PHP MySQL create condition for query

Question

I have code like this at my PHP code:

    <?php
    require('../../server.php');
    $role = strtoupper($_POST['role']);
    $pool = strtoupper($_POST['pool']);
    $psh = strtoupper($_POST['comp']);

    if($role = "POOL")
    {
        $query2 = "INSERT INTO m_login (email, password, role, company_id)
                            VALUES ('$email', '$pass', '$role', '$pool')";
    }
    else
    {
        $query2 = "INSERT INTO m_login (email, password, role, company_id)
                            VALUES ('$email', '$pass', '$role', '$psh')";
    }

    if (mysql_query($query2))
    {
        $whatdo = strtoupper("add user ").$id;
        include_once('../../serverlog.php');
        $querys = "INSERT INTO m_log (user_id, description, waktu) VALUES ('$user', '$whatdo', '$input')";
        if(mysql_query($querys))
        {
          echo'<script>alert("Penambahan data berhasil!");</script>
          <meta http-equiv="refresh" content="0; url=index.php" />';
        }
        else
        {
          echo mysql_error();
        }
    }
    else
    {
        echo'<script>alert("Failed!");</script> <br/>'.mysql_error().'<meta http-equiv="refresh" content="10; url=index.php" />';
    }
?>

my question is, am I wrong to create condition for query2? because when I ran the program, my data always get POOL result for the role, although I have select Admin or Supervisor, it always return POOL

I'm using for choosing the role at registration form. So when I choose option admin, it return pool, when I choose spv, it return pool.

Anyone can give me solution?

Sorry for my bad English

Well, I will try to fix that.. I will learn it before (because I don't understand about that) Thanks for advice.. — Cross Vander
– Cross Vander, Commented Nov 12, 2012 at 7:45

Mr. Alien · Accepted Answer · 2012-11-09 03:58:40Z

4

This is wrong, it should be

if($role == "POOL") {
  /*Code goes here*/
}

because = will assign value POOL to your variable $role, so use == to compare, or === to compare similar data types

answered Nov 9, 2012 at 3:58

Mr. Alien

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Comments

Niet the Dark Absol · Accepted Answer · 2012-11-09 03:58:46Z

3

= is assignment. == is comparison. Put if($role == "POOL") and it should work fine.

answered Nov 9, 2012 at 3:58

Niet the Dark Absol

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Comments

Mayank K. Swami · Accepted Answer · 2012-11-09 04:01:47Z

2

you care assigning the value in if($role = "POOL") use instead == should be if($role == "POOL")

= is the assignment operator, == equality operator.

= is the assignment operator. b = 1 will set the variable b equal to the value 1.
== is the equality operator. it returns true if the left side is equal to the right side, and returns false if they are not equal

answered Nov 9, 2012 at 4:01

Mayank K. Swami

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