Question

我有两个表A（group_id, id, subject）和B（id, date）。以下是ID上的表A和B的联合表。我尝试使用distinct和partition删除仅在group_id（field）中的重复项，但没有运气：

我的代码：

select 
    a.group_id, a.id, a.subject, b.date 
from
    A a 
inner join
    (select 
         b.*, 
         row_number() over (partition by group_id order by date asc) as seqnum
     from 
         B b) b on a.id = b.id and seqnum = 1
order by
    date desc;

运行代码时出现此错误：

不能在第1行的“从B到seqnum的按group_id的按日期排序asc）的查询附近单独使用分区

这是我的预期结果：

先谢谢您！

Answer 1

您似乎想要显示的表中每一行的最早日期。您的问题提到了两个表，但您只显示了一个。

我建议在大多数数据库中使用相关子查询：

select b.*
from b
where b.date = (select min(b2.date)
                from b b2
                where b2.group_id = b.group_id
               );

我明白了。您需要先join，然后再使用row_number()：

select ab.*
from (select a.group_id, a.id, a.subject, b.date,
             row_number() over (partition by a.group_id order by b.date) as seqnum
      from A a join
           B b
           on a.id = b.id
     ) ab
where seqnum = 1
order by date desc;

Answer 2

您快到了。但是您尝试用于分区的列（即group_id）来自表a，该表在子查询中不可用。

您需要JOIN并在子查询中分配行号，并在外部查询中过滤 then 。

select * 
from (
    select 
        a.group_id, 
        a.id, 
        a.subject, 
        b.date,
        row_number() over (partition by a.group_id order by b.date asc) as seqnum
    from a
    inner join b on ON a.id = b.id 
)
where seqnum = 1
ORDER BY date desc;

Answer 3

另一种实现目标的方法，虽然可能不是有效的方法

SELECT
  A.group_id, A.id, B.Date, A.subject
FROM A
INNER JOIN B
ON A.Id = B.Id
INNER JOIN
(
  SELECT
    A.Group_id, MIN(B.Date) AS Date
  FROM A
  INNER JOIN B
  ON A.Id = B.Id
  GROUP BY A.group_id
) AS supportTable
ON A.group_id = supportTable.group_id
AND B.Date = supportTable.Date

从SQL中的列中删除重复的值

3 个答案: