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

Need help with unexpected T-STRING error
I'm a newbie and I'm still learning PHP. However this error has me stumped. I've googled, searched t

rdns with php
I am trying to build a script that will show all of the different domains that are hosted on the sam

Data storage spaces in varchar2
Trying to understand what's happening.
I am selecting a value from a table that is defined as c

generating all possible random letters
hi'

how can i randomize the letters a,b,c,d,e all possible ways, and i want to print the res

defining website tags
Hi,

What would be the best way to define tags for my site, such as website title, url etc.

Bandwidth monitoring?
Hi guys,

I need a little information I have written a php app and I occurred to that I need

C - Reading a file into a byte array
Hi,

I'm trying to read a file into a byte array in C. I have to use C as this is for a loadru

Problem executing bash script using shell_exec
Hi there,
I created a bash script file using following code to convert doc documents to pdf using

ScriptResource.axd gives an error on fresh install of ASP.NET Ajax 1.0
Hello,I have a fresh install of Microsoft ASP.NET Ajax 1.0. When I create an Ajax enabled website in

Remove values in array2 from array1
I have two arrays.

Array 1 is where the array key holds various different numbers. For exampl