Question

我的项目有这样的建模：

public abstract class Actor{}
public class Person extends Actor {}
public class Organization extends Actor {}

public abstract class Role{ @ManyToOne @JoinColumn(name="ID_ACTOR") }
public class Customer extends Role{}
public class Employee extends Role{}

我想获得一个扮演特定演员的List<Role>：

public List<Role> getRoles(Actor actor) {
  CriteriaBuilder builder = em.getCriteriaBuilder();
  CriteriaQuery<Role> criteria = builder.createQuery(Role.class);

  //How to do this filter using Criteria API (JPA 2.1)?

  return query.getResultList();
}

下面的SQL可以完成这项工作：

SELECT 'Customer' as role, c.id as id, c.id_actor as actor
FROM customer c
left join person p on p.id = c.id_actor
left join organization o on o.id = c.id_actor
where c.id_actor = ?

UNION 

SELECT 'Employee' as role, e.id as id, e.id_actor as actor
FROM employee e
left join person p on p.id = e.id_actor
left join organization o on o.id = e.id_actor
where e.id_actor = ?

Answer 1

我解决了在Actor和Role类之间建立双向关系的问题：

public abstract class Actor{ @OneToMany(mappedBy="actor") List<Role> roles }
public class Person extends Actor {}
public class Organization extends Actor {}

public abstract class Role{ @ManyToOne @JoinColumn(name="ID_ACTOR") Actor actor }
public class Customer extends Role{}
public class Employee extends Role{}

这样我就能得到扮演特定演员的List<role>：

public List<Role> getRoles(Actor actor) {
    return actor.getRoles();
}

最初我想避免双向关系，但我意识到在这种特殊情况下，有直接的好处

标准多态性如何加入JPA？

1 个答案: