Unable to display contents in Second Drop Down Box


Posted on 16th Feb 2014 07:03 pm by admin

Hi All,

What I am trying to do is 2 dependent drop down boxes and when user selects submit button the values are to be passed to the database to run a insert query.

Right now, I am stuck with the second drop down box because it does not show up the values based on the selection in the first drop down box.

Important point here is " My second drop down box displays columns from a single row(returned from second query)"

Attached is my code please do suggest me the required changes to make and also I pasted the code here,


<?php
include("../include/dbcommon.php");

if(isset($_GET["country"]) && is_numeric($_GET["country"]))
{
$country = $_GET["country"];
$emp_number=$country;
echo($emp_number);
}
if(isset($_GET["state"]) && is_numeric($_GET["state"]))
{
$state = $_GET["state"];
$emp_positionheld=$state;
echo ($emp_positionheld);
}
?>

<script language="JavaScript">
function autoSubmit()
{
var formObject = document.forms['theForm'];
formObject.submit();
}
</script>
<form name="theForm" method="get">
<select name="country" onChange="autoSubmit();">
<option value="null"></option>
<?php
//POPULATE DROP DOWN MENU WITH Employee Names
$sql = "SELECT EmployeeNumber,LastName FROM employees";
$countries = mysql_query($sql,$conn);
while($row = mysql_fetch_array($countries))
{
echo ("<option value="$row[EmployeeNumber]" " . ($country == $row["EmployeeNumber"] ? " selected" : "") . ">$row[LastName]</option>");
}
?>
</select>



<?php
if($country != null && is_numeric($country) )
{
?>
<select name="state" onChange="autoSubmit();">
<option value="null"></option>
<?php
//POPULATE DROP DOWN MENU WITH Job Position helds For a Given Employee
$sql = "SELECT * FROM employee_positionheld WHERE EmployeeNumber = $country ";
$states = mysql_query($sql);
$row = mysql_fetch_array($states);
for($k=1;$k<=6;$k++)
{
echo("inside for");
$temp[$k]='Position held' . ' '.$k;
$queryvar=$temp[$k];
//echo($queryvar);
echo ("<option value="$row[$queryvar]" " . ($state == $row[$queryvar] ? " selected" : "") . ">$row[$queryvar]</option>");
}
?>
</select>
<?php
}
?>
</form>

MY EMPLOYEE_POSITION HELD TABLE looks like this

EmployeeNumber Employee Name Position held 1 Position held 2 Position held 3 Position held 4 Position held 5 Position held 6

34550 Suraj Entryleveltech1 Seniortech1 programmer1 0 0 0 0

No comments posted yet

Your Answer:

Login to answer
272 Like 26 Dislike
Previous forums Next forums
Other forums

Strange PHP/mySQL error ... am I just tired?
Code: <?

## CONNECT TO DB FUNCTION!
function ConnectTo($db2con)
{
$hostNam

Help Import Animoto and Youtube
CAn someone help me urgently want to allow users on my website to import youtube/revver/dailymotion

Creating an invoice
Using fpdf I am trying to create an invoice to send to clients. The products are sold in pricing tie

Hom to make one url to open together with another url
I have a chat, which i want to be opened, as soon as the users login to the site. As it is now, when

I have a parse error in this query help..
Code: $query1="INSERT INTO `rating` (`item_name`, `rating`, `ip_address`, `date_rated`) VALUES

Transform value 0 in no and 1 in yes help needed
Hello,
I have this in the table: "value" type int 0 to represent false and 1 to represe

mysql timestamp manipulation
How could I use a timestamp (e.g 2009-10-30 13:20:35 ), and with php find out if it is:

from

Filtering Tables
Hello Thanks for any help any one can give Im very new to PHP..

I need to pull data for clien

Notice Undefined index: reset/ Attempting to redirect, works, but get a notice
Full Header.php is:
Code: [Select]<?php
$reset= false;
$reset= strip_tags($_GET[&quo

Save User Inputs while javascript reloading!
hi..
i have written a PHP code in which i reload the page when user selects a value in dropdown..

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash