不知道这个右外连接的结果是什么

Question

右外连接类似于维恩图的Union对吗？
我的意思是A right outer Join B我们应该得到B的所有行和A中的任何匹配行。
出于某种原因，我很困惑与以下：
假设表Orders：

mysql> select * from orders; 
+------------+------------+---------+----------+---------+ 
| orderedon | name  | partnum | quantity | remarks | 
+------------+------------+---------+----------+---------+ 
| 1996-05-19 | TRUE-WHEEL |  76 |  3 | PAID | 
| 1996-09-02 | TRUE-WHEEL |  10 |  1 | PAID | 
| 1996-06-30 | TRUE-WHEEL |  42 |  8 | PAID | 
| 1996-06-30 | BIKE SPEC |  54 |  10 | PAID | 
| 1996-05-30 | BIKE SPEC |  23 |  8 | PAID | 
| 1996-01-17 | BIKE SPEC |  76 |  11 | PAID | 
| 1996-01-17 | LE SHOPPE |  76 |  5 | PAID | 
| 1996-06-01 | LE SHOPPE |  10 |  3 | PAID | 
| 1996-06-01 | AAA BIKE |  10 |  1 | PAID | 
| 1996-07-01 | AAA BIKE |  76 |  4 | PAID | 
| 1996-07-01 | AAA BIKE |  46 |  14 | PAID | 
| 1996-07-11 | JACKS BIKE |  76 |  14 | PAID | 
| 1996-05-15 | TRUE-WHEEL |  23 |  6 | PAID | 
| 1996-05-30 | BIKE SPEC |  20 |  2 | PAID | 
+------------+------------+---------+----------+---------+ 
14 rows in set (0.00 sec) 

和表Part：

mysql> select * from part; 
+---------+---------------+---------+ 
| partnum | description | price | 
+---------+---------------+---------+ 
|  54 | PEDALS  | 54.25 | 
|  42 | SEATS   | 24.50 | 
|  46 | TIRES   | 15.25 | 
|  23 | MOUNTAIN BIKE | 350.45 | 
|  76 | ROAD BIKE  | 530.00 | 
|  10 | TANDEM  | 1200.00 | 
+---------+---------------+---------+ 
6 rows in set (0.00 sec) 

我期待的是，下面的查询：
select p.partnum p_partnum,p.description p_desc,p.price p_price,o.name o_name,o.partnum o_partnum from part p right outer join orders o on o.partnum=54;
会给我所有的行Orders，只有part那行ve partnum=54。
但我得到这个：

mysql> select p.partnum p_partnum,p.description p_desc,p.price p_price,o.name o_name,o.partnum o_partnum from part p right outer join orders o on o.partnum=54; 
+-----------+---------------+---------+------------+-----------+ 
| p_partnum | p_desc  | p_price | o_name  | o_partnum | 
+-----------+---------------+---------+------------+-----------+ 
|  NULL | NULL   | NULL | TRUE-WHEEL |  76 | 
|  NULL | NULL   | NULL | TRUE-WHEEL |  10 | 
|  NULL | NULL   | NULL | TRUE-WHEEL |  42 | 
|  54 | PEDALS  | 54.25 | BIKE SPEC |  54 | 
|  42 | SEATS   | 24.50 | BIKE SPEC |  54 | 
|  46 | TIRES   | 15.25 | BIKE SPEC |  54 | 
|  23 | MOUNTAIN BIKE | 350.45 | BIKE SPEC |  54 |  
|  76 | ROAD BIKE  | 530.00 | BIKE SPEC |  54 | 
|  10 | TANDEM  | 1200.00 | BIKE SPEC |  54 | 
|  NULL | NULL   | NULL | BIKE SPEC |  23 |  
|  NULL | NULL   | NULL | BIKE SPEC |  76 |  
|  NULL | NULL   | NULL | LE SHOPPE |  76 |  
|  NULL | NULL   | NULL | LE SHOPPE |  10 |  
|  NULL | NULL   | NULL | AAA BIKE |  10 |  
|  NULL | NULL   | NULL | AAA BIKE |  76 |  
|  NULL | NULL   | NULL | AAA BIKE |  46 |  
|  NULL | NULL   | NULL | JACKS BIKE |  76 | 
|  NULL | NULL   | NULL | TRUE-WHEEL |  23 |  
|  NULL | NULL   | NULL | BIKE SPEC |  20 |  
+-----------+---------------+---------+------------+-----------+ 
19 rows in set (0.00 sec) 

为什么我会得到额外的行？为什么它将Order和partnum=54这一行合并到零件的所有行？

Answer 1

FROM part p 
RIGHT OUTER JOIN orders o 
    ON o.partnum=54 

...只订购了一个条件，你需要补充的是一个条件，该部分也对应于订单或数据库将考虑任何部分匹配;

FROM part p 
RIGHT OUTER JOIN orders o 
    ON o.partnum=54 
AND o.partnum = p.partnum 

当然，如果你只想显示其中partnum = 54，你最好移动o.partnum=54到WHERE条件，而不是行。 JOIN条件通常用于连接表，WHERE通常用于过滤。

Answer 2

因为你必须要RIGHT JOIN与orders加盟条件partnum=54你会得到所有行orders加入了与parts时partnum=54。您有一行partnum=54，该行与parts的所有行（交叉连接）相连。

Answer 3

您在ON condition上使用o.partnum = 54，这就是为什么你会在结果中获得额外的行 。

您需要将条件o.partnum = 54放在where子句中。

这个怎么样

select p.partnum p_partnum,p.description p_desc,p.price p_price,o.name o_name,o.partnum o_partnum 
from part p right outer join orders o 
on o.partnum = p.partnum 
where o.partnum=54; 

编辑：您可参照本beautiful article结果集如何得到打扰当你在ON子句和Where 放状态。

Answer 4

您的查询

select p.partnum p_partnum,p.description p_desc,p.price p_price,o.name o_name,o.partnum o_partnum 
from part p right outer join orders o on o.partnum=54; 

你要为右连接所以即使没有匹配的加盟表格右侧会显示记录，正是在这样的情况发生您的查询案例

同样是左连接所有表格的左边的记录即使不匹配也会显示

参考http://www.w3schools.com/sql/sql_join_right.asp

详细解释

希望这有助于