题目需求
现有各直播间的用户访问记录表（live_events）如下，表中每行数据表达的信息为，一个用户何时进入了一个直播间，又在何时离开了该直播间。
| user_id (用户id) | live_id (直播间id) | in_datetime(进入直播间的时间) | out_datetime (离开直播间的时间) |
|---|---|---|---|
| 100 | 1 | 2021-12-1 19:30:00 | 2021-12-1 19:53:00 |
| 100 | 2 | 2021-12-1 21:01:00 | 2021-12-1 22:00:00 |
| 101 | 1 | 2021-12-1 19:05:00 | 2021-12-1 20:55:00 |
现要求统计各直播间最大同时在线人数，期望结果如下：
| live_id | max_user_count |
|---|---|
| 1 | 4 |
| 2 | 3 |
| 3 | 2 |
数据准备
drop table if exists live_events;
create table if not exists live_events
(
user_id int comment '用户id',
live_id int comment '直播id',
in_datetime string comment '进入直播间时间',
out_datetime string comment '离开直播间时间'
)
comment '直播间访问记录';
INSERT overwrite table live_events
VALUES (100, 1, '2021-12-01 19:00:00', '2021-12-01 19:28:00'),
(100, 1, '2021-12-01 19:30:00', '2021-12-01 19:53:00'),
(100, 2, '2021-12-01 21:01:00', '2021-12-01 22:00:00'),
(101, 1, '2021-12-01 19:05:00', '2021-12-01 20:55:00'),
(101, 2, '2021-12-01 21:05:00', '2021-12-01 21:58:00'),
(102, 1, '2021-12-01 19:10:00', '2021-12-01 19:25:00'),
(102, 2, '2021-12-01 19:55:00', '2021-12-01 21:00:00'),
(102, 3, '2021-12-01 21:05:00', '2021-12-01 22:05:00'),
(104, 1, '2021-12-01 19:00:00', '2021-12-01 20:59:00'),
(104, 2, '2021-12-01 21:57:00', '2021-12-01 22:56:00'),
(105, 2, '2021-12-01 19:10:00', '2021-12-01 19:18:00'),
(106, 3, '2021-12-01 19:01:00', '2021-12-01 21:10:00');
思路分析
此题与牛客——SQL179 各城市最大同时等车人数相似思路
-
进入直播间，人数+1；离开直播间，人数-1
select live_id, in_datetime as c_datetime, 1 as action from live_events union all select live_id, out_datetime as c_datetime, -1 as action from live_eventst ; -
对直播间分组，按日期降序开窗，sum进行统计即可
sum(action) over (partition by live_id order by c_datetime) as n
完整代码
select
live_id,
max(n) as max_user_count
from (select *,
sum(action) over (partition by live_id order by c_datetime) as n
from (select live_id,
in_datetime as c_datetime,
1 as action
from live_events
union all
select live_id,
out_datetime as c_datetime,
-1 as action
from live_events) t
) t1
group by live_id
;