Question

这是查询的结果。有许多组合，但最多三个与一个人相关联。

NAME     ID    GROUPA
=====================
James    20    A
James    20    B
James    20    D
Michael  30    A
Michael  30    B
Michael  30    C
Jordan   40    I

我想要的是......

NAME     ID    GROUP1     GROUP2     GROUP3
===========================================
James    20    A          B          D
Michael  30    A          B          C
Jordan   40    I

我尝试了这个查询，但它不起作用......

SELECT NAME, ID, GROUPA
FROM 
    (SELECT NAME, ID, 
        ROW_NUMBER() OVER (PARTITION BY GROUPA ORDER BY EMPLOYEE_ID) AS GROUP
    FROM TABLE1)
GROUP BY NAME, ID, GROUPA

感谢您的指导。

Answer 1

select name,id, 
 max(case when groupa = 'A' then groupa end) as group1,
  max(case when groupa = 'B' then groupa end) as group2,
  max( case when groupa = 'C' then groupa end) as group3
   from tablename
   group by name, id

如果groupa的数量是固定的，则上述查询将起作用。

编辑：使用数据透视

 select * from
  (select name, id , groupa from tablename)
  pivot xml (
  max(groupa) for groupa in
   (select distinct groupa from tablename)
   )

感谢您的解决方案。我快到了。这是我在运行此查询后得到的。实际上有53种不同的GROUPS，因此它增加了53列，但分配给用户的最大组数为5。

NAME      ID      A      B      C     D      E      F      G      H
James     20      A      null   null  null   null   null   null   H
Michael   30      A      B      null  null   E      null   null   null

我如何得到这样的结果......

NAME      ID      GROUP_1  GROUP_2  GROUP_3
James     20      A        H
Michael   30      A        B        E

如我在问题中所提到的，如何得到我的结果？谢谢，

Answer 2

您可以使用条件聚合和row_number()：

执行此操作

select name, id,
       max(case when seqnum = 1 then groupa end) as group1,
       max(case when seqnum = 2 then groupa end) as group2,
       max(case when seqnum = 3 then groupa end) as group3
from (select t.*,
             row_number() over (partition by name order by employee_id) as seqnum
      from table1 t
     ) t
group by name, id;

ORACLE中的列到列

2 个答案: