就像我做的那样
SELECT [Date]
FROM [FRIIB].[dbo].[ArchiveAnalog]
GROUP BY [Date]
如何指定分组时段?我使用的是MS SQL 2008。
我已经试过了，使用% 10和/ 10。
SELECT MIN([Date]) AS RecT, AVG(Value)
FROM [FRIIB].[dbo].[ArchiveAnalog]
GROUP BY (DATEPART(MINUTE, [Date]) / 10)
ORDER BY RecT
是否有可能使日期输出没有毫秒?
当前回答
declare @interval tinyint
set @interval = 30
select dateadd(minute,(datediff(minute,0,[DateInsert])/@interval)*@interval,0), sum(Value_Transaction)
from Transactions
group by dateadd(minute,(datediff(minute,0,[DateInsert])/@interval)*@interval,0)
其他回答
在T-SQL中，您可以:
SELECT [Date]
FROM [FRIIB].[dbo].[ArchiveAnalog]
GROUP BY [Date], DATEPART(hh, [Date])
or
Date (mi， [Date])
or
使用DATEPART(mi， [Date]) / 10(像Timothy建议的那样)
终于讲完了
GROUP BY
DATEPART(YEAR, DT.[Date]),
DATEPART(MONTH, DT.[Date]),
DATEPART(DAY, DT.[Date]),
DATEPART(HOUR, DT.[Date]),
(DATEPART(MINUTE, DT.[Date]) / 10)
select dateadd(minute, datediff(minute, 0, Date), 0),
sum(SnapShotValue)
FROM [FRIIB].[dbo].[ArchiveAnalog]
group by dateadd(minute, datediff(minute, 0, Date), 0)
declare @interval tinyint
set @interval = 30
select dateadd(minute,(datediff(minute,0,[DateInsert])/@interval)*@interval,0), sum(Value_Transaction)
from Transactions
group by dateadd(minute,(datediff(minute,0,[DateInsert])/@interval)*@interval,0)
作者最初给出的答案还不错。为了扩展这个想法，你可以这样做
group by datediff(minute, 0, [Date])/10
这样你就可以按更长的时间分组，比如60分钟，720分钟，也就是半天。
