我在oracle中创建一个查询的问题似乎并不想加入缺失的值
我有的表是这样的:
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这样的:
a.refnum, b.refnum, c.refnum
我认为这会奏效:
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
所以价值应该是:
1, null, 6 2, 4, 7 3, 5, 8
但它唯一的回归:
2, 4, 7 3, 5, 8
我认为左连接将显示左侧的所有值并为右侧创建一个null.
救命 :(
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