Mysql statement inside a function doesn't work

Question

I have a MySQL statement in a function but somehow it doesn't work. I get the following error:

Fatal error: Call to a member function prepare() on a non-object in D:\xampp\test.php on line 7

function isloggedin($getcookiedata) {
    $data = explode("|", $getcookiedata);
    $userid = $data[0];
    $md5password = $data[1];

    $sql = "SELECT user_id, username, email, newsletter, user_group FROM users WHERE user_id = ? AND md5password = ?";
    $stmt = $mysqli->prepare($sql);
    $stmt->bind_param('ss', $userid, $md5password);
    $stmt->execute();
    $stmt->bind_result($r_userid, $r_username, $r_email, $r_newsletter, $r_user_group);
    $stmt->store_result();
    $checker = $stmt->num_rows;
    $stmt->fetch();
    $stmt->close(); 
}

When I just do the SQL statement outside a function, it works. But not inside the function.

Please add your solution as an answer instead (yes you can answer your own questions). In about two days you can accept it then as "the answer". — hakre
– hakre, Commented Jul 9, 2011 at 10:27

Mat · Accepted Answer · 2011-07-09 10:21:44Z

2

If $mysqli is a global you're defining somewhere, you need to pull it into your function's scope with:

function isloggedin($getcookiedata) {
    global $mysqli;
    ...
}

See PHP - Variable scope

answered Jul 9, 2011 at 10:21

Mat

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