oracle 一对一左外连接
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ONE TO ONE LEFT OUTER JOIN
提问by Sheldon
I was wondering, is there a way to make a kind of one to one left outer join:
我想知道，有没有办法进行一种一对一的左外连接：
I need a join that matches say table A with table B, for each record on table A it must search for its pair on table B, but there exists only ONE record that match that condition, so when it found its pair on B, it must stop and continue with the next row at table A.
我需要一个连接来匹配表 A 和表 B，对于表 A 上的每条记录，它必须在表 B 上搜索它的对，但是只有一条记录匹配该条件，所以当它在 B 上找到它的对时，它必须停止并继续表 A 的下一行。
What I have is a simple LEFT OUTER JOIN.
我拥有的是一个简单的左外连接。
select * from A LEFT OUTER JOIN B ON A.ID = B.ID ORDER BY (NAME) ASC
Thanks in advance!
提前致谢！
采纳答案by Adam Robinson
The syntax you present in your question is correct. There is no difference in the query for joining on a one-to-one relationship than on a one-to-many.
您在问题中提出的语法是正确的。加入一对一关系的查询与加入一对多关系的查询没有区别。
回答by HLGEM
SQL doesn't work this way. In the first place it does not look at things row-by-row. In the second place what defines the record you want to match on?
SQL 不是这样工作的。首先，它不会逐行查看事物。其次，什么定义了您要匹配的记录？
Assuming you don't really care which row is selcted, something like this might work:
假设您并不真正关心选择了哪一行，这样的事情可能会起作用：
SELECT *
From tableA
left outer join
(select b.* from tableb b1
join (Select min(Id) from tableb group by id) b2 on b1.id - b2.id) b
on a.id = b.id
BUt it still is pretty iffy that you wil get the records you want when there are multiple records with the id in table b.
但是，当表 b 中有多个带有 id 的记录时，您是否会获得所需的记录仍然很不确定。
