Question

到目前为止，我尝试用max和min减去这是正常的，除非有第三条记录，其中包含0。我还在一个案例陈述中试图寻找枯萎的预测或工作，并将其中的一个设置为否定，但这也不起作用。我需要按员工，rectype，datebegin和dateend分组的差异。 0 recs可以进入那里

     Table

     Employee RecType DateBegin DateEnd Hours


     Sample

     123,Forecast,1/1/2014,1/8/2014,5
     123,Forecast,1/9/2014,1/16/2014,7
     123,Forecast,1/9/2014,1/16/2014,0
     123,Worked,1/1/2014,1/8/2014,5
     123,Worked,1/9/2014,1/16/2014,4


     Output I'm looking for

     123,Difference,1/1/2014,1/8/2014,0
     123,Difference,1/9/2014,1/16/2014,3

Answer 1

假设您将拥有Employee，RecType，DateBegin，DateEnd的匹配记录对于预测和工作，您可以通过以下方式获得结果：

Set DateFormat  mdy
declare @a table(Employee int ,RecType varchar(20), DateBegin DateTime, DateEnd dateTime, Hours float)

insert into @a 
select 123,'Forecast','1/1/2014','1/8/2014',5
UNION
select 123,'Forecast','1/9/2014','1/16/2014',7
UNION
select 123,'Forecast','1/9/2014','1/16/2014',0
UNION
select 123,'Worked','1/1/2014','1/8/2014',5
UNION
select 123,'Worked','1/9/2014','1/16/2014',4

Select x.Employee,x.DateBegin,x.DateEnd,x.Hours-y.Hours as Hours
from
(
Select Employee,DateBegin,DateEnd,SUM(Hours) as  Hours
from @a where RecType='Forecast'
Group by Employee,DateBegin,DateEnd
) x
JOIN
(
Select Employee,DateBegin,DateEnd,SUM(Hours) as  Hours
from @a where RecType='Worked'
Group by Employee,DateBegin,DateEnd
) y
on x.Employee=y.Employee and x.DateBegin=y.DateBegin and x.DateEnd=y.DateEnd

希望在1列中减去数据

1 个答案: