您必须自行加入表格。假设你的表被称为configs
(因为你没有告诉我们它叫什么),这样的事情应该工作:
select t1.item, t1.value1, t1.value2, t2.item, t2.value1, t2.value2
from configs as t1
inner join configs as t2
on t2.value1 = t1.value1 and t2.item = 'reg'
where t1.item = 'class';
如果你需要的所有行已不匹配的结果返回以及reg
行,将inner join
更改为left outer join
。如果您希望此查询性能良好,请确保您在item
和value1
上有索引。
这里是概念的快速证据显示上述作品查询:
mysql> create table configs (
-> id int unsigned primary key auto_increment,
-> item varchar(32) not null,
-> value1 varchar(32) not null,
-> value2 varchar(32) not null,
-> index (item),
-> index (value1)
->) engine=innodb charset=utf8;
Query OK, 0 rows affected (0.01 sec)
mysql> insert into configs (id, item, value1, value2) values
-> (2, 'class', 'ship', 'bow'),
-> (3, 'class', 'car', 'tires'),
-> (5, 'reg', 'ship', 'level1'),
-> (7, 'reg', 'ship', 'level2'),
-> (9, 'reg', 'car', 'level5');
Query OK, 5 rows affected (0.00 sec)
Records: 5 Duplicates: 0 Warnings: 0
mysql> select * from configs;
+----+-------+--------+--------+
| id | item | value1 | value2 |
+----+-------+--------+--------+
| 2 | class | ship | bow |
| 3 | class | car | tires |
| 5 | reg | ship | level1 |
| 7 | reg | ship | level2 |
| 9 | reg | car | level5 |
+----+-------+--------+--------+
5 rows in set (0.00 sec)
mysql> select t1.item, t1.value1, t1.value2, t2.item, t2.value1, t2.value2
-> from configs as t1
-> inner join configs as t2
-> on t2.value1 = t1.value1 and t2.item = 'reg'
-> where t1.item = 'class';
+-------+--------+--------+------+--------+--------+
| item | value1 | value2 | item | value1 | value2 |
+-------+--------+--------+------+--------+--------+
| class | ship | bow | reg | ship | level1 |
| class | ship | bow | reg | ship | level2 |
| class | car | tires | reg | car | level5 |
+-------+--------+--------+------+--------+--------+
3 rows in set (0.00 sec)
有时更容易做到这一点的东西在代码中,不是SQL – 2012-02-26 20:38:10
@Toby艾伦:是的,但是在这种情况下,SQL是非常直截了当。 – Asaph 2012-02-26 20:41:40