Get Drop Down Menu, PHP SQL HTML - variables

Question

I am a student and also working in my university. I am not studying computer science.
I mainly do support stuff, but now I should take care of our internship database.
Task: Create a drop down menu from a town which a student choses, then show all internships in this town.
I have an existing SQL database, can create a drop down menu from it, but have NO idea how I can use the selected value. A submit button would be fine, any solution which works is fine.
How can I put the chosen value from drop down menu to a PHP variable?
<?php
$dbanfrage = "SELECT DISTINCT Stadt FROM $tabelle WHERE Land = 'China' OR Land = 'Taiwan' OR Land = 'VR China' ORDER BY STADT ";
$result = mysql_db_query ($dbname, $dbanfrage, $dbverbindung);
if (!$result)
{
$message = 'Ungültige Abfrage: ' . mysql_error() . "\n";
$message .= 'Gesamte Abfrage: ' . $dbanfrage;
die($message);
}
echo "<select>";
echo "<option>Stadt auswählen</option>"; // printing the list box select command
while($nt = mysql_fetch_array($result)) //Array or records stored in $nt
{
echo "<option value = $nt[id]>$nt[Stadt]</option>"; /* Option values are added by looping through the array */
}
echo "</select>"; // Closing of list box
mysql_close ($dbverbindung);
?>

Answer 1

Just suggesting how you can do it --
<select name="country" id="country" onChange="showState();" >
<option value="">Select Country</option>
<option value="1">India</option>
<option value="2">Nepal</option>
</select>
Your Script--
<script type="text/javascript">
function showState( )
{
var value = document.getElementById('country').value;
var url = '<?php echo base_url ?>showstate.php';
$.ajax({
type: "GET",
url: url,
data:{'country':value},
success:function(results)
{
$('#div_state').html(results);
}
});
}
</script>
Your showstate.php php page --
//INCLUDE CONNECTION file to for db query
$type = $_GET['country'];
//Your DB QUERIES
This will load the content in your #div_state div. You can also use Jquery For this..i think this will help you :) an d let me know if you have any query.

Answer 2

<select name="name">
<option value="select">select</option>
<?php
$dbanfragee=mysql_query("SELECT DISTINCT Stadt FROM $tabelle WHERE Land = 'China' OR Land = 'Taiwan' OR Land = 'VR China' ORDER BY STADT ");
while($row=mysql_fetch_array($$dbanfrage)){?>
<option value="<?=$row['id']?>"><?=$row['stadt']?></option>
<?php }?>
</select>