仅在最大日期而不是在组的其他条目中显示具有特定值的组

Posted 2023-03-28

【中文标题】仅在最大日期而不是在组的其他条目中显示具有特定值的组

【英文标题】：Display group having certain value only on max date and not in other entries of group

【发布时间】：2021-06-09 05:02:17

【问题描述】：

我希望显示在 id 列上分组的组，其中 trans_cd='Audit' 仅在该组的 max trans_proc_dt 上。 max trans_proc_dt 不应该有任何其他trans_cd='Audit' execpt。

    ID  TRANS_PROC_DT   TRANS_CD    TRANS_AMT
165     5/13/2020       Renewal         553
165     10/22/2020      Cancellation    -376
165     11/24/2020      Audit   3
165     6/2/2021        Change  0
165     6/2/2021        Audit   -7
165     6/3/2021        Audit   0
497     5/1/2020        Renewal 1394
497     1/11/2021       Cancellation    -578
497     2/10/2021       Audit   -3
497     4/28/2021       Audit   76
497     5/12/2021       Audit   -73
497     6/2/2021        Change  0
511     4/27/2020       Renewal 4409
511     7/30/2020       Change  0
511     10/5/2020       Cancellation    -2558
511     2/18/2021       Audit   2806
577     5/15/2020       Renewal 829
577     2/12/2021       Audit   -123
577     4/28/2021       Audit   118
577     5/12/2021       Audit   5
577     6/2/2021        Change  0
577     6/2/2021        Audit   -5
577     6/3/2021        Audit   0
577     12/4/2020       Renewal 1996
577     6/2/2021        Change  0
751     5/13/2020       Renewal 1307
751     1/28/2021       Cancellation    -523
751     3/3/2021        Audit   481
751     4/28/2021       Audit   120
751     5/12/2021       Audit   -601
751     6/2/2021        Change  0
751     6/2/2021        Audit   601
751     6/3/2021        Audit   0
984     5/13/2020       Renewal 1081
984     11/2/2020       Change  0
984     6/3/2021        Audit   0

我的输出应该是

ID  TRANS_PROC_DT   TRANS_CD    TRANS_AMT
511     4/27/2020       Renewal 4409
511     7/30/2020       Change  0
511     10/5/2020       Cancellation    -2558
511     1/27/2021       Renewal 4409
511     2/18/2021       Audit   2806
984     5/13/2020       Renewal 1081
984     11/2/2020       Change  0
984     6/3/2021        Audit   0

我想不出如何获取我的结果集。

【问题讨论】：

我不太遵循逻辑来获得您显示的结果。你能详细说明一下吗？

【参考方案1】：

您可以使用分析函数计算每组Audit 代码的绝对最大日期和最小日期，然后将它们进行比较：如果其他日期没有更多此类事务发生，则它们应该相等。

with a(ID, TRANS_PROC_DT, TRANS_CD, TRANS_AMT) as (
select 165, '5/13/2020', 'Renewal', 553 from dual union all
select 165, '10/22/2020', 'Cancellation', -376 from dual union all
select 165, '11/24/2020', 'Audit', 3 from dual union all
select 165, '6/2/2021', 'Change', 0 from dual union all
select 165, '6/2/2021', 'Audit', -7 from dual union all
select 165, '6/3/2021', 'Audit', 0 from dual union all
select 497, '5/1/2020', 'Renewal', 1394 from dual union all
select 497, '1/11/2021', 'Cancellation', -578 from dual union all
select 497, '2/10/2021', 'Audit', -3 from dual union all
select 497, '4/28/2021', 'Audit', 76 from dual union all
select 497, '5/12/2021', 'Audit', -73 from dual union all
select 497, '6/2/2021', 'Change', 0 from dual union all
select 511, '4/27/2020', 'Renewal', 4409 from dual union all
select 511, '7/30/2020', 'Change', 0 from dual union all
select 511, '10/5/2020', 'Cancellation', -2558 from dual union all
select 511, '2/18/2021', 'Audit', 2806 from dual union all
select 577, '5/15/2020', 'Renewal', 829 from dual union all
select 577, '2/12/2021', 'Audit', -123 from dual union all
select 577, '4/28/2021', 'Audit', 118 from dual union all
select 577, '5/12/2021', 'Audit', 5 from dual union all
select 577, '6/2/2021', 'Change', 0 from dual union all
select 577, '6/2/2021', 'Audit', -5 from dual union all
select 577, '6/3/2021', 'Audit', 0 from dual union all
select 577, '12/4/2020', 'Renewal', 1996 from dual union all
select 577, '6/2/2021', 'Change', 0 from dual union all
select 751, '5/13/2020', 'Renewal', 1307 from dual union all
select 751, '1/28/2021', 'Cancellation', -523 from dual union all
select 751, '3/3/2021', 'Audit', 481 from dual union all
select 751, '4/28/2021', 'Audit', 120 from dual union all
select 751, '5/12/2021', 'Audit', -601 from dual union all
select 751, '6/2/2021', 'Change', 0 from dual union all
select 751, '6/2/2021', 'Audit', 601 from dual union all
select 751, '6/3/2021', 'Audit', 0 from dual union all
select 984, '5/13/2020', 'Renewal', 1081 from dual union all
select 984, '11/2/2020', 'Change', 0 from dual union all
select 984, '6/3/2021', 'Audit', 0 from dual
)
, last_dt as (
  select
    a.*
    , min(
       case trans_cd
       when 'Audit'
       then to_date(TRANS_PROC_DT, 'mm/dd/yyyy')
       end
    ) over(partition by id)
    as audit_dt
    , max(to_date(TRANS_PROC_DT, 'mm/dd/yyyy'))
      over(partition by id)
    as max_dt

  from a
)
select
  id
  , trans_proc_dt
  , trans_cd
  , trans_amt
from last_dt
where max_dt = audit_dt
order by id, to_date(TRANS_PROC_DT, 'mm/dd/yyyy')

身份证 | TRANS_PROC_DT | TRANS_CD | TRANS_AMT --: | :------------ | :----------- | --------: 511 | 2020 年 4 月 27 日 |更新 | 4409 511 | 2020 年 7 月 30 日 |改变 | 0 511 | 2020 年 10 月 5 日 |取消 | -2558 511 | 2021 年 2 月 18 日 |审计 | 2806 第984章2020 年 5 月 13 日 |更新 | 1081 第984章2020 年 11 月 2 日 |改变 | 0 第984章2021 年 6 月 3 日 |审计 | 0

db小提琴here

【参考方案2】：

你可以试试下面-

Select
   * 
from
   tableName tn 
where
   ID not in 
   (
      Select
         ID 
      from
         tableName t 
      where
         trans_cd = 'Audit' 
         and TRANS_PROC_DT <> (
         Select
            max(TRANS_PROC_DT) 
         from
            tableName t1 
         where
            t1.ID = t.ID) 
   )