ORACLE SQL 只返回重复值（不是原始值）

Question

我有一个包含以下信息的数据库

Customer_id, plan_id, plan_start_dte,

由于一些客户切换计划，有客户有多个重复的customer_ids，但有不同的plan_start_dte。我试图计算每天有多少次会员从任何其他计划（plan_id = 'premium'）切换到高级计划。

也就是说，我正在尝试大致这样做：返回所有具有重复 customer_id 的行，除了原始计划 (min(plan_start_dte))，其中 plan_id = 'premium'，并将它们按 plan_start_dte 分组。

我可以获取所有重复记录及其计数：

with plan_counts as (
    select c.*, count(*) over (partition by CUSTOMER_ID) ct
    from   CUSTOMERS c
)
select *
from plan_counts
where ct > 1  

其他步骤让我卡住了。首先，我尝试选择除原始计划之外的所有内容：

SELECT CUSTOMERS c
where  START_DTE not in (
    select min(PLAN_START_DTE)
    from   CUSTOMERS i
    where  c.CUSTOMER_ID = i.CUSTOMER_ID
) 

但这失败了。如果我能解决这个问题，我相信我必须添加一个附加条件，c.PLAN_ID = 'premium' 然后按日期分组并进行计数。有人有什么想法吗？

Answer 1

我想你想要lag():

select c.*
from (select c.*,
             lag(plan_id) over (partition by customer_id order by plan_start_date) as prev_plan_id
      from customers c
     ) c
where prev_plan_id <> 'premium' and plan_id = 'premium';

我不确定你想要什么输出。对于每天发生的次数：

select plan_start_date, count(*)
from (select c.*, lag(plan_id) over (partition by customer_id order by plan_start_date) as prev_plan_id
      from customers c
     ) c
where prev_plan_id <> 'premium' and plan_id = 'premium'
group by plan_start_date
order by plan_start_date;