获取envelope.ie重叠的时间跨度

Question

我有一个带有这样的在线会话的表（空行只是为了更好地显示）：

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“包围”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

我有此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, ts, 
        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，我担心性能。该表具有数亿行，时间分辨率为毫秒（如示例中所示，不是一分钟）。因此，CTEt3将会非常庞大​​。是否有人有避免自我加入和“
CONNECT BY”的解决方案？

单个智能分析功能将非常有用。

Answer 1

也尝试这个。我尽我所能进行了测试，我相信它涵盖了所有可能性，包括合并相邻的时间间隔（10:15至10:30和10:30至10:40合并为一个时间间隔，即10:15至10:40
）。它也应该相当快，用处不大。

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, NULL, max(stop_time) from ip_sessions group by ip_address
        ),
     n as
        (
         select ip_address, start_time, m_time 
         from m 
         where start_time > m_time or start_time is null or m_time is null
        ),
     f as
        (
         select ip_address, start_time,
            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
/