ขอคำแนะนำ Add/delete row ในตาราง โดยใช้ JQuery มี Select Option ที่ดึงข้อมูลมาจาก DB

$(document).ready(function() {
    $(".btnAdd").click(function() {
      $("#druglist tbody").append( 
        "<tr>"+ 
          "<td> <select required aria-required=\"true\" name='drugname' id='drugname'>"+
              "<option value=\"\">-</option>"+
              <?php
                $sql = "SELECT Drug_ID, Drug_Name + '-' + Drug_Type + ':'  as druginfo, ​Drug_Qnt_Unit, Drug_Unit FROM Drug";
                $stmt = sqlsrv_query($conn, $sql, array(), array( "Scrollable" => SQLSRV_CURSOR_KEYSET ));
                while( $result = sqlsrv_fetch_array( $stmt, SQLSRV_FETCH_ASSOC) ) { ?>+
                "<option value='<?php echo $result['Drug_ID'];?>'><?php echo $result['druginfo']." ".$result['Drug_Qnt_Unit']." ".$result['Drug_Unit']; ?></option>"<?php } ?>
            +"</select></td>"+ 
          "<td><input type='text'/></td>"+ 
          "<td><img src='../img/save.png' class='btnSave'><img src='../img/remove.png' class='btnDelete'/></td>"+ 
        "</tr>"); 
      $(".btnSave").bind("click", Save); 
      $(".btnDelete").bind("click", Delete); 
    });
  });

  function Save(){
      var par = $(this).parent().parent(); //tr 
      var tdDrugInfo = par.children("td:nth-child(1)"); 
      var tdQnt = par.children("td:nth-child(2)"); 
      var tdButtons = par.children("td:nth-child(3)"); 
      tdDrugInfo.html(tdDrugInfo.children("select").val()); 
      tdQnt.html(tdQnt.children("input[type=text]").val());
      tdButtons.html("<img src='../img/remove.png' class='btnDelete'/>"); 
      $(".btnDelete").bind("click", Delete);
  };

  function Delete(){ 
    var par = $(this).parent().parent(); //tr 
    par.remove(); 
  }; 
});