Show the error when sending a form

I have a very strange case in sql server manager

I have been ordered to work on a table that shows all the user fields in an editable form (that each data is a input )

Everything is fine but when I try to add a fifth columna with campos of formulario do not edit anything

and do not send me error .... already try with sqlsrv_error for my query

  

sqlsrv_query ($ conn, $ query) or die (print_r (sqlsrv_errors (), true)); and he does not show me anything

this is my formulario :

        $sql = "SELECT * FROM usuario ORDER BY Id_usuario ASC";
        $result = sqlsrv_query($conn, $sql);
   <form name='form_update' method='post' action='periodo_update.php'>
<?php
    while ($usuario = sqlsrv_fetch_array($result)) {
  echo '<tr>';
  echo "<td>{$usuario['idusuario']}<input type='hidden' name='idusuario[$i]' value='{$usuario['idusuario']}' /></td>";
  echo "<td>$NoEmpleado</td>";
  echo "<td><img src='$foto'  width='56'></td>";
  echo "<td>$us_nombre_real</td>";

  //echo "<td>$ruta</td>";
  echo "<td>$puesto_descripcion</td>";


echo "<td><input type='text' class='form-control' onkeypress='return /\d/.test(String.fromCharCode(((event||window.event).which||(event||window.event).which)));' name='pago[$i]' value='{$usuario['pago']}' /></td>";

echo "<td><input type='text' class='form-control' onkeypress='return /\d/.test(String.fromCharCode(((event||window.event).which||(event||window.event).which)));' name='sueldos[$i]' value='{$usuario['sueldos']}' /></td>";

echo "<td><input type='text' class='form-control' onkeypress='return /\d/.test(String.fromCharCode(((event||window.event).which||(event||window.event).which)));' name='dias_trabajados[$i]' value='{$usuario['dias_trabajados']}' /></td>";

echo "<td><input type='text' class='form-control' onkeypress='return /\d/.test(String.fromCharCode(((event||window.event).which||(event||window.event).which)));' name='dias_adicionales[$i]' value='{$usuario['dias_adicionales']}' /></td>";

  echo '</tr>';
        ++$i;
    }
</tr>
   <td></td>
   <td></td>
   <td></td>   
<td><input type='submit' value='Update' class='btn btn-info'/></td>
   </tr>
   </form>
------update

$size = count($_POST['pago']);
    $i = 0;
while ($i < $size) {
    $pago1 = $_POST['pago'][$i];
    $sueldos1 = $_POST['sueldos'][$i];
    $dias_trabajados1 = $_POST['dias_trabajados'][$i];
    $dias_adicionales1 = $_POST['dias_adicionales'][$i];
    $idusuario = $_POST['idusuario'][$i];

    $query = "UPDATE usuario SET pago = '$pago1', sueldos = '$sueldos1', dias_trabajados ='$dias_trabajados1', dias_adicionales ='$dias_adicionales1'  WHERE idusuario = '$idusuario' ";



    sqlsrv_query($conn, $query) or die ("Error in query: $query");
    //echo "$pago<br /><br /><em>Updated!</em><br /><br />";
    ++$i;
}

in advance thank you very much