I'm trying to create a class which can be used for connecting to MySQL database. This is my code:
The class:
<?php
class createCon  {
    var $host = 'localhost';
    var $user = 'root';
    var $pass = '';
    var $db = 'example';
    var $myconn;
    function connect() {
        $con = mysqli_connect($this->host, $this->user, $this->pass, $this->db);
        if (!$con) {
            die('Could not connect to database!');
        } else {
            $this->myconn = $con;
            echo 'Connection established!';}
        return $this->myconn;
    }
    function close() {
        mysqli_close($myconn);
        echo 'Connection closed!';
    }
}
And this is where I try to query the database:
<?php
include 'connect.php';
$connection = new createCon();
$connection->connect();
$query = 'SELECT * FROM  `nickname`';
$result = mysqli_query($connection, $query);
if($numrows = mysqli_num_rows($result)) {
    echo $numrows;
    while ($row = mysqli_fetch_assoc($result)) {
        $dbusername = $row['nick'];
        $dbpassword = $row['pass'];
        echo $dbusername;
        echo $dbpassword;
    }
}
I get the following error when I try to make a query:
Warning: mysqli_query() expects parameter 1 to be mysqli, object given in C:\wamp\www\uppgift 1 kompletering\test.php on line 13
PHP uses mysqli query() or mysql_query() function to create or delete a MySQL database. This function takes two parameters and returns TRUE on success or FALSE on failure.
Error: mysqli_query() expects at least 2 parameters, 1 given.
For successful queries which produce a result set, such as SELECT, SHOW, DESCRIBE or EXPLAIN , mysqli_query() will return a mysqli_result object. For other successful queries, mysqli_query() will return true .
You want to pass in $connection->myconn instead of $connection. As in:
$result = mysqli_query($connection->myconn, $query);
As it stands, you're passing in an instance of your class, rather than a mysqli, which is what the error messages are complaining about.
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