如何在oracle中将特定id的不同值放入单行？

Question

我有一个下表

y 表

id     val
--     ---
1       A 
1       B 
1       C 
2       B
2       C
3       A

需要像下面这样的输出

 sno   val1    va2       val3
-----  ----   -----     ------
  1      A     B          C   
  2      B     C          -
  3      A     -          -

我用 self join 写了查询

SELECT t1.id as sno, t1.val as val1, t2.val as va2, t3.val as val3
  FROM y t1 JOIN y t2 ON t1.id = t2.id
  JOIN y t3 ON t2.id = t3.id
  ORDER BY 1,2,3,4;

但我像笛卡尔加入一样返回。 如何获得正确的结果。

Answer 1

您可以使用row_number()和条件聚合：

select id,
       max(case when seqnum = 1 then val end) as val_1,
       max(case when seqnum = 2 then val end) as val_2,
       max(case when seqnum = 3 then val end) as val_3
from (select t.*
             row_number() over (partition by id order by val) as seqnum
      from t
     ) t
group by id;

Answer 2

根据您的 Oracle 版本，您可以通过不同的方式解决此问题。 您可以修复您的自联接（您需要左外联接，而不是像您那样的内联接 - 使用内联接，即使在修复其余代码之后，您也只会获得具有输入表中所有三个值的 ID） . 正如 Gordon Linoff 所示，您可以使用条件聚合； 这是以前在 Oracle 10.2 版之前完成的透视（您所询问的操作的名称）的方式。 从 11.1 版开始，您可以对pivot运算符执行相同操作。

从 Oracle 12.1 开始，您可以使用match_recognize子句执行相同操作。 方法如下：

create table y (id, val) as 
  select 1, 'A'  from dual union all
  select 1, 'B'  from dual union all 
  select 1, 'C'  from dual union all
  select 2, 'B'  from dual union all
  select 2, 'C'  from dual union all
  select 3, 'A'  from dual
;

select id, val1, val2, val3
from   y
match_recognize(
  partition by id
  order     by val
  measures  r1.val as val1, r2.val as val2, r3.val as val3
  pattern   (^ r1 r2? r3?)
  define    r1 as null is null
);

  ID VAL1 VAL2 VAL3
---- ---- ---- ----
   1 A    B    C
   2 B    C
   3 A

Answer 3

如果你真的想使用 self join 来实现它，那么你需要使用group by和self join如下：

SELECT t1.id as sno, min(t1.val) as val1, min(t2.val) as va2, min(t3.val) as val3
  FROM y t1 
  LEFT JOIN y t2 ON t1.id = t2.id AND t1.val < t2.val
  LEFT JOIN y t3 ON t2.id = t3.id And t2.val < t3.val
  Group by t1.id

干杯！！