Call to member function query() on Null [duplicate] - php

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Reference - What does this error mean in PHP?
(38 answers)
Closed 7 years ago.
So just trying to make a simple register/login form which i have done before, but not for hell can i figure out why this is happening. my code looks like the below.
Ive commented out the error above the line that it says its occurring.
<?php
class connection{
private $hostname = "localhost";
private $username = "root";
private $password = "";
private $database = "test";
private $conn;
public function __construct(){
$conn = new mysqli($this->hostname, $this->username, $this->password, $this->database)or die("MySQL Connection Error");
}
public function getConn(){
return $this->conn;
}
}
class queries{
private $conn;
public function __construct($conn){
$this->conn = $conn;
}
public function checkUser($username, $email){
$query = "SELECT * FROM users WHERE username = '$username' OR email = '$email'";
//Call to a member function query() on null in C:\xampp\htdocs\i\functions.php on line 28
$result = $this->conn->query("$query");
return $result;
}
public function insertUser($activated, $activation_code, $firstname, $lastname, $username, $password, $email){
$query = "INSERT INTO users (activated, activation_code, firstname, lastname, username, password, email)
VALUES ('$activated', '$activation_code', '$firstname', '$lastname', '$username', '$password', '$email')";
$result = $this->conn->query("$query");
}
}
Seems Simple enough right.. now heres my php code on the page thats being used (register.php).
if (isset($_POST['register'])) {
include 'functions.php';
$connection = new connection();
$query = new queries($connection->getConn());
$firstname = mysql_real_escape_string($_POST['firstname']);
$lastname = mysql_real_escape_string($_POST['lastname']);
$username = mysql_real_escape_string($_POST['username']);
$email = mysql_real_escape_string($_POST['email']);
$password = sha1($_POST['password']);
$user = $query->checkUser($username, $email);
if ($user->num_rows > 0) {
echo "User Exists";
}else{
echo "User Does not Exist";
}
}

Answer 1

Your code seems correct, except in the constructor of your connection class, you've forgot to use the $this, change it to something like the following:
public function __construct() {
$this->conn = new mysqli($this->hostname, $this->username, $this->password, $this->database)or die("MySQL Connection Error");
}