0

我正在寻找一种通过跨多行存在相同外键来聚合的方法。

例如，ID 号 1 和 2 应该聚合在一起，因为它们在所有行中都具有完全相同的外键，而 ID 号 3 在所有行中没有相同的外键，因此应该分开。

注意：使用 DB2

示例源数据：

ID  QUANTITY    COLOR
1   10          BLUE
1   10          RED
1   10          GREEN
2   30          BLUE
2   30          RED
2   30          GREEN
3   15          GREEN
3   15          YELLOW

期望的结果集：

TEMP_ID     SUMQTY  COLOR
1           40      BLUE
1           40      RED
1           40      GREEN
2           15      GREEN
2           15      YELLOW
4

1 回答 1

1

首先，我们需要确定两个“id”的“颜色”是否相同。以下查询对 进行完全外连接color，然后按 聚合id。如果两个ids 具有相同的颜色，则完整的外连接始终匹配——没有 NULL：

select s1.id, s2.id
from s s1 full outer join
     s s2
     on s1.color = s2.color
group by s1.id, s2.id
having sum(case when s1.color is null then 1 else 0 end) = 0 and
       sum(case when s2.color is null then 1 else 0 end) = 0

Using the same idea, we can assign the minimum s1.id as the "id" of the group. This group id then gives us the information needed for the final aggregation:

select s3.groupid, sum(s3.quantity) as quantity, s3.color as color
from (select min(s1.id) as groupid, s2.id
      from (select s1.id, s2.id
            from s s1 full outer join
                 s s2
                 on s1.color = s2.color
            group by s1.id, s2.id
            having sum(case when s1.color is null then 1 else 0 end) = 0 and
                   sum(case when s2.color is null then 1 else 0 end) = 0
           ) ss
      group by s2.id
     ) sg join
     s s3
     on sg.id = s.id
group by sg.groupid, s3.color

In DB2 you can also do this with listagg. First you need to get the list of colors to identify commonality, and then use it:

select min(s.id) as groupid, sum(s.quantity) as quantity, s.color
from (select id, listagg(color order by color) as thegroup
      from s
      group by id
     ) sg join
     s
     on sg.id = s.id
group by sg.thegroup, s.color

This is probably more efficient than the first solution.

于 2013-03-31T00:46:16.133 回答