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

Company Code for Vendors created through Business Partners
Hi, experts!

I'm trying to set up the integration between Business Partner (BP) and Vendo

want to add a break after a certain amount of characters
i was wondering how to add a break in a blog post after a certain amount of characters... any idea o

array & querys help/advice.
Ok so here goes , I have a mysql database and basically here's what I need to do.

// foreach

Help With editting and deleting form
Hallo !!

So look at this image :

http://img194.imageshack.us/img194/8272/snapshot5f.pn

need to apply an if/else statement to Tim Thumb script
Not sure how to work this. I essentially want to call a variety of image sizes based on which style

limit string by words??
hello all,

I was wonder if there is a function to limit a string by the number of words inste

login box shows up again after logging in 2nd login works
Hello,

I am having trouble on my site Lockerz.rack111.com where when you login the login box

user data not transferring to new page
I'm a PHP learner. After an index.php page, the user goes to a login.php page. Both pages seem to

Scrolling news bar and MySQL
I am not sure if this is the correct forum to post this on. If not, I apologize. I am still pretty

Forms Authentication and Refresh at Login page
Hello, When I try to press the Login button in my webapplication at my login.aspx page nothing h