我遇到了在oracle中创建查询的问题,该问题似乎并不想加入缺失值
我拥有的表是:table myTable(refnum, contid, type)
values are:
1, 10, 90000
2, 20, 90000
3, 30, 90000
4, 20, 10000
5, 30, 10000
6, 10, 20000
7, 20, 20000
8, 30, 20000
我所追求的领域的分解是这样的:
select a.refnum from myTable a where type = 90000
select b.refnum from myTable b where type = 10000 and contid in (select contid from myTable where type = 90000)
select c.refnum from myTable c where type = 20000 and contid in (select contid from myTable where type = 90000)
我之后的查询结果是:
a.refnum, b.refnum, c.refnum
我认为这样可行:
select a.refnum, b.refnum, c.refnum
from myTable a
left outer join myTable b on (a.contid = b.contid)
left outer join myTable c on (a.contid = c.contid)
where a.id_tp_cd = 90000
and b.id_tp_cd = 10000
and c.id_tp_cd = 20000
所以值应该是:
1, null, 6
2, 4, 7
3, 5, 8
但它唯一的回归:
2, 4, 7
3, 5, 8
我认为左连接会显示左边的所有值,并为右边创建一个空值。
帮助:(
答案 0 :(得分:27)
你说左边连接将返回没有匹配的右边的空值是正确的,但是当你将这个限制添加到where子句时,你不允许返回这些空值:
and b.id_tp_cd = 10000
and c.id_tp_cd = 20000
您应该能够将这些放在连接的“on”子句中,因此只返回右侧的相关行。
select a.refnum, b.refnum, c.refnum
from myTable a
left outer join myTable b on (a.contid = b.contid and b.id_tp_cd = 10000)
left outer join myTable c on (a.contid = c.contid and c.id_tp_cd = 20000)
where a.id_tp_cd = 90000
答案 1 :(得分:2)
或使用Oracle语法而不是ansi
select a.refnum, b.refnum, c.refnum
from myTable a, mytable b, mytable c
where a.contid=b.contid(+)
and a.contid=c.contid(+)
and a.type = 90000
and b.type(+) = 10000
and c.type(+) = 20000;
REFNUM REFNUM REFNUM
---------- ---------- ----------
1 6
2 4 7
3 5 8