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

MySql:

GROUP BY
DATE(`your_date_field`),
HOUR(`your_date_field`),
FLOOR( MINUTE(`your_date_field`) / 10);

其他回答

如果你想实际显示日期,有一个变量分组,并能够指定大于60分钟的时间框架:

DECLARE @minutes int
SET @minutes = 90

SELECT
    DATEADD(MINUTE, DATEDIFF(MINUTE, 0, [Date]) / @minutes * @minutes, 0) as [Date],
    AVG([Value]) as [Value]
FROM [FRIIB].[dbo].[ArchiveAnalog]
GROUP BY
    DATEDIFF(MINUTE, 0, [Date]) / @minutes

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)

应该是这样的

select timeslot, count(*)  
from 
    (
    select datepart('hh', date) timeslot
    FROM [FRIIB].[dbo].[ArchiveAnalog]  
    ) 
group by timeslot

(不是100%确定语法-我更喜欢Oracle的那种人)

在Oracle中:

SELECT timeslot, COUNT(*) 
FROM
(  
    SELECT to_char(l_time, 'YYYY-MM-DD hh24') timeslot 
    FROM
    (
        SELECT l_time FROM mytab  
    )  
) GROUP BY timeslot 

间隔10分钟,你会

GROUP BY (DATEPART(MINUTE, [Date]) / 10)

正如tzup和Pieter888已经提到的……做一个小时的间隔

GROUP BY DATEPART(HOUR, [Date])
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)