Question

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的Func（一个Person - 几个Funcs）的一对多关系。我想查询Person，它有很多Funcs - 史蒂夫（a，b，c）;约翰（a，b，d）; ele（b，d） - 我查询（a，b）函数和steve和john应该返回。

现在我只是通过所有人进行迭代并查询函数 - 这非常非常慢。你能帮我吗？

UPD

我成功了

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

但我认为那不是一个好主意，是吗？

Answer 1

将Person和Person_Func与JOIN链接起来是一种更简洁的方法：

SELECT p.id
FROM Person p INNER JOIN Person_Func pf ON p.id = pf.person_id
              INNER JOIN Person_Func pf2 ON p.id = pf2.person_id
WHERE
    pf.func_id = 'a' 
    AND pf2.func_id = 'b'

这里有SQLFiddle code

一对多关系的Doctrine查询子集

1 个答案: