sql-server – 在SQL Server 2014中如何按时间间隔分组？

在sql SERVER中如何按时间间隔分组？

为了节省时间,我提出了这个解决方案,对我而言,它的效果非常好.您可以生成任何时基,然后按任意间隔分组.非常适合做时间加权平均值.如果有人有更好的方法,我很乐意听取你的意见.

小时

declare @startdate datetime2
declare @enddate datetime2
declare @interval int
set @startdate = '2017-01-01 00:00:00'
set @enddate   = '2017-01-31 00:00:00'
set @interval = 4 --Group by Every 4 hours
;with 
ALL_INTERVALS 
AS (
    SELECT TOP (DATEDIFF(HOUR,@startdate,@enddate))
    TIMES = DATEADD(HOUR,CONVERT(INT,ROW_NUMBER() OVER (ORDER BY s1.[object_id])),@startdate),1 AS VALUE
        FROM sys.all_objects AS s1
        CROSS JOIN
        sys.all_objects AS s2
)
select DATEADD(HOUR,((DATEDIFF(HOUR,TIMES)/@interval)*@interval),@startdate) AS TIMES,SUM(VALUE) AS TESTDATA 
from ALL_INTERVALS
group by DATEADD(HOUR,@startdate)
order by DATEADD(HOUR,@startdate)

分钟

注意.
你可以将你的间隔设置为60来实现小时,1440来实现几天….

declare @startdate datetime2
declare @enddate datetime2
declare @interval int
set @startdate = '2017-01-01 00:00:00'
set @enddate   = '2017-01-31 00:00:00'
set @interval = 7
;with 
ALL_INTERVALS 
AS (
    SELECT TOP (DATEDIFF(MINUTE,@enddate))
    TIMES = DATEADD(MINUTE,1 AS VALUE
        FROM sys.all_objects AS s1
        CROSS JOIN
        sys.all_objects AS s2
)
select DATEADD(MINUTE,((DATEDIFF(MINUTE,SUM(VALUE) AS TESTDATA 
from ALL_INTERVALS
group by DATEADD(MINUTE,@startdate)
order by DATEADD(MINUTE,@startdate)

解决方法

我认为你已经过度复杂了.
您可以使用GROUP BY(DATEDIFF,’2017-01-01′,TheDateTime)/ 30进行分组,每30分钟一次.当然,我选择的日期只是一个随机日期.您可以选择,如果你想要的,你的样本数据中的第一个(或最后一个)日期.
您还可以使用此技术获取任何时间段的每个时间间隔 – 只需将关键字MINUTE更改为您要使用的任何日期部分,并将intreval 30更改为您想要的任何时间间隔.

请考虑以下示例数据：

;WITH CTE AS 
(
    SELECT CAST('2017-01-01T00:00:00' as datetime) As TheDateTime,0 as rn
    UNION ALL
    SELECT DATEADD(MINUTE,1,TheDateTime),rn + 1
    FROM CTE
    WHERE rn < 60
)

SELECT TheDateTime,rn INTO #T
FROM CTE
OPTION(MAXRECURSION 0)

#T现在包含以下数据：

TheDateTime                 rn
2017-01-01 00:00:00.000     0
2017-01-01 00:01:00.000     1
2017-01-01 00:02:00.000     2
2017-01-01 00:03:00.000     3
...
2017-01-01 00:59:00.000     59
2017-01-01 01:00:00.000     60

要将最大rn分组30分钟,您只需要：

SELECT DATEDIFF(MINUTE,'2017-01-01',TheDateTime) / 30,MAX(rn)
FROM #T
GROUP BY DATEDIFF(MINUTE,TheDateTime) / 30

结果：

interval    max_rn
0           29
1           59
2           60