SQL：差距和孤岛问题 - 日期不连续导致排名不准确

Question

这是我对the other question 的跟进。

Quarter    Segment    Customer    *Counter*
   Q1 2018    A          A1          1
   Q2 2018    A          A1          2
   Q3 2018    A          A1          3
   Q4 2018    B          A1          1
   Q1 2019    B          A1          2
   Q2 2019    A          A1          1
   Q1 2020    A          A1          *1* I want 1 not 2 here because it's not consecutive (we don't have Q3 & Q4 2019)
   Q2 2020    A          A1          *2* I want 2 not 3 here because it reset in Q1 2020.

如果日期是连续的，则以下查询有效。我将如何调整查询以获得我正在寻找的内容？我尝试添加一个滞后 1 行的新列，并尝试检查新列中的季度值是否本质上是真实的前 1 个季度（如果是，使用计数器，如果不是，使用“1”重新启动）。但它不会重置以下记录。那我不知道该怎么办了。

select quarter, customer, segment,
       row_number() over (partition by customer, segment, seqnum - seqnum_cs order by right(quarter, 4), left(quarter, 2)) as counter
from (select t.*,
             row_number() over (partition by customer order by right(quarter, 4), left(quarter, 2)) as seqnum,
             row_number() over (partition by customer, segment order by right(quarter, 4), left(quarter, 2)) as seqnum_cs
      from t
     ) t
order by customer, seqnum;

Answer 1

您需要修复数据模型！无论如何，这是通过枚举四分之一来解决的。

select quarter, customer, segment,
       row_number() over (partition by customer, segment, q_seqnum - seqnum_cs order by q_seqnum) as counter
from (select t.*,
             row_number() over (partition by customer, segment order by q_seqnum) as seqnum_cs
      from (select t.*,
                   cast(right(quarter, 4) as int) * 4 + cast(substring(quarter, 2, 1) as int) as q_seqnum
            from t
           ) t
     ) t
order by customer, q_seqnum;