SQL - 如果差异低于阈值，则合并两行

Question

我在 SQL Server 中有这样的表：

id  start_time  end_time
1   10:00:00    10:34:00
2   10:38:00    10:52:00
3   10:53:00    11:23:00
4   11:24:00    11:56:00
5   14:20:00    14:40:00
6   14:41:00    14:59:00
7   15:30:00    15:40:00

我想要的是一个根据两个连续记录之间的时间差输出合并记录的查询（行 n 的 end_time 和行 n+1 的 start_time ) .时间差小于2分钟的记录合并为一个时间条目，并保留第一条记录的ID。如果多个连续记录的时间差小于 2 分钟，这也应该合并两个以上的记录。

这将是预期的输出：

id  start_time  end_time
1   10:00:00    10:34:00
2   10:38:00    11:56:00
5   14:20:00    14:59:00
7   15:30:00    15:40:00

提前感谢有关如何构建查询的任何提示。

编辑： 我从以下代码开始计算lead_time和时间差，但不知道如何分组和合并。

WITH rows AS
        (
        SELECT  *, ROW_NUMBER() OVER (ORDER BY Id) AS rn
        FROM #temp
        )
SELECT  mc.id, mc.start_time, mc.end_time, mp.start_time lead_time, DATEDIFF(MINUTE, mc.[end_time], mp.[start_time]) as DiffToNewSession
FROM    rows mc
LEFT JOIN    rows mp
ON      mc.rn = mp.rn - 1

Answer 1

t-sql中的window函数可以实现很多数据统计，比如

create table #temp(id int identity(1,1), start_time time, end_time time)
insert into #temp(start_time, end_time)
values  ('10:00:00', '10:34:00')
      , ('10:38:00', '10:52:00')
      , ('10:53:00', '11:23:00')
      , ('11:24:00', '11:56:00')
      , ('14:20:00', '14:40:00')
      , ('14:41:00', '14:59:00')
      , ('15:30:00', '15:40:00')

;with c0 as(
select *, LAG(end_time,1,'00:00:00') over (order by id) as lag_time
from #temp
), c1 as(
select *, case when DATEDIFF(MI, lag_time, start_time) <= 2 then 1 else -0 end as gflag
from c0
), c2 as(
select *, SUM(case when gflag=0 then 1 else 0 end) over(order by id) as gid
from c1
)
select MIN(id) as id, MIN(start_time) as start_time, MAX(end_time) as end_time
from c2
group by gid

为了更好的描述数据构建的过程，我简单的用c0,c1,c2...来表示层级，可以合并一些层级进行优化。 如果不能使用 id 作为排序条件，则需要更改上述语句中的排序部分。

Answer 2

您可以使用recursive cte 来获得您想要的结果。这种方法只是简单地将当前end_time 与下一个start_time 进行比较。如果它小于2 mintues 阈值，则使用与grp_start 相同的start_time。最后，简单地在grp_start 上做一个GROUP BY

with rcte as
(
    -- anchor member
    select  *, grp_start = start_time
    from    tbl
    where   id  = 1

    union all

    -- recursive member
    select  t.id, t.start_time, t.end_time, 
            grp_start = case when datediff(second, r.end_time, t.start_time) <= 120
                             then r.grp_start
                             else t.start_time
                             end
    from    tbl t
            inner join rcte r   on  t.id    = r.id + 1
)
select  id = min(id), grp_start as start_time, max(end_time) as end_time
from    rcte 
group by grp_start

demo

Answer 3

我想这应该可以在没有递归的情况下解决问题。我再次使用了几个 ctes 以使解决方案更易于阅读。估计可以减少一点……

INSERT INTO T1 VALUES
(1,'10:00:00','10:34:00')
,(2,'10:38:00','10:52:00')
,(3,'10:53:00','11:23:00')
,(4,'11:24:00','11:56:00')
,(5,'14:20:00','14:40:00')
,(6,'14:41:00','14:59:00')
,(7,'15:30:00','15:40:00')
GO

WITH cte AS(
SELECT *
      ,ROW_NUMBER() OVER (ORDER BY id) AS rn
      ,DATEDIFF(MINUTE, ISNULL(LAG(endtime) OVER (ORDER BY id), starttime), starttime) AS diffMin
      ,COUNT(*) OVER (PARTITION BY (SELECT 1)) as maxRn
  FROM T1
),
cteFirst AS(
SELECT *
  FROM cte
  WHERE rn = 1 OR diffMin > 2
),
cteGrp AS(
SELECT *
      ,ISNULL(LEAD(rn) OVER (ORDER BY id), maxRn+1) AS nextRn
  FROM cteFirst
)
SELECT f.id, f.starttime, MAX(ISNULL(n.endtime, f.endtime)) AS endtime
  FROM cteGrp f
  LEFT JOIN cte n ON n.rn >= f.rn AND n.rn < f.nextRn
  GROUP BY f.id, f.starttime