you via fetch_object
:
$result = $mysqli->query($query);
while ($building = $result->fetch_object()) {
// access properties via $building->buildingid, etc.
}
Unless the building
class actually offers functionality via methods, and implements some abstraction, it's not needed. What you can have instead is a DAO (Data Access Object) that wraps the DB (mysqli
) and the data it retrieves is used by your model.
interface Dao {
public function get($id);
}
class BuildingDao implements Dao {
// DB can be a wrapper for mysqli
// but servers as an interface so it
// can be replaced easily
public function __construct(DB $db) {
$this->db = $db;
}
public function get($id) {
return $this->db->prepare(
"SELECT buildinglevel FROM building WHERE buildingid = ?"
)->execute($id)->fetch();
}
}
Upvotes: 1
Reputation: 1271
You should inject instance of mysqli to __construct() of building class:
$mysqli = new mysqli('user', 'password', 'localhost', 'test');
if ($mysqli->connect_errno) { printf("Connect failed: %s\n", $mysqli->connect_error); }
class building
{
private $mysql;
private $buildingid;
private $userid;
// I need to have a mysqli_query here to get the info for the correct building,
//to be able to set the "buildinglevel" for each object from the MYSQL DB, seems basic
//but none of the things ive tried has worked.
public function __construct($buildingid, $userid, $mysqli)
{
$this->buildinglevel;
$this->mysqli = $mysqli;
$this->userid = (int)$userid;
$this->buildingid= (int)$buildingid;
}
public function getLevel()
{
$query = $this->mysqli->query("SELECT ".$this->buildingid." FROM worlds WHERE city_userid=".$this->userid);
$row = $query->fetch_assoc();
if (!$query) {
return $this->mysqli->error;
}
if ($query->num_rows == 0) {
return 'no database records found';
}
return $row;
}
}
$Bulding = new building("cityHall", $user['id'], $mysqli);
$level = $Bulding->getLevel();
var_dump($level);
Upvotes: 3
Reputation: 39389
What your class there seems to be, is what’s known as a model: it represents some form of data, in your case a particular building.
One approach is to pass the MySQLi object as a constructor object, as well as the ID of the building you’re wanting to query for, and assign the result to class properties. This would look as follows:
<?php
class Building
{
private $db;
protected $id;
protected $level;
public function __construct(mysqli $db, $id = null)
{
$this->db = $db;
if (!is_null($id) && intval($id) > 0) {
$stmt = $this->db->prepare("SELECT buildingid, buildinglevel FROM building WHERE `id` = ?");
$stmt->bind_param('i', $id);
$stmt->execute();
$stmt->bind_result($this->id, $this->level);
$stmt->fetch();
}
}
public function getId()
{
return (int)$this->id;
}
public function getLevel()
{
return (int)$this->level;
}
}
You can then fetch your building’s properties like this:
$building = new Building($mysqli, 1);
printf('Building ID is %d and building level is %d', $building->getId(), $building->getLevel());
Upvotes: 0