Ежемесячный отчет об отпуске применяется

Question

У меня есть две таблицы, одна из которых tblemployee с именем сотрудника, идентификатором сотрудника, а другая таблица tblleaves с empid, Leave_Date, fromDate, toDate, Description.

Если сотрудник выбирает одну дату, оставьте ее для сохранения значения даты в Leave_Dateи если сотрудник выбирает несколько дат, он сохраняет значение от даты и до даты.

На странице вывода я хочу указать имя сотрудника, Leave Days and Leave Dates.Даты выхода имеют даты от Leave_date, FromDate и ToDate.

     <?php 


        if(isset($_POST['apply'])){

        $ym=$_POST['month'];
        list($Year, $Month) = explode("-", "$ym", 2);

        $sql = "SELECT 
       tblemployees.FirstName,
       tblemployees.LastName,
       count(tblleaves.empid) as Leave_Days,
       GROUP_CONCAT( tblleaves.Leave_Date SEPARATOR ', ' ) AS leave_dates
    FROM
       tblleaves
       JOIN tblemployees
          ON tblleaves.empid = tblemployees.id
    WHERE YEAR(Leave_Date) = $Year
       AND MONTH(Leave_Date) = $Month
    GROUP BY tblemployees.EmpId";

        $query = $dbh -> prepare($sql);
        $query->execute();
        $results=$query->fetchAll(PDO::FETCH_OBJ);

        $cnt=1;
        if($query->rowCount() > 0)
        {
        foreach($results as $result)
        {               ?>  
          <tr>
            <td> <?php echo htmlentities($cnt);?></td>
              <td><?php echo htmlentities($result->FirstName);?>&nbsp;<?php echo htmlentities($result->LastName);?></td>
               <td><?php echo htmlentities($result->Leave_Days);
     ?></td>
<td><?php echo htmlentities($result->leave_dates);

    ?></td><?php $cnt++;}}}?>
</tr>
</tbody>
</table>

Я хочу вывод страницы

employee name     Leave Days      Leave Dates 
KrishnanR            3              12-06-2019, 13-06-2019, 14-06-2019
                                     (FromDate and ToDate)
PrakashR             1              12-06-2019
                                     (Leave_Date)

SelvaK               3        12-06-2019,13-06-2019&14-06-2019,|  14-06-2019
                                      (FromDate and ToDate) |  (Leave_Date)

Answer 1

Рассмотрим следующий грубый пример ...

Пример схемы (заимствовано и адаптировано из P.Salmon):

DROP TABLE IF EXISTS employee_leave;
CREATE TABLE employee_leave
(leave_id SERIAL PRIMARY KEY
,employee_id INT NOT NULL
,leave_from DATE NOT NULL
,leave_to DATE NOT NULL
);

INSERT INTO employee_leave
(employee_id
,leave_from
,leave_to
) VALUES
(11,'2019-05-30','2019-06-02'),
(11,'2019-06-05','2019-06-05'),
(11,'2019-06-06','2019-06-06'),
(11,'2019-06-30','2019-07-11'),
(12,'2019-05-30','2019-07-11'),
(13,'2019-05-11','2019-05-12');

Пример кода:

<?php

include('path/to/connection/stateme.nts');

$query = "
SELECT employee_id
     , leave_from
     , leave_to
     , datediff(leave_to,leave_from)+1 days
  FROM employee_leave
 ORDER
    BY employee_id
     , leave_from; -- ORDER BY is not strictly necessary, as the ordering can be done in presentation code.
";

$result = mysqli_query($conn,$query);

$array = array();

while($row = mysqli_fetch_assoc($result)){
  $array[] = $row;
}

$new_array = array();

foreach($array as $k=>$v){
  if($v['days']>1){
    $days = ' days'; } else { $days = ' day'; }
  $new_array[$v['employee_id']][] = $v['leave_from'].' - '.$v['leave_to'].' ('.$v['days'].$days.')';
}

print_r($new_array);
?>

Используя схему выше, этот код выводит ...

Array
(
    [11] => Array
        (
            [0] => 2019-05-30 - 2019-06-02 (4 days)
            [1] => 2019-06-05 - 2019-06-05 (1 day)
            [2] => 2019-06-06 - 2019-06-06 (1 day)
            [3] => 2019-06-30 - 2019-07-11 (12 days)
        )

    [12] => Array
        (
            [0] => 2019-05-30 - 2019-07-11 (43 days)
        )

    [13] => Array
        (
            [0] => 2019-05-11 - 2019-05-12 (2 days)
        )

)

Обратите внимание, что этот результат рассматривает все дни как рабочие дни

Answer 2

Обновил ваш код на основе ваших требований, не проверял его.Давайте попробуем.

<?php 
if(isset($_POST['apply'])){

$ym=$_POST['month'];
list($Year, $Month) = explode("-", "$ym", 2);

$sql = "SELECT tblemployees.FirstName, tblemployees.LastName, count(tblleaves.empid) as Leave_Days,
               IF( t2.FromDate IS NOT NULL AND t2.ToDate IS NOT NULL , 
                    CONCAT(t2.FromDate, ',', t2.ToDate, ',' ,t2.Leave_Date), 
                    t2.Leave_Date) as 'Leave_Dates'
          FROM tblleaves JOIN tblemployees ON tblleaves.empid = tblemployees.id
         WHERE YEAR(Leave_Date) = $Year AND MONTH(Leave_Date) = $Month
      GROUP BY tblemployees.EmpId";

$query = $dbh -> prepare($sql);
$query->execute();
$results=$query->fetchAll(PDO::FETCH_OBJ);

$cnt=1;
if($query->rowCount() > 0)
{
foreach($results as $result)
{               
$leavedates = explode(',', $result->Leave_Dates);
$period = new DatePeriod(
     new DateTime($leavedates[0]),
     new DateInterval('P1D'),
     new DateTime($leavedates[1])
);
$listofleaves = [];
foreach ($period as $key => $value) {
    $listofleaves[] = $value->format('Y-m-d');       
}
$listofleaves[] = $leavedates[2];
    ?>  
  <tr>
    <td> <?php echo htmlentities($cnt);?></td>
      <td><?php echo htmlentities($result->FirstName);?>&nbsp;<?php echo htmlentities($result->LastName);?></td>
       <td><?php echo htmlentities($result->Leave_Days);
?></td>

<td><?php echo htmlentities(implode(',' , $listofleaves);


?></td><?php $cnt++;}}}?>
</tr>
</tbody>
</table>