Hibernate JPA IdentifierGenerationException：为具有@embeddedid 的类生成空ID

Question

在一种情况下，我在将我的数据库域模型映射到程序实体时遇到问题，其中实体本质上是一个连接表（一个周期），它结合了其他两个实体（一个时间段和一天）。然后另一个实体（一课）引用了这个时期实体，确定它何时发生。

当我尝试使用 saveOrUpdate(lesson) 保存带有新时期的课程时，hibernate 会抛出 IdentifierGenerationException

org.hibernate.id.IdentifierGenerationException: null id 为:class com.trials.domain.Period 生成

数据库如下图（不是真正的数据库，只是关键的表和列）

在 java hibernate 模型中，我使用了一个嵌入的 id 作为 period 类的主键，然后课程类引用了一个 period。

Period.java

@Entity
@Table(name = "period")
public class Period{
    @EmbeddedId
    private PeriodId periodId;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "day_idday", nullable = false, insertable = false, updatable = false)
    private Day day;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "timeslot_idtimeslot", nullable = false, insertable = false, updatable = false)
    private Timeslot timeslot;

    //constructors, getters, setters, hashcode, and equals
}

而嵌入的id只有主键列：

PeriodId.java

@Embeddable
public class PeriodId implements Serializable {
    @Column(name = "timeslot_idtimeslot")
    private int timeslotId;

    @Column(name = "day_idday")
    private int dayId;

    //constructors, getters, setters, hashcode, and equals
}

然后是使用句号定义为的课程类：

Lesson.java

@Entity
@Table(name = "lesson")
public class Lesson {
    @Id
    @Column(name = "idlesson")
    private int lessonId;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumns({@JoinColumn(name = "period_timeslot_idtimeslot", nullable = false, updatable = false), @JoinColumn(name = "period_day_idday", nullable = false, updatable = false)})
    private Period period;
    //constructors, getters, setters, hashcode, and equals
}

Timeslot 和 Day 实体类都是非常基本的 pojo，它们的 id 使用 GenerationType.AUTO。所以我的问题是：

1. 导致此 IdentifierGenerationException 的原因
2. 如何在保持相同数据库模型的同时避免这种情况

提前致谢

Answer 1

把那些家伙放上去

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "day_idday", nullable = false, insertable = false, updatable = false)
private Day day;

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "timeslot_idtimeslot", nullable = false, insertable = false, updatable = false)
private Timeslot timeslot;

在 PeriodId 类中并丢弃这些整数。我以这种方式完成了与您类似的映射，并且可以正常工作。

Answer 2

我能够为我的案例（scala 代码）创建以下映射，并且可以完全抛弃 @Embeddable 类：

@Entity
@Table(name = "payment_order_item", schema = "pg")
@IdClass(classOf[PaymentOrderItem])
final class PaymentOrderItem extends Serializable{

  @Id
  @ManyToOne
  @JoinColumn(name = "order_item_id", referencedColumnName = "id")
  var orderItem: OrderItem = _

  @Id
  @ManyToOne
  @JoinColumn(name = "payment_id", referencedColumnName = "id")
  var payment: Payment = _
}

那么下面的内容应该对你有用

@Entity
@Table(name = "period")
@IdClass(Period.class)
public class Period extends Serializable{

    @Id
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "day_idday", referencedColumnName = "id", nullable = false)
    private Day day;

    @Id
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "timeslot_idtimeslot", referencedColumnName = "id", nullable = false)
    private Timeslot timeslot;

    //constructors, getters, setters, hashcode, and equals
}

Answer 3

乍一看， 您在嵌入的 id 类中缺少生成的值注释。

@Embeddable
public class PeriodId implements Serializable {

    @GeneratedValue
    @Column(name = "timeslot_idtimeslot")
    private int timeslotId;

    @GeneratedValue    
    @Column(name = "day_idday")
    private int dayId;

    //constructors, getters, setters, hashcode, and equals
}