oracle - oracle左外连接不显示右空值

Question

我在 oracle 中创建一个似乎不想加入缺失值的查询时遇到问题

我的桌子是这样的：

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

我所追求的领域的细分是这样的：

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)

我要查询的结果是这样的：

a.refnum, b.refnum, c.refnum

我认为这会起作用：

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

但它唯一的回报：

2, 4, 7
3, 5, 8

我认为左连接会显示左侧的所有值并为右侧创建一个空值。

帮助 ：（

Answer 1

您说的是正确的，左连接将在没有匹配的地方返回空值，但是当您将此限制添加到 where 子句时，您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

Answer 2

或者使用 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