updating drop-down on change of another drop-down using php-ajax - php

Question

I want to change the value of city drop-down on change of country drop-down. Well now stuck at on point when I select the value in the country it says he variable $r is undefined.
page : index.php
<form method="post" action="" name="form1">
<table width="60%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="150">Country</td>
<td width="150"><select name="country" onChange="getCity('findcity.php?country='+this.value)">
<option value="">Select Country</option>
<option value="1">USA</option>
<option value="2">Canada</option>
</select></td>
</tr>
<tr style="">
<td>City</td>
<td ><div id="citydiv"><select name="city">
<option>Select City</option>
</select></div></td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
</table>
page: findcity.php
<?php
$country=$_REQUEST['country'];
$link = mysql_connect('localhost', 'root', '');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('db_ajax');
$query="select city from city where countryid=$country";
$result=mysql_query($query);
?>
<select name="city">
<? while($r=mysql_fetch_array($result)) {
?>
<option value=""><?php echo $r['city'];?></option>
<? } ?>
</select>
here is the script
<script>
function getXMLHTTP() { //function to return the xml http object
var xmlhttp=false;
try{
xmlhttp=new XMLHttpRequest();
}
catch(e) {
try{
xmlhttp= new ActiveXObject("Microsoft.XMLHTTP");
}
catch(e){
try{
xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
}
catch(e1){
xmlhttp=false;
}
}
}
return xmlhttp;
}
function getCity(strURL) {
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function() {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('citydiv').innerHTML=req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
</script>

Answer 1

try using mysql_fetch_assoc instead of mysql_fetch_array

Answer 2

you miss to write "php" near <?(write down php here) while($r=mysql_fetch_array($result)) { ". and perform some debugging like check that are you getting the id of country or not than check the query returns the proper result or not or print the query on browser and run that query on Database.

Answer 3

change your code to..
<?php
while($r=mysql_fetch_array($result))
{
echo '<option value="">'.$r['city'].'</option>';
}
?>