Sql 如何获取其他表中不存在的值，忽略具有最新日期的重复值

在给定ff的情况下，如何仅获取表A中不存在于表B中的水平_20值。情景：

如果带有代码\u级别\u 20的奖品在A中具有相同的级别\u 20值，请忽略它！ 如果有多个奖励值，请将该值与B中的最大日期进行比较

表A

表B

表C

预期产出：

我的错误代码：

---

由于预期的输出和表中的数据，使用此查询而不包括表b已足够：

with a(ALIAS, LEVEL_20, POINTS, "DATE") as
(
 select 'Jbax',300,325,date'2018-06-20' from dual union all
 select 'Cvg2',100,103,date'2018-06-20' from dual union all
 select 'Deg1',200,281,date'2018-06-20' from dual         
),   b(ALIAS, CODE_LEVEL, PRIZE, "DATE") as
(
 select 'Jbax','Level_20',500,date'2017-01-15' from dual union all
 select 'Jbax','Level_10',200,date'2017-03-20' from dual union all
 select 'Cvg2','Level_20',100,date'2017-04-05' from dual union all
 select 'Deg1',null,null,null  from dual union all    
 select 'Cvg2','Level_20',50,date'2017-02-01'  from dual
),   c(ALIAS,NAME) as
(
 select 'Jbax','Jessie' from dual union all 
 select 'Cvg2','Carol' from dual union all
 select 'Deg1','Danny' from dual
)  
select c.name, a.level_20
  from c
  join a on a.alias = c.alias  
 where (a.alias,a.level_20) not in 
      (select a.alias,b.prize 
         from a
         join b on b.prize = a.level_20 and b.code_level = 'Level_20');

 NAME   LEVEL_20
 ------ --------
 Jessie 300
 Danny  200

另外，表b中没有多个奖项

---

由于预期的输出和表中的数据，使用此查询而不包括表b已足够：

with a(ALIAS, LEVEL_20, POINTS, "DATE") as
(
 select 'Jbax',300,325,date'2018-06-20' from dual union all
 select 'Cvg2',100,103,date'2018-06-20' from dual union all
 select 'Deg1',200,281,date'2018-06-20' from dual         
),   b(ALIAS, CODE_LEVEL, PRIZE, "DATE") as
(
 select 'Jbax','Level_20',500,date'2017-01-15' from dual union all
 select 'Jbax','Level_10',200,date'2017-03-20' from dual union all
 select 'Cvg2','Level_20',100,date'2017-04-05' from dual union all
 select 'Deg1',null,null,null  from dual union all    
 select 'Cvg2','Level_20',50,date'2017-02-01'  from dual
),   c(ALIAS,NAME) as
(
 select 'Jbax','Jessie' from dual union all 
 select 'Cvg2','Carol' from dual union all
 select 'Deg1','Danny' from dual
)  
select c.name, a.level_20
  from c
  join a on a.alias = c.alias  
 where (a.alias,a.level_20) not in 
      (select a.alias,b.prize 
         from a
         join b on b.prize = a.level_20 and b.code_level = 'Level_20');

 NAME   LEVEL_20
 ------ --------
 Jessie 300
 Danny  200

另外，表b中没有多个奖项

---

这将为您提供预期的结果：

    with get_max_value as 
    (
     select b2.*, max(b2."DATE") over (partition by b2.ALIAS)  mdate from b b2
    )
    select my_c.name, my_a.level_20
      from c my_c join a my_a on my_a.alias = my_c.alias  
     where (my_a.alias,my_a.level_20) not in 
          (select a2.alias, gmv.prize 
             from a a2 join get_max_value gmv on gmv.prize = a2.level_20 
             and gmv.code_level = 'Level_20' 
             WHERE gmv."DATE" = gmv.mdate);

---

我想你的意思是相同的别名，而不是相同的价格。在我看来，相同的价格毫无意义，所以我只是用相同的别名分组。如果您需要最高价格，那么其他答案中的建议就可以了。

这将为您提供预期结果：

    with get_max_value as 
    (
     select b2.*, max(b2."DATE") over (partition by b2.ALIAS)  mdate from b b2
    )
    select my_c.name, my_a.level_20
      from c my_c join a my_a on my_a.alias = my_c.alias  
     where (my_a.alias,my_a.level_20) not in 
          (select a2.alias, gmv.prize 
             from a a2 join get_max_value gmv on gmv.prize = a2.level_20 
             and gmv.code_level = 'Level_20' 
             WHERE gmv."DATE" = gmv.mdate);

---

我想你的意思是相同的别名，而不是相同的价格。在我看来，相同的价格毫无意义，所以我只是用相同的别名分组。如果您需要最高价格，那么其他答案中的建议也可以。

此查询在表b中添加排序：

结果:

ALIAS   LEVEL_20 NAME   CODE_LEVEL      PRIZE DATE_               RN
----- ---------- ------ ---------- ---------- ----------- ----------
Cvg2         100 Carol  Level_20          100 2017-04-05           1
Cvg2         100 Carol  Level_20           50 2017-02-01           2
Deg1         200 Danny                                             1
Jbax         300 Jessie Level_20          500 2017-01-15           1
pqrs         100 Peter  Level_20          200 2017-04-05           1
pqrs         100 Peter  Level_20          100 2017-02-01           2

因此，现在您只能获取最新rn=1的行，这些行的prize为空，或者prize与a.level_20不同：

---

我在那里添加了一行

此查询在表b中添加排序：

结果:

ALIAS   LEVEL_20 NAME   CODE_LEVEL      PRIZE DATE_               RN
----- ---------- ------ ---------- ---------- ----------- ----------
Cvg2         100 Carol  Level_20          100 2017-04-05           1
Cvg2         100 Carol  Level_20           50 2017-02-01           2
Deg1         200 Danny                                             1
Jbax         300 Jessie Level_20          500 2017-01-15           1
pqrs         100 Peter  Level_20          200 2017-04-05           1
pqrs         100 Peter  Level_20          100 2017-02-01           2

因此，现在您只能获取最新rn=1的行，这些行的prize为空，或者prize与a.level_20不同：

---

我在那里添加了一行

谢谢，但我所说的多重奖励值是指Cvg2在表b中有100和50-我想将a中的20级值与表b中的最大日期进行比较。假设我将Cvg2从表b中的100和50更改为200和100，您的代码将排除Carol-但她必须在输出中，因为200 b的最大日期为100 aThanks，但我所说的多重奖励值，Cvg2在表b中有100和50-在这种情况下，我想将a中的水平_20值与表b中的最大日期进行比较。假设我将Cvg2从表b中原来的100和50更改为200和100，表b中的日期相同，您的代码将排除Carol-但她必须在输出中，因为200 b的最大日期为100，这正是我所需要的，谢谢，我很高兴您理解了我的要求！正是我需要的，谢谢我很高兴你理解我的要求！

    with get_max_value as 
    (
     select b2.*, max(b2."DATE") over (partition by b2.ALIAS)  mdate from b b2
    )
    select my_c.name, my_a.level_20
      from c my_c join a my_a on my_a.alias = my_c.alias  
     where (my_a.alias,my_a.level_20) not in 
          (select a2.alias, gmv.prize 
             from a a2 join get_max_value gmv on gmv.prize = a2.level_20 
             and gmv.code_level = 'Level_20' 
             WHERE gmv."DATE" = gmv.mdate);

select a.alias, a.level_20, c.name, b.code_level, b.prize, b.date_,
       row_number() over (partition by a.alias, c.name order by b.date_ desc) rn
  from a join c on c.alias = a.alias
  left join b on b.alias = a.alias and b.code_level = 'Level_20'

ALIAS   LEVEL_20 NAME   CODE_LEVEL      PRIZE DATE_               RN
----- ---------- ------ ---------- ---------- ----------- ----------
Cvg2         100 Carol  Level_20          100 2017-04-05           1
Cvg2         100 Carol  Level_20           50 2017-02-01           2
Deg1         200 Danny                                             1
Jbax         300 Jessie Level_20          500 2017-01-15           1
pqrs         100 Peter  Level_20          200 2017-04-05           1
pqrs         100 Peter  Level_20          100 2017-02-01           2

select name, level_20 
  from (
    select a.alias, a.level_20, c.name, b.code_level, b.prize, b.date_,
           row_number() over (partition by a.alias, c.name order by b.date_ desc) rn
      from a join c on c.alias = a.alias
      left join b on b.alias = a.alias and b.code_level = 'Level_20')
  where rn = 1 and (prize is null or prize <> level_20)