Error in my simple login system. Trying to read from the Database

Question

I have this error "Warning: mysql_result(): user_id not found in MySQL result index 13 in C:\xampp\htdocs\core\functions\users.php on line 14"

I think i don't know how to read from my Database, it looks like

Here is my users code

<?php
    function user_exists($username){
            $username = sanitize($username);
            return (mysql_result(mysql_query("SELECT COUNT(`uID`) FROM `user` WHERE `uUserName` = '$username'"), 0) == 1) ? true : false;
    }

    function user_active($username){
            $username = sanitize($username);
            return (mysql_result(mysql_query("SELECT COUNT(`uID`) FROM `user` WHERE `uUserName` = '$username' AND `uActive` = 1"), 0) == 1) ? true : false;
    }

    function user_id_from_username($username){
            $username = sanitize($username);
            return mysql_result(mysql_query("SELECT `uID` FROM `user` WHERE `uUserName` = '$username'"), 0, 'user_id');
    }

    function login($username,$password){
            $user_id = user_id_from_username($username);

            $username = sanitize($username);
            $password = md5($password);

            return (mysql_result(mysql_query("SELECT COUNT(`uID`) FROM `user` WHERE `uUserName` = '$username' AND `uPassword` = '$password'"), 0) == 1) ? $user_id : false;
    }

?>

This is my login system code

if(empty($_POST) === false){
        $username = $_POST['username'];
        $password = $_POST['password'];

        if(empty($username) === true || empty($password) === true){
                $errors[] = 'You need to enter a Username and Password';
        } else if(user_exists($username) == false){
                $errors[] = 'We cannot find that Username, have you registered?';

        } else if(user_active($username) === false){
                $errors[] = 'You have not activated your account.';
        } 
        else {
                $login = login($username,$password);
                if($login === false){
                        $errors[] = 'The username or password is incorrect';
                } else {
                        echo 'ok';
                }
        }

        print_r($errors);

}

Database queries are generally case-insensitive. Your comparison of the password as plain text at the DB level is a large red flag that your app will be very vulnerable to various hacks. -- edit: I see the hash now, sorry, ignore me — Nikki9696, Nov 13 '15 at 19:58
The problem lies on this line: `mysql_result(mysql_query("SELECT `uID` FROM `user` WHERE `uUserName` = '$username'"), 0, 'user_id');` You're using 'user_id' as the offset and not a number or field that mysql recognizes or sees. Try changing it to `uID` instead. — Paul Williams, Nov 13 '15 at 19:58
I know I am trying to easily distinguish between things right now. So I make it obvious. But it still cant find the password, any help with that? — Azmi Elqutob, Nov 13 '15 at 20:07

score 1 · Answer 1 · answered Nov 13 '15 at 20:07

You have an error at line 13 where you specify 'user_id' as an offset.

You can simply get the uID column and store the value if that row exists.

$result = mysql_query("SELECT `uID` FROM `user` WHERE `uUserName` = '$username'");
$user_id = 0;   // default value, meaning user not found

if ($result && mysql_num_rows($result) > 0){
    $row = mysql_fetch_assoc($result);
    $user_id = $row[0];
}

return $user_id;

Then you can modify your login function to check whether user_id > 0 or not. If it's larger than 0, then you got that user's id.

score 0 · Answer 2 · answered Nov 13 '15 at 20:06

Currently the only problem that I see is your user_id_from_username function.

You're trying to set an offset to a field that doesn't exist and mysql doesn't find it. So it's throwing an error:

function user_id_from_username($username){
        $username = sanitize($username);
        return mysql_result(mysql_query("SELECT `uID` FROM `user` WHERE `uUserName` = '$username'"), 0, 'uID');
}

Try the above or leave off the uID since it's not a mandatory but rather an optional parameter.

Insert obligatory, you should be using mysqli instead of mysql at this point if your PHP version supports it.

Error in my simple login system. Trying to read from the Database

2 Answers2