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

LSB (PHP 5.3) problem with static value!
hello,

i'm having a problem. static::$text variable gets lost at some point. can someone plea

Not reloading page after php form submit
Hello helpful souls out there. You guys have come through for me in the recent past and I'm hoping s

problems with search form numerical "between" sending by php
I am trying to display the results of a search, on a sql database, on a web page. I set up html &quo

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

browse folder only
hi,
is there a way i can browse or get the directory or folder only and not the file?
i just w

Quick Syntax Question
Hi folks,

I'm getting the following error: "unexpected T_LNUMBER". I'm trying to b

contact form - output to page and email
I have the following in my controller:

$message['name'] = htmlentities(strip_tags(trim($_POST

ctype() validation - allowing illegal characters
Hello,
I use ctype() to filter and validate a user form. However, I am trying to allow certain c

values not being entered into table
hi. I;ve created a form, so that when a user enters data into it, it gets added to a table in a data

Take info from one coloum and move to another
Hi all,

I have this:
Code: [Select]$array = "SELECT stock_id FROM stocks WHERE stock_