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Hi

I've managed to post a form in the background and I successfully get the return-data from the saving-page back. So far everything is fine. But when I click the check-box in the form more than once, I do not get the feedback again. Of course, because it's faded out. How do I delete the return-data and display the element again?

Thanks in advance for your help.



    <script language="JavaScript" type="text/javascript">
    $(document).ready(function(){
      $("input").click(function(){
           //$(this.form).ajaxSubmit();
           var cnt = $(this.form).find('input[name=cnt]').val();
           //alert(cnt);
           var trg = "#output" + cnt;
           var fdo = "#SBAppl" + cnt;
           $(this.form).ajaxSubmit({
                  target: trg
           });
         $("div").data("SBAppl", " ");
         $(trg).fadeOut(5000);
         jQuery.removeData(div, "SBAppl");
         $(trg).text($("div").data("SBAppl"));
         $(trg).fadeIn(500);
      });
    
    });
    </script>


        <td align="left" valign="top">
          <div id="SBAppl<%=z%>">
           <form method="POST" action="Save.asp" id="<%=z%>">
              <input type="checkbox" name="fApplicable" <% If vApplicable(z) = 1 Then Response.Write "checked" End If %> value="1">
              <input type="hidden" name="fSerial_Number" value="<%=vSerial_Number(z)%>">
              <input type="hidden" name="fSBID" value="<%=vSBID%>">
              <input type="hidden" name="cnt" value="<%=z%>">
           </form>          
          </div>
        </td>
        <td align="left" valign="top" width="40"><div id="output<%=z%>"></div></td>