Question

我想对工作时间2 + 3 + 4 = 9求和。以下是我尝试过的查询，但求和不起作用

MyQuery：

;with cte as
(
Select t.emp_reader_id as empId, 
cast(max(dt) as date) as Belongs_to,
Cast(dateadd(ss,Datediff(second,max(iif(EVENTID = 0, DT, null)),max(iif(EVENTID = 1, DT, null))),0)As Time(0)) as Worked,
case 
when LOWER(DATENAME(dw, t.att_date))='friday' 
then 
Cast(dateadd(ss,Datediff(second,max(iif(EVENTID = 0, DT, null)),max(iif(EVENTID = 1, DT, null))),0)As Time(0))
else '00:00'
end  as Friday_Worked,
max(iif(EVENTID = 0, dt, null)) as EntryTime, 
max(iif(EVENTID = 1, dt, null)) as ExitTime
from (  
   select
   Trnevents.emp_reader_id,Trnevents.EVENTID,Trnevents.DT,Trnevents.belongs_to,Daily_attendance_data.att_date,
    grp = sum(iif(Trnevents.EVENTID = 0, 1, 0) ) over (partition by Trnevents.emp_reader_id order by Trnevents.DT)
        --*, grp = sum(iif(Trnevents.EVENTID = 0, 1, 0) ) over (partition by Trnevents.emp_reader_id order by Trnevents.DT)
    from
        Trnevents 
        inner join Daily_attendance_data 
        on Daily_attendance_data.emp_reader_id=Trnevents.emp_reader_id --and dd.att_date=tt.Belongs_to
        where 
        Trnevents.emp_reader_id=9999 and 
        (CONVERT(VARCHAR(26),Daily_attendance_data.att_date, 23) >=CONVERT(VARCHAR(26), '2018-10-25', 23)
         and CONVERT(VARCHAR(26), Daily_attendance_data.att_date, 23) <=CONVERT(VARCHAR(26), '2018-10-25', 23))  and Trnevents.DT between Daily_attendance_data.in_time and Daily_attendance_data.out_time
) t 
  group by t.emp_reader_id,t.grp,t.att_date
  --order by t.grp asc
 ) 
 select empid,B.emp_name,Belongs_to,Worked,Friday_Worked,Entrytime,ExitTime,(DATEPART(HOUR,[worked])) as 'TotalHRS'
  from cte
  inner join employee B on cte.empid=B.emp_reader_id
 group by empid,Belongs_to,worked,friday_worked,entrytime,exittime,B.emp_name
 order by 1

我得到的输出是可以的，但是我需要为该特定ID或日期加总一小时的总和输出：

empid   emp_name Belongs_to Worked  Friday_Worked   Entrytime   ExitTime        TotalHRS
9999    Test    2018-10-25  02:00:00    00:00:00    2018-10-25 18:00:00.000 2018-10-25 20:00:00.000 2
9999    Test    2018-10-26  03:00:00    00:00:00    2018-10-25 22:00:00.000 2018-10-26 01:00:00.000 3
9999    Test    2018-10-26  04:00:00    00:00:00    2018-10-26 02:00:00.000 2018-10-26 06:00:00.000 4

Answer 1

在上一个查询中

您的GROUP BY没有意义。基本上，您GROUP BY子句中的每一列都是SELECT，这里没有任何聚合函数

您可以如下删除GROUP BY

select empid, B.emp_name, Belongs_to, Worked, Friday_Worked, Entrytime, ExitTime, 
       DATEPART(HOUR, [worked]) as 'TotalHRS'
from   cte
       inner join employee B on cte.empid = B.emp_reader_id
order by 1

对于所需的额外列，请使用SUM () OVER ( PARTITION BY )窗口函数

select empid, B.emp_name, Belongs_to, Worked, Friday_Worked, Entrytime, ExitTime, 
       DATEPART(HOUR, [worked]) as TotalHRS,
       SUM ( DATEPART(HOUR, [worked]) ) OVER (PARTITION BY empid) as SumTotalHRS
from   cte
       inner join employee B on cte.empid = B.emp_reader_id
order by 1

Answer 2

您可以尝试使用标量子查询

select empid,B.emp_name,Belongs_to,Worked,Friday_Worked,Entrytime,ExitTime, 
(DATEPART(HOUR,[worked])) as 'TotalHRS',
(select sum((DATEPART(HOUR,[worked]))) from cte b where cte.empid=b.empid) as 
 totalhoursum
from cte 
inner join employee B on cte.empid=B.emp_reader_id
group by empid,Belongs_to,worked,friday_worked,entrytime,exittime,B.emp_name
order by 1

如何计算CTE中的总小时数

2 个答案: