Can't retrieve DB info - php

Question

I need help figuring out why the following DB Query is not working. I know the DB connection is good. I also know the $referralname = $_SESSION['user_name']; is correctly rendering. It has to be something with my code.
I am getting the following errors. Maybe this will help to figure this out.
[12-Jun-2013 21:13:54 America/New_York] PHP Warning: mysql_query() expects parameter 1 to be string, object given in /x/x/public_html/americansolar/partner/classes/Referral.php on line 89
[12-Jun-2013 21:13:54 America/New_York] PHP Warning: mysql_num_rows() expects parameter 1 to be resource, null given in /x/x/public_html/americansolar/partner/classes/Referral.php on line 90
P.S. I am not sure if the while statement is necessary or not since it will always only return one result???
My Code:
// creating a database connection
$this->db_connection = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
// if no connection errors (= working database connection)
if (!$this->db_connection->connect_errno) {
$referralname = $_SESSION['user_name'];
// get the referrer's id
$query_get_referral_id = $this->db_connection->query("SELECT * From users WHERE user_name = '".$referralname."';");
$result = MYSQL_QUERY($query_get_referral_id);
$numberOfRows = MYSQL_NUM_ROWS($result);
$i = 0;
while ($i<$numberOfRows)
{
$thisId = MYSQL_RESULT($result,$i,"user_id");
$i++;
}
}
My Solution:
$query_get_referral_id = $this->db_connection->query("SELECT * From users WHERE user_name = '".$referralname."';");
while($row = mysqli_fetch_array($query_get_referral_id))
{
$thisId = $row['user_id'];
}

Answer 1

Youre mixing mysqli and mysql... they are two completely different and incompatible interfaces. Secondly, your $query_get_referral_id is not an id value... it is a mysqli_result object. You need to then extract the value from that object.
And lastly... DONT use mysql... stick with mysqli, or use PDO
Also you should use a prepared statement for this:
$stmt = $this->db_connection->query("SELECT user_id From users WHERE user_name = ?");
$stmt->bind_param('s', $referralname);
$stmt->execute();
if($stmt->num_rows) {
$stmt->bind_result($userId);
while($stmt->fetch()) {
// do something with $userId...
// each iteration of this loop is a
// row of the result set, it will automatically
// load the value of the user_id into $userId
}
}

Answer 2

I donot think you should query like
$query_get_referral_id = $this->db_connection->query("SELECT * From users WHERE user_name = '".$referralname."';");
$result = MYSQL_QUERY($query_get_referral_id);
Well, you should go
$result = $this->db_connection->query("SELECT * From users WHERE user_name = '".$referralname."';");