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

Why Are These Functions Causing MASSIVE Memory Problems? Please Help!
Hi,

I have a script with some options.

I use regex to replace patterns in strings, but

How can use this array as a key?
Say I have this array for example:


print_r($array);

Prints:
Code: [Select]Array

strptime() equivalent for php4 ?!
Greetings!

this is my first post, thank you in advance for your replies.

Well, the ti

Pass sql into pl/sql and create RMAN duplicate script.
Hi,

I'm new to pl/sql and I'm trying to write a script that will generate some RMAN comma

xApp Analytics in BI 7.0
Dear all,

I am trying to implement xApp Analytics in BI 7.0. After I installed the analyt

listing help
Hi,

at first, great new design!!! I like it,

i got a problem,

I want to list so

$_FILES array
Ok,

I just started experimenting with PHP today. My goal right now is to have people be able

When i am Canseling the Billing Document in VF11 I am getting Shortdump
Hi Experts

When i am Canseling the Billing Document in VF11 I am getting Shortdump.

Multi image upload
I have a gallery to build and want to build a dynamic upload form to allow for $var number of fields

Need help to modify php ELSE code
I'm trying to find a way to have the php code display a message if there is no inventory listed in t

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