How to acess request attribute in java script

[meenu24]

Hi

I am using ajax to submit the request and then getting the response back to display contents of a div.

this is my code.

function showDiv(setname,url) {
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 (e1){
try{
page_request = new ActiveXObject("Microsoft.XMLHTTP");
}
catch (e1){}
}
}
else
return false;
page_request.onreadystatechange = function(){ callback(page_request)};
page_request.open('GET',url, true);
page_request.send(null);
}

function callback(page_request)
{
if (page_request.readyState == 4) {
if (page_request.status == 200) {
document.getElementById("setaffiiates").innerHTML = page_request.responseText;
document.getElementById("setaffiiates").style.display = "block";
}
}
}

my div code is

<div id="setaffiiates" style="display:none;position: absolute; padding:2px; width: 225px; z-index: 500;">
<% NetworkSet[] netSets = (NetworkSet[])request.getAttribute("netSets"); %>
<table>
<tr>
<td>
<%=netSets[0].network%>
</td>
</tr>
</table>
</div>

Pl excuse if I have mixed the JSP code there. But I have to show all the code for clarity. Now the issue when i get the reponse I see the whole html page again with the div portion again. So it is like the page and again page with div code.
What should I do if I want to get the div area and not the whole page once more again

thanks
Meena

 

[BillyRayPreachersSon]

Don't return the whole page in your ajax response. Just return what you want to put inside the DIV.

Hope this helps,
Dan



Coedit Limited - Delivering standards compliant, accessible web solutions

[tt]Dan's Page [blue]@[/blue] Code Couch

http://www.codecouch.com/dan/

[/tt]

 

[meenu24]

Thanks Dan, you mean forming response in XML format from the servlet and then returning will be best solution.
I use struts, this is my servlet code.

request.setAttribute("setname", setName);
return mapping.findForward("setaffiliates");

I guess, I have to form some XML and then forward the request object.
I will try it.