具有 JAXB 特定合同的复杂 JPA 继承

Question

大家好，我已经到处搜索了这个，所以这实际上是最后的手段，我正在尝试为以下代码获取一个很好的数据库结构：

@Entity
@Table(name = "tasks")
@XmlRootElement(name = "task")
@XmlType(propOrder = { "id", "taskTitle", "notifications" })
public class Task {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;

    @Column(name = "identifier")
    private String taskTitle;

    @OneToMany(mappedBy = "???super.task???", cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    private Set<User> users;

    @OneToMany(mappedBy = "task", cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    private Set<Group> groups;

    @XmlElement(name = "notify_list")
    public Notify getNotifications() {
        return new Notify(this);
    }
... (the users and groups getters are XmlTransient)
}

@XmlAccessorType
@XmlType(propOrder = { "users", "groups" })
public class Notify {
    private Task task;
    public Notify() {
        super();
    }

    public Notify(Task task) {
        this.task = task;
    }

    @XmlElementWrapper(name = "users")
    @XmlElements({ @XmlElement(name = "user") })
    public Set<User> getUsers() {
        return task.getUsers();
    }

    @XmlElementWrapper(name = "groups")
    @XmlElements({ @XmlElement(name = "group") })
    public Set<Group> getGroups() {
        return task.getGroups();
    }
}

@MappedSuperclass
public class Person {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Integer id;

    @ManyToOne
    private Task task;

    @Column(name = "person_id")
    private Integer personId;

    @Transient
    private String firstName, lastName, email;

...(public getters/setters here)
}

@Entity
@Table(name = "persons")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "group_or_user")
@DiscriminatorValue(value = "group")
@XmlAccessorType
@XmlType(propOrder = { "personId", "firstName", "lastName", "email" })
public class Group extends Person {
    ...(inherits everything from Person and defines XmlElements)
}

@Entity
@DiscriminatorValue(value = "user")
@XmlAccessorType
@XmlType(propOrder = { "username", "personId", "firstName", "lastName",
    "email", "phone" })
public class User extends Group {

    @Transient
    private String username, phone;

    @XmlElement(name = "username")
    public String getUsername() {
        return username;
    }

    @XmlElement(name = "phone")
    public String getPhone() {
    return phone;
    }

...(the other getters/setters are inherited)
}

我也想要这样的 xml：

<task>
    <id>10001</id>
    <identifier>Title of the task</identifier>
    <notify_list>
      <users>
        <user>
          <username>firstname.lastname</username>
          <person_id>12</person_id>
          <firstname>FirstName</firstname>
          <lastname>LastName</lastname>
          <email>me@myname.com</email>
          <phone>023545848796</phone>
        </user>
      </users>
      <groups>
        <group>
          <person_id>21247415</person_id>
          <firstname>PizzaGroup</firstname>
          <lastname>Pizza</lastname>
          <email>blabla@something.com</email>
        </group>
      </groups>
    </notify_list>
</task>

我的意思是漂亮的数据库结构：1 个任务表和 1 个人员表（当然还有鉴别器列）。如您所见，我在用户集上的 OneToMany 关系遇到了问题，因为它说“它找不到对未知目标实体 i mapBy 的用户的任何引用，即在 MappedSuperclass 中定义的任务人所以我知道它应该继承它，因为 Group 继承了它。所以问题实际上是：我做错了什么？

编辑：

我忘了提到我只能为用户和组保留 person_id，其余的我从另一个数据库获得，但这是另一回事 :)

Answer 1

mappedBy 属性值应为“任务”。问题是 Person 中的任务字段的类型是 MaintenanceTask 而不是 Task。

另外，请确保所有实体都列在休眠配置文件中。

但我认为这种映射行不通，因为您尝试在一侧定义两个不同的关联，由另一个大小上的单个（和单个列）映射。由于组的集合也将包含用户（作为用户扩展组），您可能应该删除任务中设置的用户。只需根据组的实际类型提供两个返回过滤集合的 getter。