Oracle：按条件排序并保留历史记录

Question

我有这个示例表：

我需要创建一个排名列，它将根据 is_true=1 对 id 进行排序，这意味着 is_true 从 0 到 1 的每次更改都会使索引增加 1。 类似于以下内容：

我尝试了类似的方法：

select id, customer_id, is_true,  
       rank() over (partition by customer_id order by (case when is_true=1 then id end)) index_rank 
from table

但它不能满足我的需要。

任何帮助将不胜感激。

Answer 1

改进 GMB 提供的查询

select x.*
       ,sum(case when (is_true=1 and lag_val=0) or (lag_val is null) then 1 end) over(partition by customer_id order by id) index_rank
  from (select *
               ,lag(is_true) over(partition by customer_id order by id) lag_val       
          from dbo.t 
        )x

旧查询

select id
       ,customer_id
       ,is_true
       ,dense_rank() over(partition by customer_id order by customer_id,max_grp)
   from(
        select id
               ,customer_id
               ,is_true
               ,max(is_rank) over(partition by customer_id order by id) max_grp
          from (select id,customer_id,is_true
                       ,case when is_true=1 and lag(is_true) over(partition by customer_id order by id)=0 
                              or  lag(is_true) over(partition by customer_id order by id) is null then
                             DENSE_RANK() over(partition by customer_id order by id)
                        end as is_rank
                   from dbo.t
                 )x
      )y

Answer 2

你可以做一个窗口求和：

select
    t.*,
    1 + sum(is_true) over(partition by cutsomer_id order by id) index_rank
from mytable t

如果你想允许连续的1s而不增加排名，那么你可以先使用lag()：

select
    t.*,
    1 + sum(case when is_true = 1 and lag_is_true = 0 then 1 else 0 end) 
        over(partition by cutsomer_id order by id) index_rank
from (
    select 
        t.*,
        lag(is_true) over(partition by cutsomer_id order by id) lag_is_true
    from mytable t
) t