申请休假的月度报告 [英] Monthly report for leave apply
问题描述
我有两个表,一个是tblemployee,具有员工名称,雇员id,另一个是tblleaves,具有Empid,Leave_Date,fromDate,toDate,Description.
I have two tables one is tblemployee having employee name, employee id and another table tblleaves having 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.
If employee choose one date leave it stores the date value to Leave_Date and if employee choose multiple dates it store value of from date and to date.
在输出页面中,我需要一个雇员姓名,请假天数和请假日期.休假日期的日期为Leave_date,FromDate和ToDate.
In the output page I want an employee name, Leave Days and Leave Dates. Leave Dates have dates from Leave_date, FromDate and 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($result->FirstName);?> <?php echo htmlentities($result->LastName);?></td>
<td><?php echo htmlentities($result->Leave_Days);
?></td>
<td><?php echo htmlentities($result->leave_dates);
?></td><?php $cnt++;}}}?>
</tr>
</tbody>
</table>
我希望页面输出是
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)
推荐答案
考虑以下粗略示例...
Consider the following crude example...
示例架构(从P.Salmon借用并改编):
Sample schema (borrowed and adapted from 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');
示例代码:
<?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);
?>
使用上面的模式,此代码输出...
Using the schema above, this code outputs...
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)
)
)
请注意,此结果将整天都视为工作日
Note that this result considers all days as working days
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