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Cómo mostrar datos en la fila de la tabla después de seleccionar datos de otra fila

Necesito mostrar el código del artículo después de seleccionar la descripción en cada fila. la primera fila funciona bien, pero después de agregar las filas 2 y 3, no funciona.

 <script> $(document).ready(function(){ /*add new row in table after clicking the addrow button*/ var count = $(".itemRow").length; $(document).on('click', '#addRows', function() { count++; var htmlRows = ''; htmlRows += '<tr>'; htmlRows += '<td><div class="custom-control custom-checkbox"><input type="checkbox" class="custom-control-input itemRow" id="itemRow_'+count+'"> <label class="custom-control-label" for="itemRow_'+count+'"></label> </div></td>'; htmlRows += '<td><select name="itemcode[]" id="itemcode" class="form-control form-control-sm loadIcode" autocomplete="off" disabled style="width:50%"><option value=""></option></select></td>'; htmlRows += '<td><select name="description[]" id="itemname" class="form-control form-control-sm loadIcode" autocomplete="off" style="width:50%"><option value="" disabled selected>Select Item Description<?php echo fill_description($connect);?></option></select></td>'; htmlRows += '</tr>'; $('#invoiceItem').append(htmlRows); $('.loadIcode').change(function(){ if($(this).val() != '') { var loadIcode = $(this).attr("id"); var query = $(this).val(); var result = ''; if(loadIcode == "itemname") { result = 'itemcode'; } $.ajax({ url:"result.php", method:"POST", data:{loadIcode:loadIcode, query:query}, success:function(data){ $('#'+result).html(data); } }) } }); }); }); </script>

ingrese la descripción de la imagen aquí

about 4 years ago · Juan Pablo Isaza
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