Question

//Successful connection to $db

function insert_users($db, $username, $password, $email)
{
    echo "FUNCTION CALLED"; //This is outputted successfully

    $query = "INSERT INTO `users` (`id`, `username`, `password`, `email`) VALUES ('', ?, ?, ?)";
    $stmt = mysqli_stmt_init($db);

    if(mysqli_stmt_prepare($stmt, $query))
    {
        echo "QUERY PREPARED"; // rest of code was snipped (will put up upon request)
    } else {
        echo "QUERY DENIED"; //This is outputted successfully
    }
}


//The $user $pass and $mail are defined and then the function is called

insert_users($db, $user, $pass, $mail);

Database structure:

testdb (database)
    -> users (table)
        -> id            //Primary key, unique key
        -> username      //unique key
        -> password
        -> email

Why doens’t the query go through with it’s operation?
It echoes “Query denied” instead of “Query prepared”. The query is invalid I believe. If more information is required, ill be happy to edit this question

EDIT

I added

mysqli_stmt_error($stmt);after the echo "QUERY DENIED";
nothing happens…

EDIT 2

$server = 'localhost';
$user = 'root';
$password = '';
$databse = 'testdb';

$db = @mysqli_connect($server, $user, $password, $database) or die("Could not connect to Database server. Please inform an administrator");

That’s my database setup. Do i put the variables in quotes?

Answer 1

If id is an auto_increment column, you can just drop it from your insert

INSERT INTO `users` (`username`, `password`, `email`) VALUES (?, ?, ?)

You have a typo in $databse = 'testdb';. Rename it to $database.