如果我有下面这张表
id time
start 08.00 AM
stop 09.00 AM
stop 09.30 AM
start 09.30 AM
start 11.00 AM
start 11.30 AM
stop 11.40 AM
stop 12.00 PM
我想要的输出只需要每次第一次启动和停止后的每次启动和最后一次停止
有什么解决办法吗？
这就是我想要的输出：
id time
start 08.00 AM
stop 09.00 AM
start 09.30 AM
stop 11.40 AM
stop 12.00 PM
Whitout 对每一行都有任何主键或一些不同的 id ......我想到的唯一解决方案是：
select rn, id , time
from
(select ROW_NUMBER() over (order by time) as rn, id, time from test
where id = 'start'
union
select ROW_NUMBER() over (order by time) as rn, id, time from test
where id = 'stop'
) d
order by rn
基本上我在开始行和停止行之间建立了一个联合，如下所示：
(select ROW_NUMBER() over (order by time) as rn, id, time from test
where id = 'start'
union
select ROW_NUMBER() over (order by time) as rn, id, time from test
where id = 'stop'
) d
返回：
1 start 08.00
2 start 11.00
3 start 12.00
4 start 13.00
1 stop 09.00
2 stop 10.00
3 stop 14.00
4 stop 15.00
从原始输入：
id time
start 08.00
stop 09.00
stop 10.00
start 11.00
start 12.00
start 13.00
stop 14.00
stop 15.00
现在您只需按他们自己的行号对它们进行排序......就是那个 rn。
最后，您将拥有：
1 start 08.00
1 stop 09.00
2 start 11.00
2 stop 10.00
3 start 12.00
3 stop 14.00
4 start 13.00
4 stop 15.00
@注意：我的示例值与您的值接近...但是虚构的...
我有一些替代方案，它将以不同的格式给出结果。但它确实返回了请求的信息。
SELECT
MIN([main].[Start]) AS [Start],
[main].[End] AS [Stop],
DATEDIFF(minute, MIN([main].[Start]), [main].[End]) AS [Minutes]
FROM
(
SELECT
[sub].[Start],
MIN([sub].[End]) AS [End]
FROM
(
SELECT
[start].[time] AS [Start],
[start].[id] AS [StartingStatus],
[end].[time] AS [End],
[end].[id] AS [EndingStatus]
FROM [Log] [start], [Log] [end]
WHERE [start].[id] = 'start'
AND [start].[time] < [end].[time]
AND [start].[id] <> [end].[id]
) AS [sub]
GROUP BY
[sub].[Start],
[sub].[StartingStatus]
) AS [main]
GROUP BY
[main].[End]
基本上，它首先选择所有记录where [id] = 'start'。然后它会查找所有具有较晚时间和 的记录where [id] <> 'start'。
在这些记录中，它按每个“开始”分组并确定第一个“停止”时间。然后将这些记录再次分组，以找到每个“停止”时间的第一个“开始”时间。
结果如下所示：
+--------------------------------+--------------------------------+---------+
| START | STOP | MINUTES |
+================================+================================+=========+
| January, 29 2013 08:00:00+0000 | January, 29 2013 09:00:00+0000 | 60 |
+--------------------------------+--------------------------------+---------+
| January, 29 2013 09:30:00+0000 | January, 29 2013 11:30:00+0000 | 120 |
+--------------------------------+--------------------------------+---------+
| January, 29 2013 11:30:00+0000 | January, 29 2013 12:00:00+0000 | 30 |
+--------------------------------+--------------------------------+---------+
小提琴是在这里找到的。
PS：这个答案是基于我在这里的回答。
;WITH cte AS
(
SELECT id AS pr, [time], ROW_NUMBER() OVER(ORDER BY [time]) AS Id,
COUNT(*) OVER() AS cnt
FROM dbo.test21 t1
), cte2 AS
(
SELECT Id, pr, [time], pr AS prStart, [time] AS StopTime, pr AS prStop, [time] AS StartTime
FROM cte
WHERE Id = 1
UNION ALL
SELECT c1.Id,
CASE WHEN c1.pr != c2.pr OR c1.Id = c1.cnt THEN c1.pr ELSE c2.pr END,
CASE WHEN c1.pr != c2.pr OR c1.Id = c1.cnt THEN c1.[time] ELSE c2.[time] END,
c1.pr, c1.[time], c2.pr, c2.[time]
FROM cte c1 JOIN cte2 c2 ON c1.Id = c2.Id + 1
)
SELECT pr, [time], MIN(StartTime) AS StartTime,
MAX(DATEDIFF(minute, StartTime, [time])) AS Interval
FROM cte2
GROUP BY pr, [time]
SQLFiddle上的演示
在评论中做出澄清后，这变成了一个经典的gaps-n-islands问题。
假设这是 SQL Server 2005+，您可以使用以下方法：
WITH partitioned AS (
SELECT
*,
grp = ROW_NUMBER() OVER (ORDER BY time, CASE id WHEN 'stop' THEN 0 ELSE 1 END)
- ROW_NUMBER() OVER (PARTITION BY id ORDER BY time)
FROM atable
)
SELECT
id,
time = MIN(time)
FROM partitioned
GROUP BY
id,
grp
ORDER BY
time
;