Hi
I am new to ajax but i have done one simple menu program using ajax i created one index page with menus if i click each menu i want that linked page in the same window now i am getting as a separate window that is i want to refresh only that particular part and not the whole page i will give you the code that i have done please help me out
this is my index.html page were i have my menus
<html>
<head>
<title>My ajax website</title>
<script language="javascript" type="text/javascript" src="ajax.js"></script>
</head>
<body>
<div id="nav">
<a href="home.html" onClick="sendRequest('home.html');">Home</a>
<a href="first.html"onClick="sendRequest('first.html');">First Page</a>
<a href="second.html onClick="sendRequest('second.html');">Second Page</a>
</div>
<br/>
<div id="content">
<script language="Javascript" type="text/javascript">
if ((window.location.href.split("#", 2)[1] == null) || (window.location.href.split("#", 2)[1] == "") || (window.location.href.split("#", 2)[1] == "index")){
sendRequest("index.html");
}else{
sendRequest(window.location.href.split("#", 2)[1] + ".html");
}
</script>
</div>
</body>
</html>
this is my ajax.js
function createRequestObject() {
var req;
if(window.XMLHttpRequest){
// Firefox, Safari, Opera...
req = new XMLHttpRequest();
} else if(window.ActiveXObject) {
// Internet Explorer 5+
req = new ActiveXObject("Microsoft.XMLHTTP");
} else {
// There is an error creating the object,
// just as an old browser is being used.
alert('Problem creating the XMLHttpRequest object');
}
return req;
}
// Make the XMLHttpRequest object
var http = createRequestObject();
function sendRequest(webpage) {
// Open PHP script for requests
http.open('get', webpage);
http.onreadystatechange = handleResponse;
http.send(null);
}
function handleResponse() {
if(http.readyState == 4 && http.status == 200){
// Text returned FROM the PHP script
var response = http.responseText;
if(response) {
// UPDATE ajaxTest content
//alert(response);
document.getElementById("content").innerHTML = response;
}
}
}