CASE WHEN 给出 2 行重复 ID

Question

我写了一个查询，它会产生如下结果。

select t1.id as id,
CASE
    WHEN t3.key_id is not null and t3.key_id='SILICA'
    THEN t3.id
    END as SILICA_ID,
CASE
    WHEN t3.key_id is not null and t3.key_id='Mutual'
    THEN t3.id
    END as Mutual_ID
from table1 t1, table2 t2, table3 t3 
where t1.id=t2.t1_id and t2.id=t3.t2_id and t3.key_id in ('SILICA','Mutual') 
order by t1.id;

id SILICA_ID Mutual_id == ========= ========= 1 空 1234 1 3456 空 2 空 7890 2 2374 空 2 4587 空 2 空 3489

但是，我需要如下结果。

id SILICA_ID Mutual_id == ========= ========= 1 3456 1234 2 2345 7890 2 4587 3489

我怎样才能做到这一点？

提前致谢

Answer 1

每个t1 可以有多个t2，每个t2 可以有多个t3。但是您希望每个 t1.id 有一个结果行。所以你想聚合你的数据。

您的结果仅显示每个t1.id 和t3.key_id 一个t3.id。这是出乎意料的，表明数据模型错误或样本数据不足。

无论如何，为了每个t1.id 聚合每个t1.id 获得一个结果行并使用条件聚合：

select t1.id as id,
  max(case when t3.key_id = 'SILICA' THEN t3.id end) as silica_id,
  max(case when t3.key_id = 'Mutual' then t3.id end) as mutual_id
from t1
join t2 on t2.t1_id = t1.id 
join t3 on t3.t2_id = t2.id and t3.key_id in ('SILICA','Mutual') 
group by t1.id
order by t1.id;

更新：您更改的请求显示结果行，其中 SILICA/Mutual 对似乎没有直接关系。您似乎更想以类似报纸的方式列出它们。您可以通过分别选择“SILICA”和“Mutual”来执行此操作，为它们提供行号并加入这些行号。

select t1_id, s.t3_id as silica_id, m.t3_id as mutual_id 
  t1.id as t1_id, 
    t3.id as t3_id
(
  select distinct 
    t1.id as t1_id, 
    t3.id as t3_id,
    row_number() over (partition by t1.id order by t3.id) as rn
  from t1
  join t2 on t2.t1_id = t1.id 
  join t3 on t3.t2_id = t2.id and t3.key_id = 'SILICA'
) s
full outer join
(
  select distinct 
    t1.id as t1_id, 
    t3.id as t3_id,
    row_number() over (partition by t1.id order by t3.id) as rn
  from t1
  join t2 on t2.t1_id = t1.id 
  join t3 on t3.t2_id = t2.id and t3.key_id = 'Mutual'
) m using (t1_id, rn);

Answer 2

您可以为此使用pivot 语法：

select *
from   (
      select     t1.id, 
                 t3.id as t3_id, 
                 t3.key_id
      from       tabl t1
      inner join tabl t2 on t1.id = t2.t1_id
      inner join tabl t3 on t2.id = t3.t2_id
      where      t3.key_id in ('SILICA', 'Mutual')
) pivot (
      max(t3_id)
      for key_id in ('SILICA' as silica_id, 'Mutual' as mutual_id)
);

注意：使用join 子句来表达连接条件是一个好习惯，而不是将它们放在where 子句中。

Answer 3

您需要一个聚合解决方案，例如

select id
       , max(silica_id) as silica_id
       , max(Mutual_ID) as Mutual_ID
from (
    select t1.id as id,
    CASE
        WHEN t3.key_id is not null and t3.key_id='SILICA'
        THEN t3.id
        END as SILICA_ID,
    CASE
        WHEN t3.key_id is not null and t3.key_id='Mutual'
        THEN t3.id
        END as Mutual_ID
    from table t1, table t2, table t3 
    where t1.id=t2.t1_id and t2.id=t3.t2_id and t3.key_id in ('SILICA','Mutual')
    ) 
group by id
order by id;