title: Render a specific part of html after success Ajax
body: <p>Myproblem is when I click my first selectbox, after there is a value, it will trigger the second selectbox. I implement it in Ajax, but successfully render ,but my other textfield value is gone. How could I just render a specific part of the responde html(success ajax call)?</p>
  
  <pre><code>$(document).ready(function(){ 
  
      if ($('#product_category').val() == 'Choose Category')
          document.getElementById('product_subcategory').disabled = true;
  
      $('#product_category').change(function () {   
          if ($('#product_category').val() == 'Choose Category')
              document.getElementById('product_subcategory').disabled = true;
          else
              document.getElementById('product_subcategory').disabled = false;
  
  
          data = $('#product_category').val();
          //alert(data);
  
          var param = 'category_name=' + data;
          $.ajax({
            url: MYURL,
            data: param,
            success: function(result) {
              alert('Choose product subcategory');
              alert(param);
              $('body').html('');
              $('body').html(result);
            }
          });
         //   window.location = MYURL?category_name="+data;
      });
  
      $('#product_subcategory').change(function () {    
          data = $('#product_subcategory').val();
       //     paramCategory = $(document).getUrlParam('category_name');
        //    alert(paramCategory);
  
          $.get(MYURL, function(data){
              alert("Data Loaded: " + data);
              });
          //window.location = MYURL?subcategory_name=" +  data;
      });
  
  });
  </code></pre>
  
  <p>in my form, i use $_GET['category_name'] to get my value Ajax return value. I Debug in firebug, and it is successfully. I tried to render again the html, but my previous textarea's value and textfiel's value is gone since what i did is $('body').html(''); $('body').html(result);, So,how could I manage to get the success ajax return value, and use it in the PHP.</p>
  
  <p>any confusion ,please tell me...
  Thank you for spending ur time.</p>
  
  <hr>
  
  <p>Hmm, I'm using a div and show the Div when it was return success ajax call.</p>