hibernate 根据外键作为条件查询

List rooms = getHibernateTemplate().find("from Room room where room.roomName = '" + room.getRoomName()+ "'" + " and room.house_id = "+ room.getHouse().getId());

设计是：room->house ManyToOne 单向的关系

hibernate会自动在room表中生成house_id外键，所以最开始用的是room.house_id，会报错：

org.springframework.orm.hibernate3.HibernateQueryException: could not resolve property: house_id of: com.StringBeans.entity.core.Room [from
com.StringBeans.entity.core.Room room where room.roomName = '1232131' and room.house_id = 1]; nested exception is org.hibernate.QueryException: could not resolve property: house_id of: com.StringBeans.entity.core.Room [from com.StringBeans.entity.core.Room
room where room.roomName = '1232131' and room.house_id = 1]

应该用面向对象的思想，即room.house.id