加入Hibernate查询

Question

我有课:

class User{
String id;
ArrayList<UserAttribute> attributeList=new ArrayList<UserAttribute>();
}

class UserAttribute{
User user;
String value;
String key;
}

我有与表相关的表.

表用户

column id

table user_attribute

column user_id
column key
column value

示例用户表内容

user_id="123"

user_id="234"

示例user_attribute表内容

user_id="123", key="address", value="green house street"

user_id=234, key="address", value="blue house street"

user_id="123", key="phone", value="234324"

user_id="234", key="phone", value="786988"

我想让有"绿屋街"地址的用户和(不是或)有电话"234324"

我在sql编辑器中编写sql(oracle):

 select u.* from user u left join user_attribute prop on u.id=prop.user_id inner join user_attribute prop2 on prop.user_id = prop2.user_id where prop.value = 'green house street' and prop.key = 'address' and prop2.key ='phone'  and prop2.value = '234324';

我在hibernate方面尝试了很多东西:但是因为get而无法使用join

"未经考虑的令牌"

使用"on"子句时出错.

最后我写道

entityManager.createQuery(
select u from User u, u.attributeList prop, u.attributeList prop2 where prop.value = :address and prop.key = :key and prop2.key =:key2  and prop2.value = :phone and prop.user.id=u.id and prop2.user.id=u.id and prop2.user.id=prop.user.id
)..setParameter("key", "address")
                .setParameter("phone", "234324")
                .setParameter("key2", "phone")
                .setParameter("address", "green house street").getSingleResult();

仍然没有机会,

我努力寻找一个单一的查询解决方案,但到目前为止还无法实现.谢谢你的想法.

Answer 1

您需要进行两个连接,但不需要指定任何on子句,因为Hibernate知道,由于关联的映射,表是如何链接在一起的.所以以下查询应该有效:

select u from User u
inner join u.attributeList a1
inner join u.attributeList a2
where a1.value = 'green house street' and a1.key = 'address' 
and a2.key ='phone' and a2.value = '234324'