一个Group By的聚合和求和

Question

我有一个视图，我从我的数据库中获取一些值和差异。我使用这段代码：

Select DATEDIFF(minute, 0, DATEADD(day, 0, t1.HorasEfe)) as Soma, t1.IDDiligencia
from DiligenciaSub t1
group by t1.IDDiligencia,t1.HorasEfe
order by t1.HorasEfe

我得到这个作为输出：

我需要什么：

总结IDDiligencia等于的Soma的值！

是否可以调整我的实际查询来执行此操作？

Answer 1

只需将HorasEfe从群组中删除，然后添加SUM：

Select SUM(DATEDIFF(minute, 0, DATEADD(day, 0, t1.HorasEfe))) as Soma, t1.IDDiligencia
from DiligenciaSub t1
group by t1.IDDiligencia

Answer 2

Select DATEDIFF(minute, 0, DATEADD(day, 0, t1.HorasEfe)) as Soma, t1.IDDiligencia
from DiligenciaSub t1
group by t1.IDDiligencia
order by DATEDIFF(minute, 0, DATEADD(day, 0, t1.HorasEfe))                                 

Answer 3

Select 
SUM(Select (DATEDIFF(minute, 0, DATEADD(day, 0, t1.HorasEfe))) from DiligenciaSub t1) as Soma,
t1.IDDiligencia
from DiligenciaSub t1
group by t1.IDDiligencia,t1.HorasEfe
order by t1.HorasEfe

Answer 4

从原始查询中删除sum和group by

将结果插入临时表#t

然后使用此查询

select sum(soma) as sum_soma,IDDiligencia as idd
from #t
group by IDDiligencia

您将获得以下结果