sql - 从组 SQL 中选择特定事件

Question

数据集看起来像

id    statusid    statusdate    
100    22         04/12/2016    
100    22         04/14/2016
100    25         04/16/2016
100    25         04/17/2016
100    25         04/19/2016
100    22         04/22/2016
100    22         05/14/2016
100    27         05/19/2016
100    27         06/14/2016
100    25         06/18/2016
100    22         07/14/2016
100    22         07/18/2016

任务是选择第一次记录每个状态。记录每个状态的唯一次数。

示例：对于状态 22

• 首次状态日期：04/12/2016
• 上次状态第一次日期：07/14/2016
• 进入该状态的唯一次数：3

Answer 1

select  ID,
        statusid,
        Status,
        max(case when rn_grp_asc=1 then statusdate end) as frst_time_first_status,
        max(case when rn_grp_desc=1 then statusdate end) as last_time_first_status,
        count(distinct grp) as unique_times_in_status
from    (   select  t.*,
                    row_number() over(partition by id,statusid order by grp,statusdate) as rn_grp_asc,
                    row_number() over(partition by id,statusid order by grp desc,statusdate) as rn_grp_desc
            from    (   Select  ID,
                                StatusID,
                                Status,
                                StatusDate,
                                row_number() over(partition by id order by Systemdate)-row_number() over(partition by id,statusid order by Systemdate) as grp
                       from     t
                      where ID=100
           ) t
      ) t
group by id,statusid,Status

上面的查询返回正确的结果，但下面的查询没有。

select  ID,
        statusid,
        Status,
        max(case when rn_grp_asc=1 then statusdate end) as frst_time_first_status,
        max(case when rn_grp_desc=1 then statusdate end) as last_time_first_status,
        count(distinct grp) as unique_times_in_status
from    (   select  t.*,
                    row_number() over(partition by id,statusid order by grp,statusdate) as rn_grp_asc,
                    row_number() over(partition by id,statusid order by grp desc,statusdate) as rn_grp_desc
            from    (   Select  ID,
                                StatusID,
                                Status,
                                StatusDate,
                                row_number() over(partition by id order by Systemdate)-row_number() over(partition by id,statusid order by Systemdate) as grp
                       from     t    
         ) t
) t
 where  ID=100
group by id,statusid,Status

Answer 2

这比看起来复杂。我假设您想要给定状态的唯一连续时间。使用行号差异方法将连续行分类。然后获取这些组中的行号。最后聚合以获得第一组和最后一组中的第一天以及不同组的数量。

select statusid
,max(case when rn_grp_asc=1 then statusdate end) as last_time_first_status
,max(case when rn_grp_desc=1 then statusdate end) as last_time_first_status
,count(distinct grp) as unique_times_in_status
from (select t.*
      ,row_number() over(partition by statusid order by grp,statusdate) as rn_grp_asc 
      ,row_number() over(partition by statusid order by grp desc,statusdate) as rn_grp_desc
      from (select t.*,row_number() over(order by statusdate)
           -row_number() over(partition by statusid order by statusdate) as grp
            from tbl t
           ) t
      ) t
group by statusid