【发布时间】:2013-01-04 21:23:56
【问题描述】:
我有以下表格:
mysql> show columns from Person;
+------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+------------+--------------+------+-----+---------+-------+
|guid | varchar(255) | NO | PRI | NULL | |
+------------+--------------+------+-----+---------+-------+
mysql> show columns from Person_Func;
+-----------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-----------+--------------+------+-----+---------+-------+
| Person_id | varchar(255) | NO | PRI | NULL | |
| Func_id | varchar(255) | NO | PRI | NULL | |
+-----------+--------------+------+-----+---------+-------+
mysql> show columns from Func;
+-------------+--------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------------+--------------+------+-----+---------+-------+
| entry | varchar(255) | NO | | NULL | |
| description | varchar(255) | NO | | NULL | |
| Guid | varchar(255) | NO | PRI | NULL | |
+-------------+--------------+------+-----+---------+-------+
Symfony 类 Person 包含使用连接表 Person_Func 与 Func 的一对多关系(一个人 - 几个 Func)。我想查询拥有多个 Funcs 的 Person - 史蒂夫(a,b,c);约翰 (a, b, d); ele (b, d) - 我查询 (a, b) 函数,应该返回 steve 和 john。
现在我只是遍历所有人员并查询函数 - 这非常非常慢。你能帮帮我吗?
UPD 我已经成功了
SELECT DISTINCT d1.guid from (select p.guid, f.entry from Person p, Person_Func jt, Func f where p.Guid = jt.person_id and jt.func_id = f.guid and f.entry in ('A', 'B')) as d1,
(select p.guid, f.entry from Person p, Person_Func jt, Func f where p.Guid = jt.person_id and jt.func_id = f.guid and f.entry in ('A', 'B')) as d2
where d1.guid=d2.guid and d1.entry != d2.entry
但我认为这不是一个好主意,是吗?
【问题讨论】:
-
运气好能找到最佳实践吗?
标签: sql doctrine one-to-many dql