Undefined variable: mysqli - PHP OOP - mysqli

Question

How can I implement mysqli in an extended class?
I am uploading an image and storing it in a MySQL database, but I get this error:
Notice: Undefined variable: mysqli in ...ecc/ecc/ on line 33
Fatal error: Call to a member function query() on a non-object in ...ecc/ecc/ on line 33
Here is my test code:
<?php
interface ICheckImage {
public function checkImage();
public function sendImage();
}
abstract class ACheckImage implements ICheckImage {
public $image;
private $mysqli;
public function _construct(){
$this->image = $_POST['image'];
$this->mysqli = new mysqli('localhost','test','test','test');
}
}
class Check extends ACheckImage {
public function checkImage() {
if($this->image > 102400) {
echo "File troppo grande";
}
}
public function sendImage() {
//This is the line 33 give me the error
if ($mysqli->query("INSERT INTO images (image) VALUES ('$this->image')")) {
echo "Upload avvenuto &nbsp";
} else {
echo "Errore &nbsp" . $mysqli->error;
}
}
}
$form = new Check();
$form->checkImage();
$form->sendImage();
?>

Answer 1

There are some errors in your code.
The $mysqli member is private inside the abstract class. It will not be inherited by the Check class, so it does not exist there. Make it protected.
Access to the members of a class always needs $this-> in front, specifically $this->mysqli in this instance.
The constructor function must be named __construct with two underscores in front.
The image check looks wrong. $_POST['image'] does contain something that you expect to store in the database, but you also compare it with an integer value and seem to echo an error message if it is bigger. While the data handling will work, e.g. you can compare a string from POST data with an integer, it looks like you want something else.