Dear prog wizards,

Please help as I am noob at this. I got source code from other website in which I made small changes to cater my need. I have created a database called 'a_database' and table called 'property' using XAMPP 1.7.7 . The columns are 'id', 'Type', 'Price' and 'Location'.

My first dropdown box filters Type
My second dropdown box filters Location

For example : Type is House and Factory
Location is Bangkok and Singapore

When I select House, I will get the desired table.

Then, I select Bangkok and expect to filter House which are located in Bangkok. Unfortunately, this did not happen. Can someone please help me on this ?

My ajax code :
Code:
<html> <head> 
<script type="text/javascript"> 

function showtype(str){
	var ajaxRequest;  // The variable that makes <strong class="highlight">Ajax</strong> possible!
	if (str=="") {   
		document.getElementById("txtHint").innerHTML="";   
		return;   
	}  	
	try{
		// Opera 8.0+, Firefox, Safari
		ajaxRequest = new XMLHttpRequest();
	} catch (e){
		// Internet Explorer Browsers
		try{
			ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
		} catch (e) {
			try{
				ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
			} catch (e){
				// Something went wrong
				alert("Your browser broke!");
				return false;
			}
		}
	}
	// Create a function that will receive data sent from the server
	ajaxRequest.onreadystatechange = function(){
		if(ajaxRequest.readyState == 4){
			var ajaxDisplay = document.getElementById('txtHint');
			ajaxDisplay.innerHTML = ajaxRequest.responseText;
		}
	}
	ajaxRequest.open("GET","gettype.php?q="+str,true); 
	ajaxRequest.send(null); 	
}
function showlocation(str){
	var ajaxRequest;  // The variable that makes <strong class="highlight">Ajax</strong> possible!
	if (str=="") {   
		document.getElementById("txtHint").innerHTML="";   
		return;   
	}  	
	try{
		// Opera 8.0+, Firefox, Safari
		ajaxRequest = new XMLHttpRequest();
	} catch (e){
		// Internet Explorer Browsers
		try{
			ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
		} catch (e) {
			try{
				ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
			} catch (e){
				// Something went wrong
				alert("Your browser broke!");
				return false;
			}
		}
	}
	// Create a function that will receive data sent from the server
	ajaxRequest.onreadystatechange = function(){
		if(ajaxRequest.readyState == 4){
			var ajaxDisplay = document.getElementById('txtHint');
			ajaxDisplay.innerHTML = ajaxRequest.responseText;
		}
	}
	ajaxRequest.open("GET","getlocation.php?q="+str,true); 
	ajaxRequest.send(null); 	
}
//-->	
</script> 
</head> 
<body>  
<form> 
<select name="type" onchange="showtype(this.value)"> 
<option value="">Select property type:</option> 
<option value="1">All</option> 
<option value="2">factory</option> 
<option value="3">house</option> 
</select><br />
<select name="Location" onchange="showlocation(this.value)"> 
<option value="">Select property location:</option> 
<option value="1">All</option> 
<option value="2">bangkok</option> 
<option value="3">Patpong</option> 
</select> </form> <br /> 
<div id="txtHint"><b>Person info will be listed here.</b></div>  </body> </html>
My getlocation.php

Code:
<?php 
	$q=$_GET["q"]; 
	// check for your q values, 1 = all, 2 = bangkok, 3 = Patpong
		$con = mysql_connect('localhost', 'root', ''); 
	if (!$con) { 
		die('Could not connect: ' . mysql_error()); 
	} 
	mysql_select_db("a_database", $con); 
	// based on the value passed in we may have to change query, if all run this...
	if ($q == 1) {
		$sql="SELECT * FROM property"; 
	} else if($q == 2){
		$sql="SELECT * FROM property WHERE Location = 'bangkok'"; 
	} else if($q == 3){
		$sql="SELECT * FROM property WHERE Location = 'Patpong'"; 		
	}
	$result = mysql_query($sql); 
	echo "<table border='1'> <tr> <th>Type</th> <th>Price</th> <th>Location</th> </tr>"; 
	while($row = mysql_fetch_array($result)) { 
		echo "<tr>"; 
		echo "<td>" . $row['Type'] . "</td>"; 
		echo "<td>" . $row['Price'] . "</td>"; 
		echo "<td>" . $row['Location'] . "</td>"; 
		echo "</tr>"; } 
		echo "</table>"; 
		mysql_close($con); 
	?>
My gettype.php

[CODE]<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'root', '');
if (!$con) {
die('Could not connect: ' . mysql_error());
}

mysql_select_db("a_database", $con);



if ($q == 1) {
$sql="SELECT * FROM property";
} else if($q == 2){
$sql="SELECT * FROM property WHERE Type = 'factory'";
} else if($q == 3){
$sql="SELECT * FROM property WHERE Type = 'house'";
}

$result = mysql_query($sql);
echo "<table border='1'> <tr> <th>Type</th> <th>Price</th> <th>Location</th> </tr>";
while($row = mysql_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['Type'] . "</td>";
echo "<td>" . $row['Price'] . "</td>";
echo "<td>" . $row['Location'] . "</td>";
echo "</tr>"; }
echo "</table>";
mysql_close($con);
?>[CODE]