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 Like 26 Dislike
Previous forums Next forums
Other forums

User feedback after MySQL query has been executed
Hi all, I've just registered on PHPFreaks because I've got a question that I simply can't work out b

Email to a friend script problems
I have this send-to-a-friend script with 2 issues:


1) When you open the form popup on a p

How to show the difference between two data field in a database with php.
Hello php gurus,

how r u all... i'm not so well facing a typical problem please help me...<

Firefox displaying PHP source code??
Currently testing a site thats almost built, am going to be including php on a sidebar on all pages

Passing JS to PHP
So I have the following
Code: <?php
getdetails(?><script type="text/

Find occurences of unicode characters in string
I need to prohibit filenames with everything but English characters and numbers but regexp and strin

Hit counter updating once per IP - IP HIT COUNTER
I have a hit counter, for the amount of views on a tutorial.
It'll do the query and then do..

Just a white page
Okay so, my website, when I click SignUp on it it takes me to /join.php but its a complete white pag

Error querying database.
I get the above error when trying to insert some values to a datatable.
Here's the code :

Need help-Error istream header declaration
Hi Everyone ,

I am having issue with istream declaration. I am new to C programming.
can a

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