上篇日志《ORACLE子句GROUP BY CUBE》中的问题稍作变化，现在每个字段有3个状态，SQL该如何写呢？这时CUBE的机制恐怕没有办法构成3的八次方行数据了，只能另寻办法。以下代码是以3列为例的一个比较啰嗦的方案，仅供参考。
(select 0 as a1, 0 as a2, 0 as a3
from dual
union all
select 1 as a1, 0 as a2, 0 as a3
from dual
union all
select 9 as a1, 0 as a2, 0 as a3 from dual)
select b.a1,decode(c.le,1,0,2,1,3,9) as a2,decode(d.le,1,0,2,1,3,9) as a3
from base b,
(select level as le
from dual
connect by level < (select count(1) + 1 from base)) c,
(select level as le
from dual
connect by level < (select count(1) + 1 from base)) d
order by a1,a2,a3
惭愧，有点低级错误啦。修改后代码如下：
with base as
(select 0 as a1, 0 as a2, 0 as a3
from dual
union all
select 1 as a1, 0 as a2, 0 as a3
from dual
union all
select 9 as a1, 0 as a2, 0 as a3 from dual),
mult as
(select level as le
from dual
connect by level < (select count(1) + 1 from base))
select b.a1,
decode(m1.le, 1, 0, 2, 1, 3, 9) as a2,
decode(m2.le, 1, 0, 2, 1, 3, 9) as a3
from base b, mult m1, mult m2
order by a1, a2, a3
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