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
No comments posted yet
Your Answer:
Login to answer
272 26
Other forums word wrap in emails help needed Hello, I understand how wordwrap works in php and have used it well before. However when I used wor
Conditions of info record - Error When i created GR (901) and PO create automatic my PBXX is obtain the net price of info record but
mySQL and PHP search Hello, I am trying to code a project and ran into a brick wall with one of my pages. I am pretty
Help with translating C code into assembler code Hi im doing a project that moves a robot around a maze avoiding walls and need some help with conver
imap: how to save a copy of sent emails to sent elements Hello my friends, I am writing an online emailing application with inbox outbox/sent elements
php/mysql problem Hey all, Problem: im trying to setup a shop where people can use a drop-down list to select t
How to extract/download content from HTTPS page? Hello to all the Members of this forum, Im Shoiab, A novice programmer in php.. for my first job I h
How do i do multi uploads?. Hey i would like to do this : http://hosting.mrkrabz.net/ I've got the base down you can uplo
Renaming a file that a user uploads to site? My site allows for registered users to upload images to the site under their own gallery. Currently
Onclick problem in Firefox Hi, I am using a . It doesn't seem to