以 15 秒为间隔聚合（组）时间 // Microsoft SQL Server 2019答案

【问题标题】：Aggregating (group) time by 15 seconds intervals // Microsoft SQL Server 2019以 15 秒为间隔聚合（组）时间 // Microsoft SQL Server 2019
【发布时间】：2020-04-13 09:17:08
【问题描述】：

我是 Microsoft SQL Server 2019 的新手。目前我正在尝试以 15 秒的间隔聚合 Timestamp（键入 datetime）。

想要的结果：

Timestamp               Timestamp2
2019-01-01 04:00:00.487 2019-01-01 04:00:15.000
2019-01-01 04:00:01.487 2019-01-01 04:00:15.000
2019-01-01 04:00:02.487 2019-01-01 04:00:15.000
2019-01-01 04:00:03.487 2019-01-01 04:00:15.000
2019-01-01 04:00:04.487 2019-01-01 04:00:15.000
2019-01-01 04:00:05.487 2019-01-01 04:00:15.000
2019-01-01 04:00:06.487 2019-01-01 04:00:15.000
2019-01-01 04:00:07.487 2019-01-01 04:00:15.000
2019-01-01 04:00:08.487 2019-01-01 04:00:15.000
2019-01-01 04:00:09.487 2019-01-01 04:00:15.000
2019-01-01 04:00:10.487 2019-01-01 04:00:15.000
2019-01-01 04:00:11.487 2019-01-01 04:00:15.000
2019-01-01 04:00:12.487 2019-01-01 04:00:15.000
2019-01-01 04:00:13.487 2019-01-01 04:00:15.000
2019-01-01 04:00:14.487 2019-01-01 04:00:15.000
2019-01-01 04:00:15.487 2019-01-01 04:00:30.000
2019-01-01 04:00:16.487 2019-01-01 04:00:30.000
2019-01-01 04:00:17.487 2019-01-01 04:00:30.000
2019-01-01 04:00:18.487 2019-01-01 04:00:30.000
2019-01-01 04:00:19.487 2019-01-01 04:00:30.000

我尝试转换“1 分钟间隔” 来自

Select 
Timestamp, 
dateadd(MINUTE, 1+datediff(MINUTE, 0, [Timestamp]), 0) AS Timestamp2 
FROM tags.dbo.jan
where ValueID = '349' and Timestamp < '2019-01-01 04:02:00.000'

到

Select   
Timestamp, 
dateadd(second, 15+datediff(second, 0, [Timestamp]), 0) AS Timestamp2 
FROM tags.dbo.jan
where ValueID = '349' and Timestamp < '2019-01-01 04:02:00.000'

遇到一个问题

The datediff function resulted in an overflow. The number of dateparts separating two date/time instances is too large. Try to use datediff with a less precise datepart.

然后我尝试使用datediff_big

Select   
Timestamp, 
dateadd(second, 15+datediff_big(second, 0, [Timestamp]), 0) AS Timestamp2 
FROM tags.dbo.jan
where ValueID = '349' and Timestamp < '2019-01-01 04:02:00.000'

得到错误

Arithmetic overflow error converting expression to data type int.

我尝试过使用

Select Timestamp,
       datetimefromparts(year(Timestamp), month(Timestamp), day(Timestamp),
                         datepart(hour, Timestamp),
                         datepart(minute, Timestamp),
                         (ceiling(datepart(second, Timestamp)) / 15) * 15,
                         0
                        ) as timestamp2
FROM jan.dbo.jan
where ValueID = '349' and
      Timestamp < '2019-01-01 04:02:00.000';

但我有这些结果

2019-01-01 04:00:12.487 2019-01-01 04:00:00.000
2019-01-01 04:00:13.487 2019-01-01 04:00:00.000
2019-01-01 04:00:14.487 2019-01-01 04:00:00.000
2019-01-01 04:00:15.487 2019-01-01 04:00:15.000
2019-01-01 04:00:16.487 2019-01-01 04:00:15.000
2019-01-01 04:00:17.487 2019-01-01 04:00:15.000

而不是想要的

2019-01-01 04:00:12.487 2019-01-01 04:00:15.000
2019-01-01 04:00:13.487 2019-01-01 04:00:15.000
2019-01-01 04:00:14.487 2019-01-01 04:00:15.000
2019-01-01 04:00:15.487 2019-01-01 04:00:30.000
2019-01-01 04:00:16.487 2019-01-01 04:00:30.000
2019-01-01 04:00:17.487 2019-01-01 04:00:30.000

应该从前15秒组开始

【问题讨论】：

此解决方案可能对您有所帮助.. stackoverflow.com/questions/830792/…
它有效，但不是我想要的方式

标签： sql sql-server date datetime aggregate

【解决方案1】：

你可以使用datetimefromparts():

Select Timestamp,
       datetimefromparts(year(Timestamp), month(Timestamp), day(Timestamp),
                         datepart(hour, Timestamp),
                         datepart(minute, Timestamp),
                         (ceiling(datepart(second, Timestamp)) / 15) * 15,
                         0
                        ) as timestamp2
FROM tags.dbo.jan
where ValueID = '349' and
      Timestamp < '2019-01-01 04:02:00.000';

【讨论】：

感谢您的回答！我会更新问题 1。我认为 datepart(second, Timestamp) 错过了 2。分组应该从前 15 秒组开始

【解决方案2】：

declare @t table(thetimestamp datetime);
insert into @t(thetimestamp) 
values('20190101 04:00:00.487'), ('20190101 04:00:02.487'),
('20190101 04:00:15.487'), ('20190101 04:00:20.487'),
('20190101 04:00:30.487'), ('20190101 04:00:35.487'),
('20190101 04:00:45.487'), ('20190101 04:00:57.487'), 
--
('20190101 04:00:00.000'), ('20190101 04:00:15.000'),  ('20190101 04:00:30.000'), ('20190101 04:00:45.000');

select *, 
    --add ms to next 15sec boundary
    dateadd(millisecond, 
    (15000-((1000*datepart(second, thetimestamp)+ datepart(millisecond, thetimestamp))%15000))%15000,
    thetimestamp) as upper15sec,
    --subtract ms from previous 15sec boundary
    dateadd(millisecond, 
    -(15000+((1000*datepart(second, thetimestamp)+ datepart(millisecond, thetimestamp))%15000))%15000,
    thetimestamp) as lower15sec
from @t;

【讨论】：