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

Echo multiple lines of html code
Hi guys,

I would like to know if there is a way to echo multiple lines of html codes.

Multiple websites question.
Hi, does php have any functionality that enables a programmer to extract information from a website

Preserving user-entered linebreaks
Hi, I'm building a web 2.0 thingy from scratch and I wanna display text which users input. It's all

retrieving more than one max key from an array?
so i have an array of 20 numerical values (0-100) that i need to order from highest to lowest and th

utl_file open error
i have file in the unix path

Path /popdev01/pop/popdevb/tfi/

File name

how to load a Sys file with system load and call images?
normally you need register the module as a service with CreateService, and start the service

explode() function problem maybe
Hi I'm having trouble searching my database. When I type two words in the search field it only searc

Required to login help
I'm trying to set up my site so users have to be logged into the forum to access the site.I've been

quick question
Hi ..

i have a question
how do i set a var so it displays via an echo
Code: $logo = '&a

Undefined index on my form
ok im getting Undefined index on this line.. print_r($_REQUEST['form']);

below is the full sc