连续区间聚合SQL写法

求每个连续的相同状态的片段的开始及结束边界

postgresql:

with a as (
	select 0 hour_index,'启' state
	union all  select 1 hour_index,'停' state
	union all  select 2 hour_index,'停' state
	union all  select 3 hour_index,'启' state
	union all  select 4 hour_index,'启' state
	union all  select 5 hour_index,'启' state
	union all  select 6 hour_index,'停' state
	union all  select 7 hour_index,'启' state
	union all  select 8 hour_index,'停' state
	union all  select 9 hour_index,'停' state
	union all  select 10 hour_index,'启' state
	union all  select 11 hour_index,'启' state
),b as (
	SELECT hour_index,state,lag(state) over(order by hour_index) lag_state
	from a
),c as (
	select hour_index,state, lag_state, case when state=lag_state then 0 else 1 end is_first
	from b
),d as (
	select hour_index,state, lag_state,is_first
	from c
)
select *,sum(is_first) over( order by hour_index) group_id
from d
order by 1

结果：

hour_index	state	lag_state	is_first	group_id
0	启	null	1	1
1	停	启	1	2
2	停	停	0	2
3	启	停	1	3
4	启	启	0	3
5	启	启	0	3
6	停	启	1	4
7	启	停	1	5
8	停	启	1	6
9	停	停	0	6
10	启	停	1	7
11	启	启	0	7

这样就可以通过分组求 MIN()& MAX()得到起止位置了。