HTML form submit using dropdown list

Question

I am a bit new to PHP development. I have a form that I need to submit and insert values into a specific table.

For example:

I have H&M value in a drop down list, and when I choose this option in a drop down list all the values are inserted in all other tables and in a table that I have chosen.

I have this part first:

$projectIndex = 0;
switch ('ddlProjekat') 
{
    case "0":
        $projectIndex = 0;
        break;
    case "H&M":
        $projectIndex = 1;
        break;
    case "Takko Fashion":
        $projectIndex = 2;
        break;
    case "Deichmann":
        $projectIndex = 3;
        break;
    case "CortefielSRB":
        $projectIndex = 4;
        break;
    case "CortefielBiH":
        $projectIndex = 5;
        break;
    case "CortefielMNE":
        $projectIndex = 6;
        break;
    case "CortefielCRO":
        $projectIndex = 7;
        break;          
    case "C&A Fashion":
        $projectIndex = 8;
        break;
    case "Sephora":
        $projectIndex = 9;
        break;
    case "Lurdy":
        $projectIndex = 10;
        break;
}

Then goes the HTML form with all the fields, and then this part:

if(isset($_POST['btnSubmit']))
{
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
  die('Could not connect: ' . mysql_error());
}

if(! get_magic_quotes_gpc() )
{
   $prodavnica = addslashes ($_POST['tbBrProd']);
   $grad = addslashes($_POST['tbGrad']);
   $adresa = addslashes($_POST['tbAddress']);
   $telefon = addslashes($_POST['taTelefon']);
   $rvreme = addslashes($_POST['taRVreme']);
   $aglk = addslashes($_POST['taAGLK']);
   $datumtp = addslashes($_POST['tbDatum']);
   $usb = addslashes($_POST['tbUSB']);
   $iprange = addslashes($_POST['tbIPRange']);
   $projekat = addslashes($_POST['ddlProjekat']);
   }
else
{
   $prodavnica = addslashes ($_POST['tbBrProd']);
   $grad = addslashes($_POST['tbGrad']);
   $adresa = addslashes($_POST['tbAddress']);
   $telefon = addslashes($_POST['taTelefon']);
   $rvreme = addslashes($_POST['taRVreme']);
   $aglk = addslashes($_POST['taAGLK']);
   $datumtp = addslashes($_POST['tbDatum']);
   $usb = addslashes($_POST['tbUSB']);
   $iprange = addslashes($_POST['tbIPRange']);
   $projekat = addslashes($_POST['ddlProjekat']);
   }
if($projectIndex = 1){


$sql = "INSERT INTO hm ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange)".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 2){


$sql = "INSERT INTO takko ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 3){


$sql = "INSERT INTO deichmann ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, MS4Sync) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 4){


$sql = "INSERT INTO cortefiel ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 5){


$sql = "INSERT INTO cortefiel_bih ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 6){


$sql = "INSERT INTO cortefiel_mne ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 7){


$sql = "INSERT INTO cortefiel_cro ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, BeetleModel, IPRange) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>
<?php
if($projectIndex = 8){


$sql = "INSERT INTO ca ".
       "(StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange) ".
       "VALUES('$prodavnica','$grad','$adresa', '$telefon', '$rvreme', '$aglk', '$datumtp', '$usb', '$iprange')";


mysql_select_db('fiskali');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
  die('Could not enter data: ' . mysql_error());
}
}
?>

I hope this was understandable.

I am open to all kind of questions and also critics.

Thanks in advance

What's your specific problem? Is it that your code isn't working? Please be more specific. — David Brossard
– David Brossard, Commented May 27, 2014 at 15:03
My code is working. Form is submitted, values are inserted. But drop down list choice is not working. When I choose an option, it is inserted in the right table, but also in all other tables. — user3680236
– user3680236, Commented May 27, 2014 at 15:20
Your if statements are assigning, rather than equality checking: Change if($projectIndex = 1){ to if($projectIndex == 1){. (All 8.) These assignments return true therefore each condition is executed. — bloodyKnuckles
– bloodyKnuckles, Commented May 27, 2014 at 15:52
The switch ('ddlProjekat') condition is looking either for a boolean true or string 'ddlProjekat', so $projectIndex remains 0. — bloodyKnuckles
– bloodyKnuckles, Commented May 27, 2014 at 16:03

bloodyKnuckles · Accepted Answer · 2014-05-27 16:41:54Z

2

Two problems that I can see, 1) the switch statement value was incorrectly formatted (string given, ought to be a POST variable), and 2) the conditions contained assignments, not a equality checks. Those and there is excessive redundancy. Consider this:

<?php

if(isset($_POST['btnSubmit']))
{

    $conn = new mysqli('localhost ', 'root', '', 'fiskali');
    if ($conn->connect_error)
    {
      trigger_error('Database connection failed: '  . $conn->connect_error, E_USER_ERROR);
    }

    switch ( $_POST['ddlProjekat'] )
    {
        case 'H&M'          : $tablename = 'hm';    break;
        case 'Takko Fashion': $tablename = 'takko'; break;
        case 'Deichmann'    : $tablename = 'takko'; break;
        // remaining case statements here
    }

    $sql = "INSERT INTO $tablename (StoreID, Grad, Adresa, Telefon, RadnoVreme, AGLK, DatumTP, BeetleModel, IPRange) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?)";

    $stmt = $conn->prepare($sql);
    $stmt->bind_param('issssssss', $_POST['tbBrProd'], $_POST['tbGrad'], $_POST['tbAddress'], $_POST['taTelefon'], $_POST['taRVreme'], $_POST['taAGLK'], $_POST['tbDatum'], $_POST['tbUSB'], $_POST['tbIPRange']);
    $stmt->execute();

    if($stmt === false)
    {
        trigger_error('Error: ' . $sql . $conn->error, E_USER_ERROR);
    }
    $stmt->close();

}

?>

answered May 27, 2014 at 16:41

bloodyKnuckles

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1 Comment

bloodyKnuckles Over a year ago

You ought to determine the types of the values inserted into the database and enter the correct indicator in the bind_param. php.net/manual/en/mysqli-stmt.bind-param.php

user3663181 · Accepted Answer · 2014-05-27 14:52:48Z

0

I think your switch case not working because in switch 'ddlProjekat' is static. No case follow this, Try to make a variable it. eg, $ddlProjekat. The value of $projectIndex is always 0.

answered May 27, 2014 at 14:52

user3663181

16 bronze badges

2 Comments

user3680236 Over a year ago

'ddlProjekat' is the name of HTML dropdown list. I think it should stay static. Maybe I am wrong

bloodyKnuckles Over a year ago

My first guess is what you're checking is $_POST['ddlProjekat'] .

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