SQL 将日期缩小到范围 "start - end"

SQL reduce dates to range "start - end"

我在 table 中有多个日期的重复行:

 ID      STATE       DATE
----------------------------
id01   connected  2015-04-04
id01   connected  2015-04-05
id01   connected  2015-04-08
id01   disconect  2015-04-11
id01   disconect  2015-04-12
id01   connected  2015-04-13

我想要一个包含 "start date" 和 "end date" 的查询,结果如下:

 ID      STATE    START DATE   END DATE
----------------------------------------
id01   connected  2015-04-04  2015-04-10
id01   disconect  2015-04-11  2015-04-12
id01   connected  2015-04-13  XXXXXXXXXX

最后一个"end date"不重要(last value, null, now()...)

最重要的是检测更改日期(在此示例中,2015 年 4 月 10 日没有行,2015 年 4 月 13 日发生了相同的状态)。

可能的解决方案? (无效)

SELECT ID, STATE, MIN(date), MAX(date) 
   FROM TABLE
   GROUP BY ID, STATE;

无效,因为合并间隔:

 ID      STATE    START DATE   END DATE
----------------------------------------
id01   connected  2015-04-04  XXXXXXXXXX
id01   disconect  2015-04-11  2015-04-12

查询在Impala中有运行(类似SQL92)

Impala 支持 window 函数。这个问题是"gap-and-islands"问题,所以可以通过行数的不同来解决:

select id, state, min(date) as start_date, max(date) as end_date
from (select t.*,
             row_number() over (partition by id order by date) as seqnum_id,
             row_number() over (partition by id, state order by date) as seqnum_isd
      from table t
     ) t
group by id, state, (seqnum_id - seqnum_isd);

区别的逻辑并不难,但是当你第一次学习它时会很棘手。它有助于 运行 子查询并查看行号值是什么——以及为什么差异定义了每个组。

(代表OP发表).

来自,将"gap-and-islands"问题转化为我的研究案例,有解决方案:

select 
    id,
    state,
    start_date,
    date_add(lag(start_date, 1) over (partition by id order by start_date desc), -1) as end_date
from 
    (select id, state, min(date) as start_date, max(date) as end_date
        from (select t.*,
                row_number() over (partition by id order by date) as seqnum_id,
                row_number() over (partition by id, state order by date) as seqnum_isd
            from test t
        ) t
    group by id, state, (seqnum_id - seqnum_isd)) t_range
order by start_date;