Question

this is my code for taking external page into div using ajax
what i tried is i clicked on button i must display the response in div
but i tried several times but doesn’t works.
my javascript code is

var rootdomain="http://"+window.location.hostname

function ajaxinclude(url) {
    var url=rootdomain+url;
    alert(url);
var page_request = false
if (window.XMLHttpRequest) // if Mozilla, Safari etc
page_request = new XMLHttpRequest()
else if (window.ActiveXObject){ // if IE
try {
page_request = new ActiveXObject("Msxml2.XMLHTTP")
} 
catch (e){
try{
page_request = new ActiveXObject("Microsoft.XMLHTTP")
}
catch (e){}
}
}
else
return false
page_request.open('GET', url, false) //get page synchronously 
page_request.send(null)
writecontent(page_request)
}

function writecontent(page_request){
if (window.location.href.indexOf("http")==-1 || page_request.status==200)
document.getElementById("eee").innerHTML=(page_request.responseText);
}

and this is my body section :—–

<input type="button" onclick="ajaxinclude('/songcake/index.php')" value="Click !" />
<div id="eee" style=" width:400px; height:800px;">
</div>

please help

Thanks.

Answer 1

attach your method to onreadystatechange which not there in your code

 page_request.onreadystatechange = writecontent;

 function writecontent() {
   if (page_request.readyState != 4)  { return; }
   document.getElementById("eee").innerHTML=(page_request.responseText);

 }