php multiple form handling on one page

Question

I've gone through the similar questions and I think I've implemented all the ideas. I have two forms on the login page, one for admins login and one for employees login

The first form has a button name="login"

the second form has a button name "login2"

The login for admins is working perfectly. this is the code for it

 if (isset($_POST['login']))

    { $username = mysqli_real_escape_string($con,$_POST['username']);
      $pwd      = md5($_POST['password']);

      if (isset($username) == true && isset($pwd) == true){

         $login = login($username, $pwd);

        var_dump($login);
        if($login != null) {

            $_SESSION ['user_id']  = $login ['buss_id'];
            $_SESSION ['usernameholder'] = $login ['username'];
            ob_start();

            if ($login['level'] =='1')
                {?> 
                <script>
                        window.location.href = "sadmin/index.php";
                    </script> <?php   } 
                    else if($login['level'] =='2') { ?>
                    <script>
                        window.location.href = "user_admin/index.php";
                    </script>

                    <?php } else if($login['level'] =='3') { ?>
                    <script>
                        window.location.href = "reps/index.php";
                    </script>
                    <?php
                    } /* end of login levels. */


        } /* /if $login !=null */
      } /* /if $login = login */
      } /* / if isset el $_POST */

and the function login is this

    function login($username, $password)
{   $db_host="localhost";
    $db_username="root";
    $db_password="";
    $db_name="dbname";
    $con=mysqli_connect($db_host, $db_username,$db_password, $db_name);

    $qry = "SELECT * FROM `businesses` WHERE `username` = '$username' AND `password` = '$password' AND `active` = 1 LIMIT 1";
    $sql = mysqli_query($con,$qry);
    while($row = mysqli_fetch_array($sql))
    {
        return $row;
    } }

What I did was to simply copy paste the login code and make the names like this $username2

$pwd2 etc...

and changed the query in the function to this

$qry2 = "SELECT * FROM `employees` WHERE `username` = '$username2' AND `password` = '$password2' AND `active` = 1 LIMIT 1";

As you might have noticed I did var_dump($login) and so I did var_dump($login2) and that keeps returning a NULL value. What's wrong with my code please?!!!

I'm gonna add the login2 code for whoever wanna have a look at it

if (isset($_POST['login2']))

    { $username2 = mysqli_real_escape_string($con,$_POST['username2']);
      $pwd2      = md5($_POST['password2']);

      if (isset($username2) == true && isset($pwd2) == true){

         $login2 = login_employee($username2, $pwd2);

        var_dump($login2);
        if($login2 != null) {

            $_SESSION ['works_for']  = $login2 ['buss_id_fk'];
            $_SESSION ['emp_id'] = $login2 ['emp_id'];
            $_SESSION ['user_name'] = $login2 ['username'];
            ob_start();

            if ($login_employee['level'] =='1')
                {?> 
                <script>
                        window.location.href = "sadmin/index.php";
                    </script> <?php   } 
                    else if($login_employee['level'] =='2') { ?>
                    <script>
                        window.location.href = "user_admin/index.php";
                    </script>

                    <?php } else if($login_employee['level'] =='3') { ?>
                    <script>
                        window.location.href = "reps/index.php";
                    </script>
                    <?php
                    } /* end of login levels. */


        } /* /if $login !=null */
      } /* /if $login = login */
     } /* / if isset el $_POST */

And this is the function login_employee code

function login_employee($username2, $password2)
{   $db_host="localhost";
    $db_username="root";
    $db_password="";
    $db_name="leadapp";
    $con=mysqli_connect($db_host, $db_username,$db_password, $db_name);

    $qry2 = "SELECT * FROM `employees` WHERE `username` = '$username2' AND `password` = '$password2' AND `active` = 1 LIMIT 1";
    $sql2 = mysqli_query($con,$qry2);
    while($row2 = mysqli_fetch_array($sql2))
    {
        return $row2;
    } }

    ?>

And just for the sake of clearance i'm adding a screenshot not the code, of my forms.

do you rename function login($username, $password)... ?

bxN5
– bxN5

2017-03-02 21:39:24 +00:00
Commented Mar 2, 2017 at 21:39 — bxN5
– bxN5, Commented Mar 2, 2017 at 21:39
Yes I made it login_employee($username2, $password2)

Mar1ak
– Mar1ak

2017-03-02 21:40:34 +00:00
Commented Mar 2, 2017 at 21:40 — Mar1ak
– Mar1ak, Commented Mar 2, 2017 at 21:40
hard to guess, better show non working code

bxN5
– bxN5

2017-03-02 21:42:36 +00:00
Commented Mar 2, 2017 at 21:42 — bxN5
– bxN5, Commented Mar 2, 2017 at 21:42
Done I just added the non working function's code.

Mar1ak
– Mar1ak

2017-03-02 21:45:28 +00:00
Commented Mar 2, 2017 at 21:45 — Mar1ak
– Mar1ak, Commented Mar 2, 2017 at 21:45

bxN5 · Accepted Answer · 2017-03-02 22:07:09Z

1

let's get some debag of login_employee function

 function login_employee($username2, $password2)
{   $db_host="localhost";
    $db_username="root";
    $db_password="";
    $db_name="leadapp";
    $con=mysqli_connect($db_host, $db_username,$db_password, $db_name);
    var_dump($username2);  var_dump($password2); 
    $qry2 = "SELECT * FROM `employees` WHERE `username` = '$username2' AND `password` = '$password2' AND `active` = 1 LIMIT 1";
    $sql2 = mysqli_query($con,$qry2);
    var_dump(mysqli_fetch_array($sql2));
    while($row2 = mysqli_fetch_array($sql2))
    {
        return $row2;
    } }

    ?>

so first two var dumps you get name and password, after that check if this value really in db employees

if var_dump(mysqli_fetch_array($sql2)); do not return your db row - it's mean you have null because there is no match row in database

answered Mar 2, 2017 at 22:07

bxN5

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4 Comments

Mar1ak Over a year ago

that's what's driving me crazy, it returns NULL. The table does exist, just like the table that holds the admins. It's the same code, same database, only different tables, and the two tables do exist. the admins are in table businesses. gyazo.com/ebe0c65d4b8b77304d39d61b6460e456

bxN5 Over a year ago

do var_dump($qry2) paste here pls and result copy in query in phpmyadmin

Mar1ak Over a year ago

Ok now we're getting somewhere, this is what I get. string(126) "SELECT * FROM employees WHERE username = 'tr' AND password = 'c4ca4238a0b923820dcc509a6f75849b' AND active = 1 LIMIT 1" It's selecting the correct hashed password but adding 9b at the end, any idea why?

Mar1ak Over a year ago

Ok apparently md5 uses 32 character input and the table's password field was set to 30 characters. So it was causing problems. I set it to 50 characters and it's fixed. Thanks a lot man, your suggestion opened my eyes!

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