two drop down menu mysql and php - php

Question

I am creating a php search. I select the 2 menus result from the database. The first menu works fine, but the second one no. Can you please help me?
This is my home.php
<?php
require("ConfigPage.php");
$sql="SELECT LocationName
FROM tblplacelocation
Join tblplace ON tblplace.PlaceID=tblplacelocation.PlaceID
Join tbllocation ON tbllocation.LocationID=tblplacelocation.LocationID
where PlaceType='Hotels'" ;
$result = mysql_query($sql,$con);
echo '<form><P style="color:white;">Region</P>
<select name="PlaceID" onchange="showUser(this.value)">
<option value="">Region:</option>';
while($data = mysql_fetch_array($result))
{
$id= $data['PlaceID'];
$lname = $data['LocationName'];
echo "<option value='". $lname. "'>".$lname."</option>";
}
echo "</select></form>";
echo "<div id='txtHint'></div>";
echo "<div id='divImage'></div>";?>
this is the getuser.php
<?php
$q = intval($_GET['q']);
include("ConfigPage.php");
$sql="SELECT PlaceName
FROM tblplacelocation
Join tblplace ON tblplace.PlaceID=tblplacelocation.PlaceID
Join tbllocation ON tbllocation.LocationID=tblplacelocation.LocationID
WHERE LocationName = '".$q."' AND PlaceType='Hotels'";
$result = mysql_query($sql,$con);
echo '<form><P style="color:white;">Place Name</P>
<select name="adminID" onchange="showImages(this.value)">
<option value="">Select a place:</option>';
while($data = mysql_fetch_array($result))
{
$pid= $data['LocationName'];
$pname = $data['PlaceName'];
echo "<option value='". $pid. "'>".$pname."</option>";
}
echo "</select></form>";
mysql_close($con);
?>
and this is the java script:
function showUser(str)
{
if (str=="")
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
I don't know what's wrong. The first menu displays the result but the second menu contains nothing.

Answer 1

I would suggest you to use Firebug for debugging.
check if you have any javascript errors (maybe syntax error)
check if your ajax request is being send, if yes, check what's inside the response
check if your dom element is being selected correctly ... in firebug you can execute javascript in the console area. check if your document.getElementById selects the correct element by inserting something yourself
if all of this works, check if your select element has empty options. if this is the case, your variables should be empty
if your select element has no options, something with your sql select may be wrong
If you're just experimenting, it's ok, but otherwise let me say that your code looks pretty ugly. This is not the way you should programm things.
You should think about dividing your code by concerns. Check out the Model-View-Controller Pattern and beginner tutorials for any PHP-Framework.
MVC:
http://en.wikipedia.org/wiki/Model%E2%80%93view%E2%80%93controller
A Framework which is easy to learn:
http://www.yiiframework.com/