sql – 获取envelope.i.e重叠时间跨度

我有一个像这样的在线会话的表(空行只是为了更好的可见性)：

ip_address  | start_time       | stop_time
------------|------------------|------------------
10.10.10.10 | 2016-04-02 08:00 | 2016-04-02 08:12
10.10.10.10 | 2016-04-02 08:11 | 2016-04-02 08:20

10.10.10.10 | 2016-04-02 09:00 | 2016-04-02 09:10
10.10.10.10 | 2016-04-02 09:05 | 2016-04-02 09:08
10.10.10.10 | 2016-04-02 09:05 | 2016-04-02 09:11
10.10.10.10 | 2016-04-02 09:02 | 2016-04-02 09:15
10.10.10.10 | 2016-04-02 09:10 | 2016-04-02 09:12

10.66.44.22 | 2016-04-02 08:05 | 2016-04-02 08:07
10.66.44.22 | 2016-04-02 08:03 | 2016-04-02 08:11

我需要“信封”在线时间跨度：

ip_address  | full_start_time  | full_stop_time
------------|------------------|------------------
10.10.10.10 | 2016-04-02 08:00 | 2016-04-02 08:20
10.10.10.10 | 2016-04-02 09:00 | 2016-04-02 09:15
10.66.44.22 | 2016-04-02 08:03 | 2016-04-02 08:11

我有这个查询返回所需的结果：

WITH t AS 
    -- Determine full time-range of each IP
    (SELECT ip_address,MIN(start_time) AS min_start_time,MAX(stop_time) AS max_stop_time FROM IP_SESSIONS GROUP BY ip_address),t2 AS
    -- compose ticks
    (SELECT disTINCT ip_address,min_start_time + (LEVEL-1) * INTERVAL '1' MINUTE AS ts
    FROM t
    CONNECT BY min_start_time + (LEVEL-1) * INTERVAL '1' MINUTE <= max_stop_time),t3 AS 
    -- get all "online" ticks
    (SELECT disTINCT ip_address,ts
    FROM t2
        JOIN IP_SESSIONS USING (ip_address)
    WHERE ts BETWEEN start_time AND stop_time),t4 AS
    (SELECT ip_address,ts,LAG(ts) OVER (PARTITION BY ip_address ORDER BY ts) AS prevIoUs_ts
    FROM t3),t5 AS 
    (SELECT ip_address,SUM(DECODE(prevIoUs_ts,NULL,1,0 + (CASE WHEN prevIoUs_ts + INTERVAL '1' MINUTE <> ts THEN 1 ELSE 0 END))) 
            OVER (PARTITION BY ip_address ORDER BY ts ROWS UNBOUNDED PRECEDING) session_no
    FROM t4)
SELECT ip_address,MIN(ts) AS full_start_time,MAX(ts) AS full_stop_time
FROM t5
GROUP BY ip_address,session_no
ORDER BY 1,2;

但是,我对表现感到担忧.该表有数亿行,时间分辨率为毫秒(不是示例中给出的一分钟).因此,CTE t3将是巨大的.有没有人有避免自我加入和“连接”的解决方案？

单个智能Analytic Function会很棒.

解决方法

试试这个吧.我尽我所能地测试了它,我相信它涵盖了所有的可能性,包括合并相邻的间隔(10:15到10:30和10:30到10:40结合成一个间隔,10：15到10:40 ).它应该也很快,它使用不多.

with m as
        (
         select ip_address,start_time,max(stop_time) over (partition by ip_address order by start_time 
                             rows between unbounded preceding and 1 preceding) as m_time
         from ip_sessions
         union all
         select ip_address,max(stop_time) from ip_sessions group by ip_address
        ),n as
        (
         select ip_address,m_time 
         from m 
         where start_time > m_time or start_time is null or m_time is null
        ),f as
        (
         select ip_address,lead(m_time) over (partition by ip_address order by start_time) as stop_time
         from n
        )
select * from f where start_time is not null
/