Question

我有返回小时数据的查询。但是我想从该查询中获取每日数据，因此每天的所有每小时数据都将平均为每日数据。

declare @Days int

set @Days = -1

select      

dateadd(hour,datepart(hour,Timestamp),cast(CAST((Timestamp) as date) as datetime)) as [Time]  
,[value] 



from  [Employee]

where dateadd(hour,datepart(hour,Timestamp),cast(CAST((Timestamp) as date) as datetime)) >= CONVERT(date, DATEADD(DAY, @Days, GETDATE()))

Answer 1

假设您要对其他列进行平均和按日期分组，则可以尝试以下操作：

DECLARE @Days int = -1;
SELECT 
      CAST(Timestamp AS date) AS date
    , AVG(Value) AS Value
FROM  [Employee]
WHERE Timestamp >= DATEADD(day, @Days, CAST(GETDATE() AS date))
GROUP BY CAST(Timestamp AS date)
ORDER BY date;

请注意，重构的WHERE子句避免将函数应用于列值。这样可以有效地使用Timestamp上的索引（可表达）。

每小时获取每日数据

1 个答案: