I've got a problem. this is what I have on my page somewhere in header

 $(document).ready(function(){
       $(".ajax").click(function(e){
       var url = $(this).attr("href");
       var url = url.split("##/");

       if (!url[1]){url[1]=" ";}
       var page = "http://<?php echo $_SERVER[SERVER_NAME]; ?>page.php?ajax=1&what="+url[0]+url[1];
       $("#content").load(page);

       var kkk = 'Asdasdasd';
       });
 });

so, after clicking on  <a href="##/new_page" class="ajax">new</a>, content is being filled with this:

<script> alert(kkk); </script>
<a href="##/old_page" class="ajax">old page</a>

However, alert is not taking in consideration as new page DOES NOT see kkk and neither link which worked on the original page works. Why?

I've got a problem. this is what I have on my page somewhere in header

 $(document).ready(function(){
       $(".ajax").click(function(e){
       var url = $(this).attr("href");
       var url = url.split("##/");

       if (!url[1]){url[1]=" ";}
       var page = "http://<?php echo $_SERVER[SERVER_NAME]; ?>page.php?ajax=1&what="+url[0]+url[1];
       $("#content").load(page);

       var kkk = 'Asdasdasd';
       });
 });

so, after clicking on  <a href="##/new_page" class="ajax">new</a>, content is being filled with this:

<script> alert(kkk); </script>
<a href="##/old_page" class="ajax">old page</a>

However, alert is not taking in consideration as new page DOES NOT see kkk and neither link which worked on the original page works. Why?

• javascript
• jquery
• ajax

• Link to review what actually happening can be helpful – tejash Commented Jun 7, 2011 at 5:20
• 1 your variable had to named kkk? not cool -_- – Ibu Commented Jun 7, 2011 at 5:21
• @Ghyath Serhal: I've got page, there is a link which looks like this ##/statistics I click it, and script (first script) in my quesstion is being proccessed (content is being loaded with content of page.php?ajax=1&what=statistics ... now, in statistics, there is a link ##/profile <-- but this link does not work! Because first script does not affect new added code... got it ? – genesis Commented Jun 7, 2011 at 5:33

2 Answers 2

In order for the link to work after loading it again, you probably want a live event. That will watch for any new objects with the ajax class as well instead of just the ones which exist at the time the original document is loaded. And if you need to set anything else up, which would normally be inside document ready, put that in an anonymous function as the second parameter to load() because it isn't guaranteed to have loaded the new content yet otherwise. eg:

   $(".ajax").live('click', function(e){
   var url = $(this).attr("href");
   var url = url.split("##/");

   if (!url[1]){url[1]=" ";}
   var page = "http://<?php echo $_SERVER[SERVER_NAME]; ?>page.php?ajax=1&what="+url[0]+url[1];
   $("#content").load(page, function () { var kkk = 'Asdasdasd'; });

   });

Or as @Levi points out, a .delegate call would acplish the same thing while being a bit less wasteful: $('content').delegate('.ajax', 'click', function() (...)

The kkk variable is being enclosed in the scope of the document ready function. Move it outside of this function instead:

var kkk = '...';

$(document).ready(function () {
    ...
});