onChange refresh particular dropdown list in PHP - php

Question

THE ISSUE?
I need help regarding PHP's onChange command. Basically i'm making a hotel room reservation system. What im trying to do is that there are tow combo boxes.
One for HOTEL NAME (COMBO BOX NO.1)
the other for FLOOR NO (COMBO BOX NO.2)
What I want to do is that if I select the hotel name form the combo box 1, then the number of floors for that particular hotel, should get auto-populated in combo box 2. (This number of floors information should come from the respective table/field in the database)
WHAT i'VE TRIED TO DO -- THE CODE
<td>Hotel:
<select name="hotel_name" id="hotel_name" title="<?php echo $row_rs_hotel['hotel_name']; ?>" onchange="<?php
while ($row_rs_floor = mysql_fetch_assoc($rs_floor));
$rows = mysql_num_rows($rs_floor);
if($rows > 0) {
mysql_data_seek($rs_floor, 0);
$row_rs_floor = mysql_fetch_assoc($rs_floor);
}
?>">
<?php
do {
?>
<option value="<?php echo $row_rs_hotel['hotel_name']?>"><?php echo $row_rs_hotel['hotel_name']?></option>
<?php
} while ($row_rs_hotel = mysql_fetch_assoc($rs_hotel));
$rows = mysql_num_rows($rs_hotel);
if($rows > 0) {
mysql_data_seek($rs_hotel, 0);
$row_rs_hotel = mysql_fetch_assoc($rs_hotel);
}
?>
</select></td>
</tr>
<tr>
<td>Floor No:
<select name="floor_no" id="floor_no" title="<?php echo $row_rs_floor['floor_no']; ?>" onchange="var id=$('hotels_name').val();
$('floors_no').load('ajax.php?id='+id);
">
<?php
do {
?>
<option value="<?php echo $row_rs_floor['floor_no']?>"><?php echo enter code here$row_rs_floor['floor_no']?></option>
<?php
} while ($row_rs_floor = mysql_fetch_assoc($rs_floor));
$rows = mysql_num_rows($rs_floor);
if($rows > 0) {
mysql_data_seek($rs_floor, 0);
$row_rs_floor = mysql_fetch_assoc($rs_floor);
}
?>
</select></td>
WHAT I'M GETTING FROM THE ABOVE CODE??
The combo box number 2 is auto-loaded with the first record in the table when the page is opened. No matter how many times i select the hotel in COMBO BOX 1, the values in combo box 2 remain unchanged.
PLEASE HELP..!!!

Answer 1

You must combine both javascript and php, as you want to retrieve the information from the database (server side using php+a bit of mysql) with a dynamic page without refresh (client side using javascript). It will certainly become a long difficult job, but I think it's the best approach.
Read this article about double drop down menu and study the code this article (or search for similar projects) and when you understand the javascript, you must fill it where it's appropriate with php.
For example, this is a bit of the javascript (or html) source code:
if (Indx==1)
{
options[0]=new Option("Choose a JavaScript Page","");
options[1]=new Option("Alerts","alerts.htm");
options[2]=new Option("Back and Forward Buttons","BackForward.htm");
options[3]=new Option("Contents","index.html");
}
For your project it should be something like this (not complete and not bug-free, but it's just to give you an idea:
if (Indx==1)
{
<?php
$i=0;
while ($row = mysql_fetch_array($result))
{
?>
options[<?php echo $i;?>]=new Option("<?php echo $row['name']."\",\"".$row['page']; ?>");
<?php $i=$i+1; } ?>
}
That should give you the same code (assuming that 'name' and 'page' are values of your columns in your database). And that needs to be done in all convenient javascript bits, but I won't do all your work... give it your best and, if it still doesn't work, come back and ask a more specific question about the code.
If you make it work, I'd like to encourage you to post it here editing your question so everyone can learn from it. Keep in mind that this is how I would approach this problem, but there are surely other ways of doing it.