表M数据如下：
| SEQ_NO | TYPE_NO | MONEY |
| A | 1 | 10 |
| A | 2 | 20 |
| B | 5 | 30 |
表N数据如下：
| TYPE_NO |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
M表和N表通过type_no关联，想得到以下结果：
| SEQ_NO | TYPE_NO | MONEY |
| A | 1 | 10 |
| A | 2 | 20 |
| A | 3 | 0 |
| A | 4 | 0 |
| A | 5 | 0 |
| B | 1 | 0 |
| B | 2 | 0 |
| B | 3 | 0 |
| B | 4 | 0 |
| B | 5 | 30 |
select nvl(t1.seqno, 'A'), t2.typeno, nvl(t1.money, 0)
from (select * from t1 where t1.seqno = 'A') t1
full join T2
on t1.typeno = t2.typeno union
select nvl(t1.seqno, 'B'), t2.typeno, nvl(t1.money, 0)
from (select * from t1 where t1.seqno = 'B') t1
full join T2
on t1.typeno = t2.typeno;
基本算是能实现结果，不知是否满足