Question

我正在尝试创建一个数据集，其中所有的Starts到最近的Stop。问题在于，第一次启动和下一次停止之间的启动次数会有所不同。

源数据集如下：

RowNum Timestamp        Action 
==============================
   1   01/01/18 12:00   Start 
   2   01/01/18 01:00   Start
   3   01/01/18 02:00   Stop
   4   01/01/18 03:00   Start
   5   01/01/18 05:00   Stop
   6   01/01/18 13:00   Start
   7   01/01/18 15:00   Start
   8   01/01/18 17:00   Start
   9   01/01/18 21:00   Stop

我希望我的最终结果是这样的：

Start           Stop
================================
01/01/18 12:00  01/01/18 02:00
01/01/18 03:00  01/01/18 05:00
01/01/18 13:00  01/01/18 21:00

或者即使有一个记录，每个起点到最近的终点也将是很好的记录。

非常感谢您提供任何指导。

Answer 1

您可以利用累积最小值查找下一个“停止”事件：

select *,
   min(case when Action = 'Stop' then Timestamp end) -- next Stop
   over (--partition by ???
         order by Timestamp
         rows between current row and unbounded following) as Stop
from tab

基于此，它是一个简单的聚合：

with cte as
 ( select *,
      min(case when Action = 'Stop' then Timestamp end) -- next Stop
      over (--partition by ???
            order by Timestamp
            rows between current row and unbounded following) as Stop
   from tab
 )
select
   min(Timestamp) as start, 
   Stop
from cte
group by Stop
order by 1

Answer 2

首先在表中的每个起点都停下来

select
A.Timestamp as Start
min(B.Timestamp) as Stop
from
  mytable A 
left join 
  mytable B
    on A.Action = 'Start' 
    and B.Action = 'Stop'
    and A.Timestamp < B.Timestamp
group by A.Timestamp

然后您可以使用该结果（以下查询中的别名表R1 int）获取最终表

select
min(Start),
Stop
from
(
select
A.Timestamp as Start
min(B.Timestamp) as Stop
from
  mytable A 
left join 
  mytable B
    on A.Action = 'Start' 
    and B.Action = 'Stop'
    and A.Timestamp < B.Timestamp
group by A.Timestamp
) as R1
group by Stop

Answer 3

解决方案：
方案A是您希望看到所有开始的站点的情况：找到在开始之后的最早站点。
方案B是您只想看到最早/最新开始的止损的方法：首先从Senario A的结果中获取一个集合，然后查找在Stop之前的最早/最新的起始。
此解决方案不考虑何时存在重复项，而您想将其保留在结果中-这将涉及第三个字段，如RowNum。
一种可能的实现：

DECLARE @Table TABLE (
    Timestamp DATETIME,
    Action VARCHAR(5)
    )

INSERT @Table
VALUES
   ('01/01/18 12:00', 'Start'),
   ('01/01/18 01:00', 'Start'),
   ('01/01/18 02:00', 'Stop'),
   ('01/01/18 03:00', 'Start'),
   ('01/01/18 05:00', 'Stop'),
   ('01/01/18 13:00', 'Start'),
   ('01/01/18 15:00', 'Start'),
   ('01/01/18 17:00', 'Start'),
   ('01/01/18 21:00', 'Stop'),
   ('01/01/18 22:00', 'Start')

SELECT * FROM @Table WHERE Action = 'Start' ORDER BY Timestamp
SELECT * FROM @Table WHERE Action = 'Stop' ORDER BY Timestamp


-- Scenario A:
SELECT Starts.Timestamp as Start, MIN (Stops.Timestamp) as Stop
FROM
(SELECT * FROM @Table WHERE Action = 'Start') as Starts
LEFT OUTER JOIN
(SELECT * FROM @Table WHERE Action = 'Stop') as Stops
on Stops.Timestamp >= Starts.Timestamp
GROUP BY Starts.Timestamp
ORDER BY Starts.Timestamp

-- Scenario B:
-- same block as above with a temp table to hold the results
SELECT Starts.Timestamp as Start, MIN (Stops.Timestamp) as Stop
INTO #allstops
FROM
(SELECT * FROM @Table WHERE Action = 'Start') as Starts
LEFT OUTER JOIN
(SELECT * FROM @Table WHERE Action = 'Stop') as Stops
on Stops.Timestamp >= Starts.Timestamp
GROUP BY Starts.Timestamp

SELECT allstops.Start, LatestStart.Stop
FROM #allstops as allstops
LEFT OUTER JOIN (
    SELECT MIN (Start) as Start, Stop -- this returns the earliest Start, switch to MAX to get the latest
    FROM #allstops
    GROUP BY Stop
    ) as LatestStart
on allstops.Start = LatestStart.Start

Answer 4

您可以将lag和ceiling函数用作：

select max(Start) as Start, max(Stop) as Stop
  from
  (
    select row_number() over ( order by rownum ) as rn,
           ( case when action = 'Start' then q.timestamp end ) as Start,
           ( case when action = 'Stop' then q.timestamp end ) as Stop
      from
    (
    select t.*,       
           lag(action) over (order by rownum) as lg
      from tab t
    ) q
    where q.action != coalesce(lg,'Stop')
   ) r
  group by ceiling(rn*.5)   
  order by ceiling(rn*.5);

        Start                 Stop
01.01.2018 12:00:00   01.01.2018 02:00:00
01.01.2018 03:00:00   01.01.2018 05:00:00
01.01.2018 13:00:00   01.01.2018 21:00:00

Rextester Demo

P.S。对于每个二进制步骤对，我们确定START和STOP行，其中行non-null，而行对的另一成员为null。由于这种逻辑，我需要mod(...,2)作为行号，并满足ceil(rn*.5)的要求，即对于1或1都产生2，对于2 3和4，以及{{1}分别为3和5的{{1}}，即最近的连续高位整数。

创建一站式多次启动的开始/停止日期范围数据集

4 个答案: