Oracle 9.2 透视不同值

Question

pivot 函数可从 Oracle 11 获得，我将需要使用 Oracle 9.2 的类似结果。

主要论点是我需要在这样的表中旋转一些具有不同结果的值：

id      col3
1       a
1       b
--
2       a   
2       a
2       b
--
3       a
3       b
3       c  

我的结果应该是这样的

id      a       b       c
1       1       1       0
2       1       1       0
3       1       1       1

要创建“手动”枢轴，我正在使用案例/何时，但我无法理解如何获得不同的价值。

现在查询是这样的：

with t as 
    ( select 1 as id, 'a' as col1 from dual union all
      select 1 as id, 'b' from dual union all
      select 2 as id, 'a' from dual union all
      select 2 as id, 'a' from dual union all
      select 2 as id, 'b' from dual union all
      select 3 as id, 'a' from dual union all
      select 3 as id, 'b' from dual union all
      select 3 as id, 'c' from dual
)
select t.id, 
       count(case when t.col1 = 'a' then 1 end) a,
       count(case when t.col1 = 'b' then 1 end) b,
       count(case when t.col1 = 'c' then 1 end) c

这会产生正确的值，但显然它只是“计算”总 a/b/c 值而不是不同的值。

感谢支持

Answer 1

如果我正确理解您的需求，您可以尝试以下方法；它按id 聚合并计算col3 的不同值：

with t as 
( select 1 as id, 'a' as col1 from dual union all
  select 1 as id, 'b' from dual union all
  select 2 as id, 'a' from dual union all
  select 2 as id, 'a' from dual union all
  select 2 as id, 'b' from dual union all
  select 3 as id, 'a' from dual union all
  select 3 as id, 'b' from dual union all
  select 3 as id, 'c' from dual
)
select id,
       count(distinct decode (col1, 'a', id, null)) a,
       count(distinct decode (col1, 'b', id, null)) b,
       count(distinct decode (col1, 'c', id, null)) c
from t
group by id

当然查询取决于col3的不同值的个数，但这和pivot是同一个问题。