PHP Problem accessing the global variable - php

Question

I'm having problem with global variable in PHP. I have mysqli config file which contains only following data:
$mysqli = new mysqli("localhost", "user", "pass", "db");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
I have the following class on another file:
class user{
function username_exists($username){
global $mysqli;
if ($stmt = $mysqli->prepare("SELECT username FROM users WHERE username=?")) {
$stmt->bind_param("s", $username);
$stmt->execute();
$stmt->store_result();
$count=$stmt->num_rows;
$stmt->close();
}
return ($count > 0 ? true : false);
}
...
some more functions
...
}
Now this works fine, but in my previous question on SO, i was told that it is a bad practice to access global variable like I'm doing in above class. So, I'm trying to pass the global variable in the constructor, in following way:
private $mysqli;
function __construct()
{
global $mysqli;
$this->mysqli = $mysqli;
}
function username_exists($username){
//global $mysqli;
if ($stmt = $this->mysqli->prepare("SELECT username FROM users WHERE username=?")) {
And I get the following error:
Fatal error: Call to a member function prepare() on a non-object in...(line number)
Can you please tell me whats problem with it and how this can be fixed? Thanks.
Edit: Sorry for the spelling mistake of __construct. It was only mistake typing here, and the error isnt because of that.

Answer 1

Well... having global in your constructor kindof beats the point. Consider passing it in as a parameter __construct($mysqli).
public function __construct($mysqli)
{
$this->mysqli = $mysqli;
}
What you're trying to do here is called dependency injection.

Answer 2

I think the problem is you misstyped __construct try changing your __cuntruct to the right name for the constructor.
The global in username_exists is also useless.
You should also write a constructor which takes the variable as argument and avoid using global completly :
class User {
var $mysqli;
function __construct($mysqli) {
$this->mysqli = $mysqli;
}
[ ... some functions ... ]
}
You must create your object like this :
$myuser = new User($mysqli);
$myUser->prepare();

Answer 3

Your constructor is not getting called because it is not the constructor at all
__cuntruct
should be
__construct

Answer 4

Couple of things. I think it would work OK if you changed __cuntruct to __construct.
You're still using the global declaration inside the username_exists function. Why not just pass the $mysqli variable in the constructor?
function _construct($mysqli) {
$this->mysqli = $mysqli;
}
then you have no globals in the class.

Answer 5

The code as written was not really what the others on SO were attempting to encourage you to do.
function __construct($mysql_handler){
$this->mysql = $mysql_handler;
}
This is passing in the parameter into the object scope at construction. When you create an instance of your object, you would pass in the MySQL handle.
$mysqli = new mysqli("localhost", "user", "pass", "db");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$u = new User($mysqlli);
Then you should be able to call mysqli member functions on the property itself.
Your constructor is also misspelled. It will only work properly with the magic method name __construct().

Answer 6

Remove global $mysqli; in the function username_exists(), it does not makes sense.
The global $mysqli; is not required/does not makes sense since you want this variable to store the reference the connection IN THE CONTEXT of your object.

Answer 7

change __cuntruct() to __construct()