甲骨文左外連接沒有顯示正確的NULL值

Question

我遇到了在Oracle中創建一個查詢其犯規的問題似乎想加入失蹤值

表我已經是這樣的：

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 

休息後，我是場下是這樣的：

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) 

查詢後，我敢的結果是這樣的：

a.refnum, b.refnum, c.refnum 

我認爲這會工作：

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 

因此它們的值應該是：

1, null, 6 
2, 4, 7 
3, 5, 8 

，但其唯一的返回：

2, 4, 7 
3, 5, 8 

我想離開加入會顯示所有值在左側，併爲右側創建一個空值。

幫助:(

Answer 1

你是說，左正確的連接將返回空值對於存在不匹配的權利，但你是不是讓你當這個限制添加到您的where子句中返回這些零點：

and b.id_tp_cd = 10000 
and c.id_tp_cd = 20000 

你應該能夠把這些在「上」的聯接，而不是，返回右側所以只有相關行的條款

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語法代替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