Question

我试图根据员工迟到的人数来扣除工资。所以我的条件是：如果一名员工连续3天迟到（周五和周六除外 - 每周假期），那么将扣除一天的工资。示例如下，这就是我现在对给定查询的内容：

Name  PunchDate             Attendance   PerDaySal          Status    
John  2017-05-12 00:00:00.000   W        461.538461538462   Weekly
John  2017-05-13 00:00:00.000   W        461.538461538462   Weekly
John  2017-05-14 09:00:00.000   P        461.538461538462   On Time
John  2017-05-15 09:16:00.000   P        461.538461538462   Late
John  2017-05-16 09:18:00.000   P        461.538461538462   Late
John  2017-05-17 09:20:00.000   P        0                  Late -- On 3rd consecutive day
John  2017-05-18 09:26:00.000   P        461.538461538462   Late
John  2017-05-19 00:00:00.000   W        461.538461538462   Weekly
John  2017-05-20 00:00:00.000   W        461.538461538462   Weekly
John  2017-05-21 09:18:00.000   P        461.538461538462   Late
John  2017-05-22 09:28:00.000   P        0                  Late -- On the next 3rd consecutive day (Possibly 6th)           
John  2017-05-23 09:28:00.000   P        461.538461538462   Late

在上文中，当连续第三天的出勤延迟计数时，则进行扣除。再次，让我提醒一下，这必须在星期五和星期六之外排除。我可以显示已计算的晚数和每周假期，但不确定如何继续扣除。我打算做一张桌子＆＃39; Deduction＆＃39;如果我没有弄错的话，为了扣除目的，将日期存储如下：

ID    Days  
1     3 ---- On 3rd late count, PerDaySal 0
2     6 ---- On 6th late count, PerDaySal 0 and so on

但是对于特定月份（即使每3天），如何验证逻辑？我尝试了一个愚蠢的（虽然没有解决方案）：

SELECT k.NAME AS Employee, m.PunchDate,

(CASE WHEN o.WeekName = 'Friday' OR o.WeekName = 'Saturday' THEN 'W' ELSE m.Status END) 
AS Attendance,

(CASE WHEN o.WeekName = 'Friday' OR o.WeekName = 'Saturday' THEN 0 ELSE (k.SALARY / 26) END) 
AS PerDaySal,

(CASE WHEN CONVERT(CHAR(8), m.PunchDate, 108) > '09:15:00' THEN 'Late'  
 WHEN CONVERT(CHAR(8), m.PunchDate, 108) >= '09:00:00' AND CONVERT(CHAR(8), m.PunchDate, 108) <= '09:15:00' THEN 'On Time'
 WHEN o.WeekName = 'Friday' OR o.WeekName = 'Saturday' AND CONVERT(CHAR(8), m.PunchDate, 108) <= '00:00:00' THEN 'Weekly' END) AS Status

--The silly one - (CASE WHEN COUNT(CONVERT(CHAR(8), m.PunchDate, 108) > '09:15:00') > 3 THEN 0 ELSE (k.SALARY / 26) END) AS PerDaySal

FROM @Attendances m INNER JOIN @Employee k ON k.ID = m.EmpId 
LEFT JOIN @Weekly o ON o.WeekDate = m.PunchDate
GROUP BY k.NAME, m.PunchDate, m.Status, k.Salary, o.WeekName

以下是包含示例数据的表结构：

declare @Attendances table (Id int not null identity(1,1) primary key,EmpId int,PunchDate datetime,Status nvarchar(4));
insert into @Attendances([EmpId],[PunchDate],[Status]) values
 (2,cast(0x0000A77200000000 as datetime),N'A')
,(2,cast(0x0000A77100000000 as datetime),N'A')
,(2,cast(0x0000A776009A5BA0 as datetime),N'P')
,(2,cast(0x0000A775009450C0 as datetime),N'P')
,(2,cast(0x0000A77400982920 as datetime),N'P')
,(2,cast(0x0000A773009450C0 as datetime),N'P');

declare @Employee table (ID int not null primary key,NAME nvarchar(50),ADDRESS nvarchar(max),SALARY float);
insert @Employee([ID], [NAME], [ADDRESS], [SALARY]) values
 (1, N'John', N'Germany', 12000)
,(2, N'Jack', N'France', 14000);

declare @Weekly table (WeekID int not null primary key,WeekNAME nvarchar(20),WeekDate datetime,Status nvarchar(10));
insert @Weekly([WeekID], [WeekName], [WeekDate], [Status]) values
 (1, N'Friday', CAST(0x0000A77100000000 AS DateTime), N'W')
,(2, N'Saturday', CAST(0x0000A77200000000 AS DateTime), N'W');

Answer 1

使用您的示例表而不是多个表，使用公共表表达式和具有case表达式的窗口函数，如果在星期五或星期六按照问题规范，则不计算一天的延迟时间写这篇文章。

;with cte as (
select * 
  , rn = row_number() over (partition by Name order by punchDate)
  , DayName = datename(weekday,punchdate)
  , LateDays = (select sum(
    case when datename(weekday,punchdate) not in ('Friday','Saturday') and [Status]='Late'
        then 1 else 0 end
  )
  from t i
  where i.Name = t.Name and i.PunchDate <= t.PunchDate
  )
from t
)
select 
    cte.Name
  , cte.PunchDate
  , cte.DayName
  , cte.Attendance
  , PerDaysSal = case when z.rn is not null then 0 
      else cte.PerDaySal end
  , cte.Status
  , cte.rn
  , cte.LateDays
from cte
  left join (
    select i.Name, rn=min(i.rn)
    from cte i
    where i.LateDays > 0
      and i.LateDays % 3 = 0
    group by Name, LateDays
      ) z
  on cte.Name = z.Name
 and cte.rn = z.rn

rextester演示：http://rextester.com/ETOI35350

返回：

+------+---------------------+-----------+------------+---------------+---------+----+----------+
| Name |      PunchDate      |  DayName  | Attendance |  PerDaysSal   | Status  | rn | LateDays |
+------+---------------------+-----------+------------+---------------+---------+----+----------+
| John | 12.05.2017 00:00:00 | Friday    | W          | 461,538461538 | Weekly  |  1 |        0 |
| John | 13.05.2017 00:00:00 | Saturday  | W          | 461,538461538 | Weekly  |  2 |        0 |
| John | 14.05.2017 00:00:00 | Sunday    | P          | 461,538461538 | On Time |  3 |        0 |
| John | 15.05.2017 00:00:00 | Monday    | P          | 461,538461538 | Late    |  4 |        1 |
| John | 16.05.2017 00:00:00 | Tuesday   | P          | 461,538461538 | Late    |  5 |        2 |
| John | 17.05.2017 00:00:00 | Wednesday | P          | 0,000000000   | Late    |  6 |        3 |
| John | 18.05.2017 00:00:00 | Thursday  | P          | 461,538461538 | Late    |  7 |        4 |
| John | 19.05.2017 00:00:00 | Friday    | W          | 461,538461538 | Weekly  |  8 |        4 |
| John | 20.05.2017 00:00:00 | Saturday  | P          | 461,538461538 | Late    |  9 |        4 |
| John | 21.05.2017 00:00:00 | Sunday    | P          | 461,538461538 | On Time | 10 |        4 |
| John | 22.05.2017 00:00:00 | Monday    | P          | 461,000000000 | Late    | 11 |        5 |
| John | 23.05.2017 00:00:00 | Tuesday   | P          | 0,000000000   | Late    | 12 |        6 |
+------+---------------------+-----------+------------+---------------+---------+----+----------+

在使用实际表时，应使用Name对EmpId进行分区，而不是按id进行分区。您还可以使用出席时的row_number()代替用于对行进行编号的If GetProcessTimes(lProcessID, lcreationtime.Value, lexittime.Value, lkerneltime.Value, lusertime.Value) Then。

条件下的员工工资扣除

1 个答案: