Redshift rank() window 函数基于不同的维度值和时间戳排序

Question

我正在尝试应用 rank() window 函数来捕捉边缘情况。

所以我有 CTE：

with  customer_details_type as 
(
select cd.*, cp.type
from db.ustomer_details as cd
left join db.customer_private as cp on cp.id = cd.customer_id
)

我想在上面的 CTE 中添加一个 rank()，它将只对具有不同 customer_type 的行进行排名，而不是对具有相同的行进行排名。所以在下面的数据中我想要这样的东西：

我尝试了以下方法：

with  customer_details_type as 
(
select cd.*, cp.type, 
, rank() over (partition by  cd.order_id,cd.customer_type order by cd.created desc nulls last) as rank
from db.ustomer_details as cd
left join db.customer_private as cp on cp.id = cd.customer_id
)

但是它按照我感兴趣的排名顺序分配了 rank=1。

所以我要解决的问题是：对于具有 count(distinct customer_type)> 的行，如何在 order_id 分区上使用 rank() window 函数仅 1，按创建的时间戳排序？上述设置中的 rank() 可以做到这一点，还是我需要为该任务添加第二个 CTE？

Answer 1

我认为你想要：

with customer_details_type as (
      select cd.*, cp.type, 
             dense_rank() over (partition by cd.order_id order by cd.customer_type) as rank
     from db.ustomer_details cd left join
          db.customer_private cp
           on cp.id = cd.customer_id
    )

这为每个订单列举了不同的客户类型。

如果您确实希望它们按第一次出现的顺序排列，则计算每种类型的最小值 created 并将其用于排序：

with customer_details_type as (
      select cd.*,
             dense_rank() over (partition by cd.order_id order by type_created, customer_type) as rank
      from (select cd.*, cp.type, 
                   min(created) over (partition by cd.order_id, cd.customer_type) as type_created
            from db.customer_details cd left join
                 db.customer_private cp
                 on cp.id = cd.customer_id
           ) cd
    )

Redshift rank() window 函数基于不同的维度值和时间戳排序

Redshift rank() window function based on distinct dimension values and timestamp ordering

sql

rank

window-functions

amazon-redshift