您如何按任何基于时间的间隔进行分组？

Question

在 SQL SERVER 中，如何按任何基于时间的间隔进行分组？

为了节省时间，我想出了这个解决方案，对我来说效果很好。您可以生成任何时基，然后按任何间隔分组。非常适合做时间加权平均。如果有人有更好的方法，我很乐意听取您的意见。

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, @startdate,TIMES)/@interval)*@interval),@startdate) AS TIMES,SUM(VALUE) AS TESTDATA 
from ALL_INTERVALS
group by DATEADD(HOUR,((DATEDIFF(HOUR, @startdate,TIMES)/@interval)*@interval),@startdate)
order by DATEADD(HOUR,((DATEDIFF(HOUR, @startdate,TIMES)/@interval)*@interval),@startdate)

分钟

注意。 您可以将间隔设置为 60 以实现小时，1440 以实现天....

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,@startdate,@enddate))
    TIMES = DATEADD(MINUTE,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(MINUTE,((DATEDIFF(MINUTE, @startdate,TIMES)/@interval)*@interval),@startdate) AS TIMES,SUM(VALUE) AS TESTDATA 
from ALL_INTERVALS
group by DATEADD(MINUTE,((DATEDIFF(MINUTE, @startdate,TIMES)/@interval)*@interval),@startdate)
order by DATEADD(MINUTE,((DATEDIFF(MINUTE, @startdate,TIMES)/@interval)*@interval),@startdate)

Answer 1

我认为你把事情复杂化了。
您可以使用GROUP BY (DATEDIFF(MINUTE, '2017-01-01', TheDateTime) / 30 每 30 分钟进行一次分组。当然，我选择的日期只是一个随机日期。如果需要，您可以选择示例数据中的第一个（或最后一个）日期。
您还可以使用此技术获取任何时间部分的每个间隔 - 只需将关键字 MINUTE 更改为您想要使用的任何日期部分，并将 intreval 30 更改为您想要的任何间隔。

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

要获得按 30 分钟分组的最大 rn，您只需要这样：

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

结果：

interval    max_rn
0           29
1           59
2           60