SQL查询根据类别列查找连续的本地最大值，日期最小值

如何解决SQL查询根据类别列查找连续的本地最大值，日期最小值

我有以下数据集

    Customer_ID Category    FROM_DATE    TO_DATE
      1            5        1/1/2000    12/31/2001
      1            6        1/1/2002    12/31/2003
      1            5        1/1/2004    12/31/2005 
      2            7        1/1/2010    12/31/2011
      2            7        1/1/2012    12/31/2013
      2            5        1/1/2014    12/31/2015
      3            7        1/1/2010    12/31/2011
      3            7        1/5/2012    12/31/2013
      3            5        1/1/2014    12/31/2015

我想要实现的结果是为相同类别的客户查找连续的本地最小/最大日期，并确定日期之间的任何差距：

    Customer_ID      FROM_Date      TO_Date       Category  
      1              1/1/2000      12/31/2001         5  
      1              1/1/2002      12/31/2003         6
      1              1/1/2004      12/31/2005         5
      2              1/1/2010      12/31/2013         7  
      2              1/1/2014      12/31/2015         5
      3              1/1/2010      12/31/2011         7
      3              1/5/2012      12/31/2013         7
      3              1/1/2014      12/31/2015         5

我的代码适用于客户1（返回所有3行）和客户2（返回2行，每个类别具有最小和最大日期），但是对于客户3，它无法确定2011年12月31日与1之间的差距2012年5月5日，类别7。

    Customer_ID      FROM_Date      TO_Date       Category  
      3              1/1/2010      12/31/2013         7  
      3              1/1/2014      12/31/2015         5

这是我的代码：

SELECT Customer_ID,Category,min(From_Date),max(To_Date) FROM 
(
SELECT Customer_ID,From_Date,To_Date,row_number() over (order by member_id,To_Date) - row_number() over (partition by Customer_ID order by Category) as p
  FROM FFS_SAMP
  )  X
 group by Customer_ID,p
 order by Customer_ID,Max(To_Date)

解决方法

这是一种间隙和孤岛问题。可能最安全的方法是使用累积max()查找与先前记录的重叠。如果没有重叠，则记录的“孤岛”开始。所以：

select customer_id,min(from_date),max(to_date),category
from (select t.*,sum(case when prev_to_date >= from_date then 0 else 1 end) over
                 (partition by customer_id,category
                  order by from_date
                 ) as grp
      from (select t.*,max(to_date) over (partition by customer_id,category
                                      order by from_date
                                      rows between unbounded preceding and 1 preceding
                                     ) as prev_to_date
            from t
           ) t
     ) t
group by customer_id,category,grp;

,

您的尝试非常接近。您只需要修复窗口函数的over()子句：

select customer_id,max(to_date) 
from (
    select 
        fs.*,row_number() over (partition by customer_id order from_date) 
            - row_number() over (partition by customer_id,category order by from_date) as grp
  from ffs_samp fs
) x
group by customer_id,grp
order by customer_id,min(from_date)

请注意，如示例数据所示，此方法在给定客户的时间段内不存在差距或重叠。