注：参考文章：
0 需求描述
1 数据准备
sql
create table if not exists table23
(
user_id int comment '用户id',
room_num string comment '房间号',
in_time string comment '入住时间',
out_time string comment '离店时间'
)
comment '旅客入住离店表';
insert overwrite table table23
values (7, '2004', '2021-03-05','2021-03-07'),
(23,'2010', '2021-03-05','2021-03-06'),
(7, '1003', '2021-03-07','2021-03-08'),
(8, '2014', '2021-03-07','2021-03-08'),
(14, '3001','2021-03-07','2021-03-10'),
(18, '3002','2021-03-08','2021-03-10'),
(23, '3020','2021-03-08','2021-03-09'),
(25, '2006','2021-03-09','2021-03-12');2 数据分析
需求：求出每个时间段，有客人在住的房间数量。
如果只考虑一人一房，可以借助于【直播间同时在线人数】统计的思路，相关sql逻辑指路：
sql
select
start_time,
end_time,
acc_cnt
from (select
`time` as start_time,
lead(`time`) over ( order by `time`) as end_time,
acc_cnt
from (select
`time`,
sum(flag) over (order by `time`) as acc_cnt
from (
select
in_time as `time`,
1 as flag
from table23
union all
select
out_time as `time`,
-1 as flag
from table23
) t1
) t2
group by `time`, acc_cnt
) t
where end_time is not null;
上述代码需要考虑一个问题：如果有多个人共住一房间 ，今天退了一个人，明天又退了一个人，后天的时候才退完，虽然这期间一直有人在退，但房间还是有人住的，这种情况是不是也算【有客人在住的房间】? 如果考虑上述情况，需要对累加的状态进行调整，此时需要考虑每个房间中截止当前时间的人数情况。
第一步 ：先求出每个房间截至当前时间 的人数累计值 ，作为状态判断辅助条件
sql
select
`time`,
room_num,
sum( user_cnt) over (partition by room_num order by `time`) user_cnt
from (
select
in_time as `time`,
room_num,
count(user_id) user_cnt
from table23
group by in_time, room_num
union all
select
out_time as `time`,
room_num,
-1 * count(user_id) user_cnt
from table23
group by out_time, room_num
) t1
第二步 ：基于累计的每个房间人数进行判断：如果房间有人就标记1，没有人时候就标记为-1。代码为：case when user_cnt > 0 时标记1，否则标记-1
sql
select
`time`,
room_num,
user_cnt,
case when user_cnt > 0 then 1 else -1 end flag
from (select
`time`,
room_num,
sum(user_cnt) over (partition by room_num order by `time`) user_cnt
from (
select
in_time as `time`,
room_num,
count(user_id) user_cnt
from table23
group by in_time, room_num
union all
select
out_time as `time`,
room_num,
-1 * count(user_id) user_cnt
from table23
group by out_time, room_num
) t1
) t2;
第三步： 基于第二步的结果，计算截止当前时间点的有人入住的房间数量 acc_cnt，SQL如下：
sql
select
`time`,
room_num,
user_cnt,
case when user_cnt > 0 then 1 else -1 end flag,
sum(case when user_cnt > 0 then 1 else -1 end) over (order by `time`) acc_cnt
from (select
`time`,
room_num,
sum(user_cnt) over (partition by room_num order by `time`) user_cnt
from (
select
in_time as `time`,
room_num,
count(user_id) user_cnt
from table23
group by in_time, room_num
union all
select
out_time as `time`,
room_num,
-1 * count(user_id) user_cnt
from table23
group by out_time, room_num
) t1
) t2;
第四步： 基于第三步的结果，对时间time 和截止当前时间点的有人入住的房间数量acc_cnt 这两个字段进行去重，SQL如下：
sql
select
`time`,
acc_cnt
from (
select
`time`,
room_num,
user_cnt,
case when user_cnt > 0 then 1 else -1 end flag,
sum(case when user_cnt > 0 then 1 else -1 end) over (order by `time`) acc_cnt
from (select
`time`,
room_num,
sum(user_cnt) over (partition by room_num order by `time`) user_cnt
from (
select
in_time as `time`,
room_num,
count(user_id) user_cnt
from table23
group by in_time, room_num
union all
select
out_time as `time`,
room_num,
-1 * count(user_id) user_cnt
from table23
group by out_time, room_num
) t1
) t2
) t3
group by `time`, acc_cnt
第五步： 基于第四步的结果，通过lead函数 （对time字段往后偏移一行）求出当前数据的结束时间end_time，SQL如下：
sql
select
`time` as start_time,
lead(`time`, 1) over (order by `time`) as end_time,
acc_cnt
from (
select
`time`,
acc_cnt
from (
select
`time`,
room_num,
user_cnt,
case when user_cnt > 0 then 1 else -1 end flag,
sum(case when user_cnt > 0 then 1 else -1 end) over (order by `time`) acc_cnt
from (select
`time`,
room_num,
sum(user_cnt) over (partition by room_num order by `time`) user_cnt
from (
select
in_time as `time`,
room_num,
count(user_id) user_cnt
from table23
group by in_time, room_num
union all
select
out_time as `time`,
room_num,
-1 * count(user_id) user_cnt
from table23
group by out_time, room_num
) t1
) t2
) t3
group by `time`, acc_cnt
) t4
：基于第五步的结果，过滤掉end_time 是null的数据，SQL如下：
sql
select
start_time,
end_time,
acc_cnt
from (
select
`time` as start_time,
lead(`time`, 1) over (order by `time`) as end_time,
acc_cnt
from (
select
`time`,
acc_cnt
from (
select
`time`,
room_num,
user_cnt,
case when user_cnt > 0 then 1 else -1 end as flag,
sum(case when user_cnt > 0 then 1 else -1 end) over (order by `time`) acc_cnt
from (select
`time`,
room_num,
sum(user_cnt) over (partition by room_num order by `time`) user_cnt
from (
select
in_time as `time`,
room_num,
count(user_id) user_cnt
from table23
group by in_time, room_num
union all
select
out_time as `time`,
room_num,
-1 * count(user_id) user_cnt
from table23
group by out_time, room_num
) t1
) t2
) t3
group by `time`, acc_cnt
) t4
) t5
where end_time is not null;
3 小结
针对【每个时间段的直播同时在线人数】 【每个时间段有客人在住的房间数量】这种类型的题目，本质是对（截至）当前时间点的状态统计。 这种问题常见的解决思路是：对当前时间点的状态打标记flag ，之后基于标记flag 做开窗计算（结合窗口函数）或聚合计算**。**