hi,
I have login form which sends the user id and password through ajax to a server side php script.The script validates the GET data.it then echo's error if not valid.the error message is displayed in a DIV on the browser.My question is how can I redirect to a different page on a successful validation?

ajax script:

var ajaxRequest; // The variable that makes Ajax possible!

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('ajaxDiv');
//if(ajaxRequest.responseText == "login success")

ajaxDisplay.innerHTML = ajaxRequest.responseText;
var redi = document.getElementById('ajaxDiv').value;
if(ajaxRequest.responseText == "login success")

//window.location.replace="http://someplace.com";
document.write('Hello World');
//}
}
}
var uid = document.getElementById('uid').value;
var pwd = document.getElementById('pwd').value;

var queryString = "?uid=" + uid + "&pwd=" + pwd ;
ajaxRequest.open("GET", "ajax-example.php" + queryString, true);
ajaxRequest.send(null);
}

the php script:

<?php
$dbhost = "localhost";
$dbuser = "some user";
$dbpass = "some password";
$dbname = "some database";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($dbname) or die(mysql_error());
// Retrieve data from Query String
$uid = $_GET['uid'];
$pwd = $_GET['pwd'];

$uid = mysql_real_escape_string($uid);
$pwd = mysql_real_escape_string($pwd);
//$wpm = mysql_real_escape_string($wpm);
//build query

$query = "SELECT uid,password FROM login WHERE uid = '$uid' and password= '$pwd' ";
$result = mysql_query($query) or die("Query failed");
# print "query executed";
$blurb = mysql_fetch_row($result);

$uid1=$blurb[0];
$password1=$blurb[1];

if (( $uid1 ¦¦ $password1 )=='')
{
//$display_string= "";
//echo ;
echo "<font face=\"Arial\" size=-1%><p style=\"color:red \"> Login failed;Please try again</p></font>";
//print<<< HTML

//_HTML_;
}
else
{
echo "login success";
//header('Location: [google.com...]
}

?>