Question

对于以下数据集（超简化）：

表1

SQL> select
  2    listagg(s.department_name, ',') within group (order by null) result
  3  from departments s, departments d;
from departments s, departments d
     *
ERROR at line 3:
ORA-01489: result of string concatenation is too long


SQL>
SQL> select
  2  rtrim(xmlagg(xmlelement (e, s.department_name || ',')).extract
  3    ('//text()').getclobval(), ',') result
  4  from departments s, departments d;

RESULT
--------------------------------------------------------------------------------
Administration,Administration,Administration,Administration,Administration,Admin

SQL>

表2

ext_id, actions
z1,     100
z2,     100
x1,     200
x2,     200

我这样加入他们：

tid, aid, eid, aum
1,   a,   z1,  100
1,   b,   z2,  100
1,   c,   z3,  100
2,   d,   x1,  200
2,   e,   x2,  200
2,   f,   x3,  200

这会为表格create table #temp as select a.*, tid, aid, aum from #t a join #t2 b on a.ext_id=b.eid提供#temp中与#t2匹配的所有记录，并映射其中包含的其他字段。

我现在要将#t1与#temp（单独的查询）进行匹配，并获取基于{{1}加入的其余#t2的所有数据}：

即。输出应该是：

aid

谢谢

Answer 1

使用外部联接。我会这样做：

select a.*, tid, aid, aum
from @t2 b left join
     #t a
    on a.ext_id = b.eid
where a.ext_id is null;

加入缺失的记录

1 个答案: