SQL - 根据日期映射行

SQL - map rows based on dates

我正在查看一些未使用 ID 等明确链接的案例数据。如果最近案例的 open_dateclose_date 匹配,我们可以判断两个案例是相关联的另一种情况,这两种情况具有相同的 type(请求)。我想检查一下过去是否有任何案例被推迟,以及被推迟了多少次。例如,下面的 case12 被推迟了两次——一次是从 case11 开始的,而 case11 本身是从 case9 推迟的。 case7 也被推迟了两次,而 case1 和 case2 从未被推迟。我认为这需要一个递归解决方案,但不确定如何实现。

case_id  open_date  close_date user_id type     status      
case12   2021-06-01 2021-08-25  user1   request  complete
case11   2021-05-01 2021-06-01  user1   request  deferred
case9    2021-03-01 2021-05-01  user1   request  deferred
case7    2020-09-15 2020-10-31  user1   request  saved
case5    2020-09-01 2020-09-15  user1   request  deferred
case3    2020-02-12 2020-09-01  user1   request  deferred
case2    2019-04-01 2019-06-01  user1   request  partial 
case1    2018-06-01 2018-06-17  user1   request  partial

考虑以下方法

select *, 
  case row_number() over(partition by user_id, type, map_id order by open_date) 
    when 1 then 'new case'
    when count(1) over(partition by user_id, type, map_id) then 'last deferred case'
    else 'deferred case'
  end as status
from (
  select * except(new_case), 
    countif(new_case) over(partition by user_id, type order by open_date) as map_id
  from (
    select *, 
      open_date != lead(close_date) over(partition by user_id, type order by open_date desc) new_case
    from your_table
  )
)
-- order by open_date desc    

 

如果应用于您问题中的示例数据

with your_table as (
  select 'case12' case_id, '2021-06-01' open_date, '2021-08-25' close_date, 'user1' user_id, 'request' type union all
  select 'case11', '2021-05-01', '2021-06-01', 'user1', 'request' union all
  select 'case9', '2021-03-01', '2021-05-01', 'user1', 'request' union all
  select 'case7', '2020-09-15', '2020-10-31', 'user1', 'request' union all
  select 'case5', '2020-09-01', '2020-09-15', 'user1', 'request' union all
  select 'case3', '2020-02-12', '2020-09-01', 'user1', 'request' union all
  select 'case2', '2019-04-01', '2019-06-01', 'user1', 'request' union all 
  select 'case1', '2018-06-01', '2018-06-17', 'user1', 'request' 
)    

输出是

希望您能将以上示例应用到您的实际用例中