自联接查询

Question

请考虑下表:

mysql> select * from phone_numbers;
+-------------+------+-----------+
| number      | type | person_id |
+-------------+------+-----------+
| 17182225465 | home |         1 |
| 19172225465 | cell |         1 |
| 12129876543 | home |         2 |
| 13049876543 | cell |         2 |
| 15064223454 | home |         3 |
| 15064223454 | cell |         3 |
| 18724356798 | home |         4 |
| 19174335465 | cell |         5 |
+-------------+------+-----------+

我试图找到那些有家用电话而不是电池的人.

此查询有效:

mysql> select h.*
    -> from phone_numbers h
    -> left join phone_numbers c
    -> on h.person_id = c.person_id
    -> and c.type = 'cell'
    -> where h.type = 'home'
    -> and c.number is null;
+-------------+------+-----------+
| number      | type | person_id |
+-------------+------+-----------+
| 18724356798 | home |         4 |
+-------------+------+-----------+

但这个没有:

mysql> select h.*
    -> from phone_numbers h
    -> left join phone_numbers c
    -> on h.person_id = c.person_id
    -> and h.type = 'home'
    -> and c.type = 'cell'
    -> where c.number is null;
+-------------+------+-----------+
| number      | type | person_id |
+-------------+------+-----------+
| 19172225465 | cell |         1 |
| 13049876543 | cell |         2 |
| 15064223454 | cell |         3 |
| 18724356798 | home |         4 |
| 19174335465 | cell |         5 |
+-------------+------+-----------+

两者之间的唯一区别是h.type = 'home'条件的位置- 在第一个where条款中,第二个on条款是条款的一部分.

为什么第二个查询返回与第一个查询相同的结果？

Answer 1

在第二个SQL中,条件h.type ='home'是外连接条件的一部分,并且不是结果的过滤器.对于h.type ='cell'的所有记录,条件h.type ='home'为FALSE,因此找不到"匹配"c行 - 所以c.number为null,这是您唯一的过滤(WHERE)条件.

在伪代码中,您的第二个SQL的工作方式如下:

for each row in phone_numbers h /* Note this is ALL home AND cell phones */
   select c.number from phone_numbers c
   where h.person_id = c.person_id
   and h.type = 'home'
   and c.type = 'cell';
   if c.number is null (i.e. no row found)
     display h.*
   end if
end loop;