Question

首先，我将向您展示表格的架构。

 Table "public.questionare"
      Column        |          Type         |                           
--------------------+-----------------------+
 id                 | integer               | 




Table "public.questionareacceptance"
         Column          |          Type         |                         
-------------------------+-----------------------+
 id                      | integer               | 
 questionare_id          | integer               |
 accept_id               | integer               |
 impact_id               | integer               |

表questionareacceptance包含：

id  | questionare_id | accept_id|       impact_id  |
----+----------------+----------+------------------+
1   |1               |1         |                  |
2   |1               |1         | 1                |
3   |1               |1         | 1                |
4   |2               |          | 1                |
5   |3               |1         | 1                |
6   |4               |1         | 1                |
7   |4               |1         | 1                |

我想要获得的是questionare ID列表，其中questionareacceptance个字段accept_id和impact_id中的NULL不是SELECT q.id AS quest, qa.id AS accepted FROM questionare q, questionareacceptance qa WHERE q.id = qa.questionare_id AND qa.accept_id IS NOT NULL AND qa.impact_id IS NOT NULL;

我的查询如下：

      quest         |          accepted     |                           
--------------------+-----------------------+
 1                  |1                      | 
 1                  |2                      | 
 1                  |3                      | 
 2                  |4                      | 
 3                  |5                      | 
 4                  |6                      | 
 4                  |7                      |

但结果就像休耕一样：

但应返回的结果仅为4，impact_id其他人的accept_id或{{1}}为空。

有人能指出我在做错的地方吗？

Answer 1

你的查询可能写得不存在：

select
    q.id as quest, qa.id as accepted
from questionare as q
    inner join questionareacceptance as qa on qa.questionare_id = q.id
where
    not exists (
        select *
        from questionareacceptance as tqa
        where
           tqa.questionare_id = q.id and
           (tqa.accept_id is null or tqa.impact_id is null)
    )

但我认为使用窗口函数会更快：

with cte as (
    select
        q.id as quest, qa.id as accepted,
        sum(case when qa.accept_id is not null and qa.impact_id is not null then 1 else 0 end) over(partition by q.id) as cnt1,
        count(*) over(partition by q.id) as cnt2
    from questionare as q
        inner join questionareacceptance as qa on qa.questionare_id = q.id
)
select quest, accepted
from cte
where cnt1 = cnt2

实际上看起来你根本不需要加入：

with cte as (
    select
        qa.questionare_id as quest, qa.id as accepted,
        sum(case when qa.accept_id is not null and qa.impact_id is not null then 1 else 0 end) over(partition by qa.questionare_id) as cnt1,
        count(*) over(partition by qa.questionare_id) as cnt2  
    from questionareacceptance as qa
)
select quest, accepted
from cte
where cnt1 = cnt2;

<强> sql fiddle demo

多对一关系，只选择祖先符合所有条件的行

1 个答案: