是否可以在 T-SQL 中按彼此的差异对行进行分组

Is it possible to group rows by difference to each other in T-SQL

是否可以编写一个 SQL 查询,根据日期时间列值与相邻行值的差异对行进行分组?

让我举个例子...我有一个 SQL 查询是这样的:

SELECT
    Id,
    StartTime,
    EndTime,
    datediff(second, max(StartTime), EndTime)) as Duration
FROM Timings
ORDER BY StartTime

returns 结果如下:

| ID | StartTime           | EndTime             | Duration
| 1  | 2017-10-06 10:59:48 | 2017-10-06 10:59:58 | 10
| 2  | 2017-10-06 11:00:02 | 2017-10-06 11:00:06 | 4
| 3  | 2017-10-06 11:00:15 | 2017-10-06 11:00:22 | 7
| 4  | 2017-10-06 11:00:30 | 2017-10-06 11:00:39 | 9
| 5  | 2017-10-06 15:34:31 | 2017-10-06 15:34:45 | 14
| 6  | 2017-10-06 15:34:48 | 2017-10-06 15:34:56 | 8
| 7  | 2017-10-06 15:34:52 | 2017-10-06 15:34:59 | 7

这里重要的是时间分为两批,前四批都在上午 11 点左右完成,后两批在下午 3 点半左右完成。

我想详细了解每批计时的start/end时间,平均时长,一组计时的个数。为此,我需要一种按批次对时间进行分组的方法,其中批次定义为按开始时间排序时,在一个结束时间和下一个开始时间之间少于 30 秒的时间组。可能吗?

真实情况的一些相关笔记...

在 SQL Server 2012+ 中:

common table expression 中使用 window 函数 lag() 获取当前行 starttimedatediff()endtime,然后 sum() over() 使用条件聚合(与硬编码值进行比较)生成 batch:

;with cte as (
select *
  , datediff(second,lag(endtime) over (order by starttime),starttime) as prev_dat
from timings
)
select id, starttime, endtime, duration
  , sum(case when coalesce(prev_dat,31)>30 then 1 else 0 end) over (
    order by starttime
    ) as batch
from cte

rextester 演示:http://rextester.com/OVNF90739

returns:

+----+---------------------+---------------------+----------+-------+
| id |      starttime      |       endtime       | duration | batch |
+----+---------------------+---------------------+----------+-------+
|  1 | 2017-10-06 10:59:48 | 2017-10-06 10:59:58 |       10 |     1 |
|  2 | 2017-10-06 11:00:02 | 2017-10-06 11:00:06 |        4 |     1 |
|  3 | 2017-10-06 11:00:15 | 2017-10-06 11:00:22 |        7 |     1 |
|  4 | 2017-10-06 11:00:30 | 2017-10-06 11:00:39 |        9 |     1 |
|  5 | 2017-10-06 15:34:31 | 2017-10-06 15:34:45 |       14 |     2 |
|  6 | 2017-10-06 15:34:48 | 2017-10-06 15:34:56 |        8 |     2 |
|  7 | 2017-10-06 15:34:52 | 2017-10-06 15:34:59 |        7 |     2 |
+----+---------------------+---------------------+----------+-------+

您可以按日期时间部分分组

group by DATEPART(StartTime, yyyy) + DATEPART(StartTime, MM) + DATEPART(StartTime, DD)

你还必须改变你 select 以匹配