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Home/ Questions/Q 7815185
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T05:26:26+00:00 2026-06-02T05:26:26+00:00

html code <table border=0 class=commentbox> <tr> <td>Some Item text </td> </tr> <tr> <td> <div

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html code

 <table border="0"  class="commentbox">
    <tr>
      <td>Some Item text
      </td>
   </tr>
   <tr>
     <td>
          <div id="<%:containerid %>"></div>
          <input type="button" class='btnReply' id="<%:rid %>" value="Reply"/>

     </td>
   </tr>
 <tr>
   <td>
      <div id="replytopost">
     </div>
   </td>
  </tr>
</table>

jquery code

   $(document).ready(function () {
    $(".commentbox .btnSave").live("click", function () {
        alert("hii");
        var itemId = $(this).attr("id").split("-")[1]
        var txt = $(this).parent().find(".txtCmnt").val();
        alert(itemId + txt);
        $.post("Handler/Topic.ashx", { reply: txt, id: itemId }, function (data) {
            alert(data);
            $("#replytopost").html(data);
            //do whatever with the response
        })
    });
});

Whenever I click dynamically created button with class .btnSave ,response is printed but If I click second time response get replaced with new value.
When I click second time response must not override. Every time response must come in new div.
How to do this.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-02T05:26:27+00:00Added an answer on June 2, 2026 at 5:26 am

    The .html() function is intended to completely replace the contents of the element. If you want to add to it, use the .append() function instead.

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