ORACLE中的行到列

Question

这是查询的结果。有很多组，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

我试过这个查询，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，上面的查询就可以工作。

编辑：使用数据透视表

 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。这是我在运行这个查询之后得到的。实际上有 53 个不同的组，所以它添加了 53 列，但分配给用户的最大组数是 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

如何获得这样的结果...

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

如何获得我在问题中提到的结果？谢谢，

Answer 2

您可以使用条件聚合和 row_number():

来做到这一点

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中的行到列

Rows to Column in ORACLE

sql

oracle

pivot