Question

I am new to using PHP in an OOP way but have found an issue with my database connection class.

I have a file with this mysqli connection class here

$db_name = 'dbname';
$db_user = 'dbuser';        
$db_password = 'dbpassword';
$db_host = 'localhost';

class database {

    public $mysqli;

    public function connect($db_host, $db_user, $db_password, $db_name){

        $this->mysqli = new mysqli($db_host, $db_user, $db_password, $db_name);

        if ($mysqli->connect_errno) {
            return "Sorry Andre, but you seem to have messed up the DB connection :(";
        }
    }
}

$newConnection = new database;
$newConnection->connect($db_host, $db_user, $db_password, $db_name);

I then want to use the variable $mysqli in a db connection in another file – this is a simple insert to a database using $mysqli variable to connect. I included the above in the connection file but it seems the $mysqli variable is not being returned when I call the method within the database class. I get the PHP error saying…

Fatal error: Call to a member function prepare() on a non-object in...

I have seen that using

global $mysqli; 

works however I want to do it the proper way as I have heard that is not good practice.

I understand that I am potentially doing things a bit wrong here as I’m new to using OOP but I assumed that by returning that variable in the connect function I can then access it from creating the class outside.

Help is appreciated,
thanks.

Answer 1

When using it outside , you access a class variable via the instance…

$newConnection = new database;
$newConnection->connect($db_host, $db_user, $db_password, $db_name);
$newConnection->mysqli /* here you have access from outside */

from inside you use the keyword $this…

// like this from inside
if ($this->mysqli->connect_errno) {
    return "Sorry Andre, but you seem to have messed up the DB connection :(";
}

In case you want to protect your variable from outside access use:

  private $mysqli; 
  // instead of 
  public $mysqli;