Question

在Oracle数据库中，我有一个表，其内容类似于以下内容：

Market_Intro_Date  Change_Date  Author
-------------------------------------------
   01.06.2025      10.07.2020   Meyer     *
   01.01.2025      30.06.2020   Harrin
   01.01.2025      01.05.2020   Floyd
   01.01.2025      15.04.2020   Peterson  *
   01.12.2024      20.03.2020   George
   01.12.2024      10.03.2020   Smith
   01.12.2024      15.01.2020   George    *
   01.01.2025      15.12.2019   Lee
   01.01.2025      01.11.2019   Alfonso
   01.01.2025      10.10.2019   Peterson  *
   01.07.2025      30.09.2019   Smith     *
   01.07.2024      20.09.2019   Lee
   01.07.2024      10.09.2019   Meyer     
   01.07.2024      01.05.2019   Smith     *

我需要一个SQL查询，该查询将返回市场引入日期的每一次更改（最后带有*标记）的首次出现以及更改日期和作者。

所以查询的结果将是：

Market_Intro_Date  Change_Date  Author
-------------------------------------------
   01.06.2025      10.07.2020   Meyer     
   01.01.2025      15.04.2020   Peterson  
   01.12.2024      15.01.2020   George    
   01.01.2025      10.10.2019   Peterson  
   01.07.2024      01.05.2019   Smith

提前谢谢

Answer 1

您可以使用扩展名为select Market_Intro_Date, min(Change_Date) as Change_Date, min(Author) keep (dense_rank first order by Change_Date) as Author from your_table group by Market_Intro_Date MARKET_INTRO_DATE | CHANGE_DATE | AUTHOR :---------------- | :---------- | :------- 01-JUL-24 | 01-MAY-19 | Smith 01-DEC-24 | 15-JAN-20 | George 01-JAN-25 | 10-OCT-19 | Peterson 01-JUN-25 | 10-JUL-20 | Meyer 01-JUL-25 | 30-SEP-19 | Smith的聚合：

select Market_Intro_Date,
  min(Change_Date) as Change_Date,
  min(Author) keep (dense_rank first order by Change_Date) as Author
from (
  select Market_Intro_Date, Change_Date, Author,
    row_number() over (partition by Market_Intro_Date order by Change_Date)
      - row_number() over (order by Change_Date) as grp
  from your_table
)
group by Market_Intro_Date, grp
order by Change_Date desc;

MARKET_INTRO_DATE | CHANGE_DATE | AUTHOR  
:---------------- | :---------- | :-------
01-JUN-25         | 10-JUL-20   | Meyer   
01-JAN-25         | 15-APR-20   | Peterson
01-DEC-24         | 15-JAN-20   | George  
01-JAN-25         | 10-OCT-19   | Peterson
01-JUL-25         | 30-SEP-19   | Smith   
01-JUL-24         | 01-MAY-19   | Smith

Read more。

我错过了其中一个日期重复的问题，所以这是一个空白和孤岛的问题。 Tabibitosan进行营救：

$JENKINS_HOME/jobs

db<>fiddle

Smith也有第二行，但这看起来是正确的；不知道您的样本数据是否遗漏了星号或01.07.2025是错字。

Answer 2

澄清后已更新

Match_recognize通常比基于解析函数的旧start_of_group解决方案更快：

select *
from your_tab
match_recognize (
   order by Change_Date
   measures
       first(Change_Date) as Change_Date,
       first(Market_Intro_Date) as Market_Intro_Date,
       first(Author     ) as Author     
   pattern (A B*)
   define
      b as Market_Intro_Date = prev(Market_Intro_Date)  and Change_Date>prev(Change_Date)
)
order by 1;

包含测试数据的完整示例：

with your_tab(Market_Intro_Date, Change_Date, Author) as (
   select to_date('01.06.2025','dd.mm.yyyy'), to_date('10.07.2020','dd.mm.yyyy'),   'Meyer   ' from dual union all
   select to_date('01.01.2025','dd.mm.yyyy'), to_date('30.06.2020','dd.mm.yyyy'),   'Harrin  ' from dual union all
   select to_date('01.01.2025','dd.mm.yyyy'), to_date('01.05.2020','dd.mm.yyyy'),   'Floyd   ' from dual union all
   select to_date('01.01.2025','dd.mm.yyyy'), to_date('15.04.2020','dd.mm.yyyy'),   'Peterson' from dual union all
   select to_date('01.12.2024','dd.mm.yyyy'), to_date('20.03.2020','dd.mm.yyyy'),   'George  ' from dual union all
   select to_date('01.12.2024','dd.mm.yyyy'), to_date('10.03.2020','dd.mm.yyyy'),   'Smith   ' from dual union all
   select to_date('01.12.2024','dd.mm.yyyy'), to_date('15.01.2020','dd.mm.yyyy'),   'George  ' from dual union all
   select to_date('01.01.2025','dd.mm.yyyy'), to_date('15.12.2019','dd.mm.yyyy'),   'Lee     ' from dual union all
   select to_date('01.01.2025','dd.mm.yyyy'), to_date('01.11.2019','dd.mm.yyyy'),   'Alfonso ' from dual union all
   select to_date('01.01.2025','dd.mm.yyyy'), to_date('10.10.2019','dd.mm.yyyy'),   'Peterson' from dual union all
   select to_date('01.07.2025','dd.mm.yyyy'), to_date('30.09.2019','dd.mm.yyyy'),   'Smith   ' from dual union all
   select to_date('01.07.2024','dd.mm.yyyy'), to_date('20.09.2019','dd.mm.yyyy'),   'Lee     ' from dual union all
   select to_date('01.07.2024','dd.mm.yyyy'), to_date('10.09.2019','dd.mm.yyyy'),   'Meyer   ' from dual union all
   select to_date('01.07.2024','dd.mm.yyyy'), to_date('01.05.2019','dd.mm.yyyy'),   'Smith   ' from dual
)
select *
from your_tab
match_recognize (
   order by Change_Date
   measures
       first(Change_Date) as Change_Date,
       first(Market_Intro_Date) as Market_Intro_Date,
       first(Author     ) as Author     
   pattern (A B*)
   define
      b as Market_Intro_Date = prev(Market_Intro_Date)  and Change_Date>prev(Change_Date)
)
order by 1;

CHANGE_DATE         MARKET_INTRO_DATE   AUTHOR
------------------- ------------------- --------
2019-05-01 00:00:00 2024-07-01 00:00:00 Smith
2019-09-30 00:00:00 2025-07-01 00:00:00 Smith
2019-10-10 00:00:00 2025-01-01 00:00:00 Peterson
2020-01-15 00:00:00 2024-12-01 00:00:00 George
2020-04-15 00:00:00 2025-01-01 00:00:00 Peterson
2020-07-10 00:00:00 2025-06-01 00:00:00 Meyer

Answer 3

类似的事情应该起作用：

select market_intro_date, change_date, replace(author,'    *','') author
from table 
where author like '%*%';

Answer 4

您可以按以下方式使用not exists：

Select t.*
  From your_table t
 Where not exists (select 1 from your_table tt
                    Where t.Market_Intro_Date = tt.Market_Intro_Date
                      And tt.changed_date < t.changed_date)

SQL查询比较表中的行（Oracle DB）

4 个答案: