Question

I’m having trouble with mysqli and prepared statements. I’ve just started learning mysqli an hour and am having trouble not understanding why I’m getting these two errors:

Notice: Undefined variable: mysqli in /opt/lampp/htdocs/lr/testingi.php on line 17

Fatal error: Call to a member function prepare() on a non-object in /opt/lampp/htdocs

/lr/testingi.php on line 17

I have a file that contains the database connection. Here it is.

$mysqli = new mysqli("localhost", "user", "password", "db");

/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

Here is the test file that is reproducing the error.

session_start();

require_once 'core/database/connect.php';

function user_id_from_username ($username) {

if ($stmt = $mysqli->prepare("SELECT `user_id` FROM `users` WHERE `username` = ?"))

$stmt->bind_param('s', $username);  
$stmt->execute();
$stmt->bind_result($user_id);
echo $user_id;  
$stmt->close(); 
}

$username = 'Jason';        

user_id_from_username ($username);  

Answer 1

Looks like you’re not passing $mysqli into the user_id_from_username function.

2 quick options:

1. A global

function user_id_from_username ($username) {
global $mysqli;
if ($stmt = $mysqli->prepare("SELECT `user_id` FROM `users` WHERE `username` = ?"))

$stmt->bind_param('s', $username);  
$stmt->execute();
$stmt->bind_result($user_id);
echo $user_id;  
$stmt->close(); 
}

2. a second parameter

function user_id_from_username ($mysqli, $username) {//..}

user_id_from_username($mysqli, $username);