如何用Hibernate条件查询选择下级实体？

Question

假设我有一个名为HQL（findRoomQuery）的文件：

select r from House h inner join h.roomList r
where h.address = :address and r.roomNo = :roomNo

House实体的映射如下：

<class name="com.example.House" table="house">      
    <id name="id" column="id" type="long">
        <generator class="assigned" />
    </id>

    <property name="address" column="address" type="string" length="100" not-null="false"/>
    <set name="roomList" cascade="none" lazy="false" fetch="join" inverse="true"> 
        <key>
            <column name="house_id"/>
        </key>
        <one-to-many class="com.example.Room"/>
    </set>
</class>

虽然Room实体是这样的：

<class name="com.example.Room" table="room">        
    <id name="id" column="id" type="long">
        <generator class="assigned" />
    </id>

    <property name="houseId" column="house_id" type="long" not-null="true"/>
    <property name="roomNo" column="room_no" type="string" length="4" not-null="false"/>
</class>

这种关系是一个房子可以有一个或多个房间。

执行查询的代码如下：

Query query = getSession().createQuery("findRoomQuery")
                .setParameter("address", address)
                .setParameter("roomNo", roomNo);
return query.list();

您可以从选择r（r是h.room列表的别名）中看到HQL返回房间实体。

如何使用Hibernate条件查询做同样的事情？可能吗？

Answer 1

试试这个

 Criteria criteria = session.createCriteria(House.class, "house"); 
              criteria.createAlias("house.roomList", "roomList");           
              criteria.add(Restrictions.eq("house.address",address));              
              criteria.add(Restrictions.eq("roomList.roomNo", roomNo)); 
              criteria.setProjection(Projections.property("roomList"));

           Room  r =  (Room) criteria.uniqueResult();

取代

 <property name="houseId" column="house_id" type="long" not-null="true"/>

自

 <many-to-one name="houseId" class="com.example.House" fetch="select">
            <column name="house_id" not-null="true" />
 </many-to-one>