Home/ Questions/Q 8358287
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-09T10:44:52+00:00

I’m performing an AJAX request to get a generated html page. I want to

  • 0

I’m performing an AJAX request to get a generated html page. I want to find a specific div in the result and inject it into the page where the request came from. I also want to include embedded script or script links which may be inside this div. Here is a simple sample of the page I want to get with AJAX

<html>
<head>
    <script>alert("don't want this")</script>
</head>
<body>
<div id='findMe'>
    <p>Some content</p>
    <script>alert("want only this")</script>
</div>
<script>alert("don't want this")</script>
</body>
</html>

So I only want to extract the div with ID findMe. Here’s what I have in the page that’s doing the request

<script>
$(document).ready(function(){

    $.ajax({
          url: "ajax.htm",
          success: function(data){  

            $(data).filter("#findMe").appendTo("#output");

          },
          dataType: "html"
    });

});
</script>

<div id='output'></div>

But the script tag is missing and no alert appears. It seems to get taken out of the div. If I do

console.log($(data))

I can see each of the script tags as a document fragement, but how to I know which one was in the div before it was popped out?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    I had to treat the result as XML. Find the element I’m looking for in the XML object using jQuery .find(), select the actual node (array pos 0), convert the contents of the node to a string, and inject into the output div. This is the only method I’ve found that executes scripts only contained in the target div.

    $.ajax({
    url: "ajax.htm",
    success: function (data){

        var node = $(data).find("#findMe")[0];

        if(node.xml)
        {
            $("#output").html(node.xml)
        }
        else if(new XMLSerializer())
        {
            $("#output").html((new XMLSerializer()).serializeToString(node));
        }
    },
    dataType: "xml"
});
    • 0