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

Your Answer:

Login to answer

Other forums

Help =( !! Upload Pics [PHP script]
Hello
I need help with a php script [MULTIPLE UPLOAD IMAGES] , where I want to add a feature (wat

PHP - MySQL Fail
My PHP code will only execute the first part of my code...

Code: <?php

sessio

Undefined variable when using $_SERVER['PHP_SELF']
Hi guyz, please suggest me something...
On first.php I have one input field NAME, and on posting

Multithreading design
Hi

I have come up with a Singleton class that manages a pool of database connections. Basical

j1inmis Ouput layout changes
Hi,

Can anybody tell how can j1inmis output layout can be changed so that vendor names ge

Java API in PHP?
I have an application that we use internally here at the office.

The software company provide

NOOB needs help with upload file
The intent of this form is to create a folder for a client and upload an image into another folder '

WELCOME SCREEN
first of all let me tell you what does my script do,

it´s a very simple query to show a e

PHP friend keyword equivalent
is there a PHP equivalent for the C++ friend keyword?

FlashVar function how to insert variables
Hi guys. Im new in this forum and also new in php programming. Can you help me about this code?
<