sql - 删除具有跨多个值的开始和结束时间戳的部分/完全重叠事件

Question

基于 this question :

如果 events 表包含具有多个值的“Room”列并且您想按房间取出重叠的 session ，那么接受的答案会如何变化？该表可能如下所示:

id   start                   end                     created_at                  room
1    2019-01-23 18:30:00.0   2019-01-23 19:00:00.0   2019-01-18 21:28:27.427612  Room1
2    2019-01-23 18:30:00.0   2019-01-23 19:00:00.0   2019-01-23 01:04:05.861876  Room1
3    2019-01-23 18:00:00.0   2019-01-23 18:45:00.0   2019-01-16 17:14:50.709552  Room1
4    2019-01-23 18:30:00.0   2019-01-23 19:30:00.0   2019-01-22 19:24:05.532491  Room1
5    2019-01-23 18:30:00.0   2019-01-23 19:30:00.0   2019-01-18 17:28:40.074205  Room1
6    2019-01-23 20:00:00.0   2019-01-23 20:30:00.0   2019-01-18 15:22:30.736888  Room1
7    2019-01-23 20:15:00.0   2019-01-23 20:45:00.0   2019-01-20 20:20:20.202020  Room1
8    2019-01-23 18:30:00.0   2019-01-23 19:00:00.0   2019-01-18 21:28:27.427612  Room2
9    2019-01-23 18:30:00.0   2019-01-23 19:00:00.0   2019-01-23 01:04:05.861877  Room2
10   2019-01-23 18:00:00.0   2019-01-23 18:45:00.0   2019-01-16 17:14:50.709552  Room2
11   2019-01-23 18:30:00.0   2019-01-23 19:30:00.0   2019-01-22 19:24:05.532491  Room2
12   2019-01-23 18:30:00.0   2019-01-23 19:30:00.0   2019-01-18 17:28:40.074205  Room2
13   2019-01-23 20:00:00.0   2019-01-23 20:30:00.0   2019-01-18 15:22:30.736888  Room2
14   2019-01-23 20:15:00.0   2019-01-23 20:45:00.0   2019-01-20 20:20:20.202020  Room2
15   2019-01-23 20:00:00.0   2019-01-23 20:30:00.0   2019-01-18 15:22:30.736888  Room3
16   2019-01-23 20:15:00.0   2019-01-23 20:45:00.0   2019-01-20 20:20:20.202021  Room3

最终结果是:

id   start                   end                     created_at                  room
2    2019-01-23 18:30:00.0   2019-01-23 19:00:00.0   2019-01-23 01:04:05.861876  Room1
7    2019-01-23 20:15:00.0   2019-01-23 20:45:00.0   2019-01-20 20:20:20.202020  Room1
9    2019-01-23 18:30:00.0   2019-01-23 19:00:00.0   2019-01-23 01:04:05.861877  Room2
14   2019-01-23 20:15:00.0   2019-01-23 20:45:00.0   2019-01-20 20:20:20.202020  Room2
16   2019-01-23 20:15:00.0   2019-01-23 20:45:00.0   2019-01-20 20:20:20.202021  Room3

下面上一个问题的答案只能给出跨所有房间的时间段内最新创建的事件，不能按房间。

select max(id), min(start), max(end), max(created_at)
from (select t.*,
             count(*) filter (where max_end < end) over (order by start) as grouping
      from (select t.*,
                   max(end) over (order by start rows between unbounded preceding and 1 preceding) as max_end
            from events t
           ) t
     ) t
group by grouping;

Answer 1

加入条件:

• 尝试在同一房间内找到另一个事件
• 但是需要和你有不同的id
• 日期必须重叠 Determine Whether Two Date Ranges Overlap
• 最后必须有一个较晚的创建日期。

如果您找不到任何其他行，则意味着您独自一人，因为不重叠或因为您是最新的约会对象。

SQL DEMO

SELECT a.*
FROM "events" a
LEFT JOIN "events" b
  ON a.room = b.room 
 AND a.id <> b.id
 AND a."start" <= b."end"
 AND a."end"   >= b."start"
 AND a.created_at < b.created_at
WHERE b.id IS NULL;

输出

| id |                start |                  end |                  created_at |  room |
|----|----------------------|----------------------|-----------------------------|-------|
|  2 | 2019-01-23T18:30:00Z | 2019-01-23T19:00:00Z | 2019-01-23T01:04:05.861876Z | Room1 |
|  7 | 2019-01-23T20:15:00Z | 2019-01-23T20:45:00Z |  2019-01-20T20:20:20.20202Z | Room1 |
|  9 | 2019-01-23T18:30:00Z | 2019-01-23T19:00:00Z | 2019-01-23T01:04:05.861877Z | Room2 |
| 14 | 2019-01-23T20:15:00Z | 2019-01-23T20:45:00Z |  2019-01-20T20:20:20.20202Z | Room2 |
| 16 | 2019-01-23T20:15:00Z | 2019-01-23T20:45:00Z | 2019-01-20T20:20:20.202021Z | Room3 |