开发者

PHP Problem accessing the global variable

开发者 https://www.devze.com 2023-02-15 19:17 出处:网络
I\'m having problem with global variable in PHP. I have mysqli config file which contains only following data:

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 : 开发者_开发技巧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.


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.


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();


Your constructor is not getting called because it is not the constructor at all

__cuntruct

should be

__construct


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.


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().


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.


change __cuntruct() to __construct()

0

精彩评论

暂无评论...
验证码 换一张
取 消