mysqli prepare statement error "MySQL server has gone away" - php

Question

I'm struggling to make the jump form Procedural to Object Orientated style so if my code is untidy or flawed please be nice - here I'm passing a couple of posts via jQuery to a class to update a record when the user checks a checkbox:
Here is the database connection
class db {
private $host ;
private $username;
private $password;
private $dbname;
protected function conn()
{
$this->host = "localhost";
$this->username = "root";
$this->password = "";
$this->dbname = "mytest";
$db = new mysqli($this->host, $this->username, $this->password, $this->dbname);
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
return $db;
}
}
Here is the update class
class updOrders extends db {
public $pid;
public $proc;
public function __construct()
{
$this->pid = isset($_POST['pid']) ? $_POST['pid'] : 0;
$this->proc = isset($_POST['proc']) ? $_POST['proc'] : 1;
// $stmt = $this->conn()->query("UPDATE tblorderhdr SET completed = ".$this->proc." WHERE orderid = ".$this->pid);
$stmt = $this->conn()->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
$stmt->bind_param('ii', $this->proc, $this->pid);
$stmt->execute();
if($stmt->error)
{
$err = $stmt->error ;
} else {
$err = 'ok';
}
/* close statement */
$stmt->close();
echo json_encode($err);
}
}
$test = new updOrders;
When I comment out the prepare statement and run the query directly (commented out) it updates, when I try and run it as a prepare statement it returns an error "MySQL server has gone away".

Answer 1

I have looked at your code and I found this.
$stmt = $this->conn()->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
I wrote nearly the same. But I saw you separated the connection from the prepare function.
$db = $this->conn();
$stmt = $db->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
I don't know either why, but it works now.

Answer 2

It would appear that the problem lies within the connection to the database. Here is the (relevant bit of the) updated code:
$db = $this->conn();
$stmt = $db->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
$stmt->bind_param('ii', $this->proc, $this->pid);
$stmt->execute();

Answer 3

When you call $this->conn(), you are creating a new connection object of class mysqli. When no more variables point to the object, PHP will trigger its destructor. The destructor for mysqli will close the connection. This means that if you do not save the return value of this method into a variable, there will be no more references pointing to the object and the connection will be closed.
To fix this, simply save the object and reuse it. Don't connect each time.
$conn = $this->conn();
$stmt = $conn->prepare("UPDATE tblorderhdr SET completed = ? WHERE orderid = ?");
On a side note, creating a class like the one you have db is absolutely pointless. This class doesn't do anything useful. You haven't added any extra functionality to mysqli. You never need to save the credentials in properties. The whole class can be replaced with this code:
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli('localhost', 'user', 'password', 'test');
$mysqli->set_charset('utf8mb4'); // always set the charset
Then your classes will expect the mysqli object as a dependency.
class updOrders {
public $pid;
public $proc;
public function __construct(mysqli $conn) {
}