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.

Did you know?Explore Trending and Topic pages for more stories like this.
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
No comments posted yet

Your Answer:

Login to answer
272 Like 26 Dislike
Previous forums Next forums
Other forums

Get value from Select menu
Hi!

Have tried to find the solution on the web. Don't know if there is an easy one. A descrip

Parse Error
Hi Guys,

I have a function in my class which returns a string link variable. The problem is i

Please help with code for SMTP authenticated PHP email form
Hello, I'm creating a PHP email form, and for this particular server, I have to use SMTP Authenticat

Undefined variables
hi
----------------------------------------------------------------------------------------------

Prevent PHP mail( ) from appending hostname to from address
Hey guys,

I want to be able to send texts to phones via email (since each phone has their own

retrieving policy name inside the function called by this particular policy
Hi there,

I've playing around with dbms_rls package, trying to set up some security repo

email CODING Problem
Email coding problem.

I need to send a section of the info to $tf and the same as a Cc to $em

Having problemswith multithreading and prime numbers
I have an assignment when I'm suppose to do the following:

Write a multithreaded Java, Pt

xml element exists
Code: [Select]<?xml version="1.0"?>
<Addresses>
<

Connecion issue
Hi,

For some reason, I keep getting this error when trying to connect:

Warning: mysql_

Sign up to write
Sign up now if you have flare of writing..
Login   |   Register
Follow Us
Indyaspeak @ Facebook Indyaspeak @ Twitter Indyaspeak @ Pinterest RSS



Play Free Quiz and Win Cash