Question

我有两个表，一个表有员工姓名，员工ID，另一个表是tblleaves带有empid，Leave_Date，fromDate，toDate，Description。

如果员工选择一个日期，则将日期值存储到Leave_Date；如果员工选择多个日期，则将日期和日期存储值。

我想要一个雇员姓名的输出页面，请假天数和请假日期。休假日期的日期为Leave_date，FromDate和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>

我想要的输出：

  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

无法帮助您使用php，但是mysql查询可以解决问题。

DROP TABLE IF EXISTS T;
CREATE TABLE T
(ID INT AUTO_INCREMENT PRIMARY KEY,NAME VARCHAR(10),LEAVE_DATE DATE, LEAVE_FROM DATE, LEAVE_TO DATE);
INSERT INTO T (NAME,LEAVE_DATE,LEAVE_FROM,LEAVE_TO) VALUES
('A','2019-05-30','2019-05-30','2019-06-02'),
('A','2019-06-05',NULL,NULL),
('A','2019-06-06',NULL,NULL),
('A','2019-06-30','2019-06-30','2019-07-11'),
('B','2019-05-30','2019-05-30','2019-07-11'),
('C','2019-05-11','2019-05-11','2019-05-12')
;

SELECT NAME,
         SUM(CASE WHEN LEAVE_FROM IS NULL THEN 1 
                ELSE DATEDIFF(IF(LEAVE_TO > '2019-06-30','2019-06-30',LEAVE_TO),IF(LEAVE_FROM < '2019-06-01','2019-06-01',LEAVE_FROM)) + 1 
              END) LEAVEDAYS,
         GROUP_CONCAT(CASE WHEN LEAVE_FROM IS NULL THEN LEAVE_DATE 
                            ELSE CONCAT(IF(LEAVE_FROM < '2019-06-01','<<2019-06-01',LEAVE_FROM),' & ',IF(LEAVE_TO > '2019-06-30','2019-06-30>>',LEAVE_TO)) 
                            END) AS LEAVEDATES 
FROM T
WHERE (LEAVE_DATE BETWEEN '2019-06-01' AND '2019-06-30') OR
        (LEAVE_FROM BETWEEN '2019-06-01' AND '2019-06-30') OR
        (LEAVE_TO   BETWEEN '2019-06-01' AND '2019-06-30') OR
        (LEAVE_FROM < '2019-06-01' AND LEAVE_TO > '2019-06-30')
GROUP BY NAME
ORDER BY ID;

请注意请假日期的上下调整，以确保我们只计算该期间的日期。还请注意，我在休假日期添加了指示符（<<或>>），以指示休假日期开始于所需的期间开始日期之前或延长至所需的期间结束日期之后。

+------+-----------+---------------------------------------------------------------------------+
| NAME | LEAVEDAYS | LEAVEDATES                                                                |
+------+-----------+---------------------------------------------------------------------------+
| A    |         5 | <<2019-06-01 & 2019-06-02,2019-06-05,2019-06-06,2019-06-30 & 2019-06-30>> |
| B    |        30 | <<2019-06-01 & 2019-06-30>>                                               |
+------+-----------+---------------------------------------------------------------------------+
2 rows in set (0.08 sec)

显示“休假日期”和“开始日期”至“

1 个答案: