Question

这是我PHP Sum a value in while loop, but with conditions

的问题的延续

我有两个表要加入，1个是用户，1个是出席。表格出勤时的列isOt表示用户被上级授予加班费。

TABLE : attendance
++++++++++++++++++++++++++++++
id   userId   totalHours  isOt  dateRecorded
1    1        0745        0     02-06-2014
2    3        0845        1     07-06-2014
3    1        0945        1     12-06-2014

TABLE : user
+++++++++++++++++++++++
id   name  departmentId
1    John  2
2    Sean  2
3    Allan 2

在很多人的帮助下，提供此查询的解决方案确实帮助我为每个用户提供SUM totalHours

但是我做了一些额外的查询行，如下所示：

$query = "SELECT u.employeeName, u.id, a.isOt, a.dateRecorded,
SUM((CAST(totalHours AS UNSIGNED) % 100)/60 + FLOOR(CAST(totalHours AS UNSIGNED)/100))  grandTotal FROM user u RIGHT JOIN attendance a
ON u.id = a.userId
WHERE u.departmentId = 2
GROUP BY u.id HAVING dateRecorded >= '01-06-2014' <= '31-06-2014'";
$result = mysqli_query($con,$query);
while($row = mysqli_fetch_array($result)) 
{
   print $row['id'] . ' ' . $row['grandTotal'];
}

上面的查询成功地显示了两行输出，就像我想要的那样但是现在对于像我这样的初学者来说还有另一个棘手的挑战

我需要通过发言来获得加时赛的总时间

(if isOt == 1)
{
minus totalHours with 0800 (min working time)
and them sum up the balance and echo it for each user
}

任何帮助表示赞赏，我仍然是初学者并且尽力学习。提前谢谢你：）

Answer 1

我会在陈述

时使用一个案例

SELECT u.employeeName, 
       u.id, 
       a.isOt, 
       a.dateRecorded,
       Case When a.totalHours > 0800 Then 0800 Else totalHours End As NormalHours
       Case When a.totalHours > 0800 Then totalHours - 0800 Else 0 End As OTHours
FROM user u RIGHT JOIN 
     attendance a
ON u.id = a.userId
WHERE u.departmentId = 2
GROUP BY u.id 
HAVING dateRecorded >= '01-06-2014' <= '31-06-2014'

假设totalHours以数字格式存储。

PHP总时数与条件持续

1 个答案: