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

Taking an HTML form and a PHP program and making it into one working file.
Hello,

So I would like to display everything INSIDE a specific directory. In other words, I

PHP5/Zend 2.0 - Resources
PHP5 Snapshots
http://snaps.php.net/

ZEND 2.0 Feature Overview and Design (PDF File

Date Question
I'm using the TIMESTAMP() function within mysql to set the date/time for certain events. However, ho

Making multiple rows with one query
I have a table called efed_handler_characters with two fields handler_id and bio_id. What query 2 is

Insert to MySQL inside foreach
Hey.
I am parsing some html, and putting inside an html db.
I need to create a randomized inte

How to schedule the a job?
Hi,
I have prepared a task ( i.e. a procedure is prepare the file and send the same to con

Sharing PHP Sessions Across Domains
I am in the process of writing a script to share a php session across various domains I have.
The

Splitting Attributes
SQL> SELECT I_NAME, substr(I_NAME,1,instr(I_NAME,'O')) "First part",

substr(I_NAME, IN

php ajax - two drop down list with a submit button
i have created a form with 2 drop-down list and a submit button. I would like to know how to pass th

Strange HTML Tag?
I recently noticed some odd HTML appear in some of the websites I host. Not all of them are run on a