Question

Hi I made a login for a website.

I made a successful one using 1 account table in a database.

Now I made another table for another account. Both tables are different in attributes that is why I separated them.

I am having a hard time log-in the second one in by using this algorithim.

The inputted login and password come from another page using a HTML Form.

<?php
    //Start session
    session_start();

    //Include database connection details
    require_once("config.php");
    require("clean.php");

    //Array to store validation errors
    $errmsg_arr = array();

    //Validation error flag
    $errflag = false;

    //Sanitize the POST values
    $login = clean($_POST['login']);
    $password = clean($_POST['password']);

    //Input Validations
    if($login == '') {
        $errmsg_arr[] = 'Login ID missing';
        $errflag = true;
    }
    if($password == '') {
        $errmsg_arr[] = 'Password missing';
        $errflag = true;
    }

    //If there are input validations, redirect back to the login form
    if($errflag) {
        $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
        session_write_close();
        header("location: login-form.php");
        exit();
    }

    //Create query for patient
    $qry="SELECT * FROM user_info WHERE login='$login' AND password='".md5($password)."'";
    $result=mysql_query($qry);

//Check whether the query was successful or not
    if($result) {
        if(mysql_num_rows($result) == 1) {
            //Login Successful User
            session_regenerate_id();
            $user_info = mysql_fetch_assoc($result);
            $_SESSION['SESS_MEMBER_ID']  = $user_info['ID_NO'];
            $_SESSION['SESS_FIRST_NAME'] = $user_info['FNAME'];
            $_SESSION['SESS_admin']      = $user_info['admin'];
            $_SESSION['SESS_itResult']      = $user_info['itResult'];
            //print "Login a success!";
            session_write_close();
            header("location: member-index.php");
            exit();
        }
        else {
                //Create query for 
                $qry="SELECT * FROM specialist_info WHERE spec_username='$login' AND spec_password='".md5($password)."'";
                $result2=mysql_query($qry);

                //Check whether the query was successful or not
                if($result2) {

                    if(mysql_num_rows($result2) == 1) {
                        //Login Successful User

                        session_regenerate_id();
                        $specialist_info = mysql_fetch_assoc($result);
                        $_SESSION['SESS_MEMBER_ID']  = $specialist_info['spec_id'];
                        $_SESSION['SESS_FIRST_NAME'] = $specialist_info['name'];
                        $_SESSION['SESS_admin'] = 1;

                        //print "Login a success!";
                        session_write_close();
                        header("location: member-index.php");
                        exit();
                        }

                    else {
                        //Login failed
                        header("location: login-failed.php");
                        exit();
                        }
                    }
            }
    }
        else {
        die("Query failed");
    }
}
?>

I was able to log-in the first(meaning there is connection between PHP and MySQL) account however I could not in the second.

From what i follow, I could not get in this one. “if(mysql_num_rows($result2) == 1)”
I know there is data in the second table in the database because i checked it(there is MD5 yes).

Maybe my logic is wrong. Suggestions? 😀 Thanks!

Answer 1

I’m not sure about the code (will review later) but you could try refactoring your tables so you have a separate table that contains user attributes, this allows you to be dynamic with your attributes so you don’t have to create a different table every time you have a different type of user.

EDIT:
Try:
var_dump(mysql_num_rows($result2)) if this shows the return value is false something is wrong with the query, otherwise it doesn’t match any rows.