Java Hibernate JPA IdentifierGenerationException：为带有@embeddedid 的类生成空 id

Hibernate JPA IdentifierGenerationException: null id generated for class with @embeddedid

提问by Japheth Ongeri - inkalimeva

I am having trouble mapping my database domain model to the program entities in one case where the entity is essentially a join table (a period) which combines two other entities (a timeslot and a day). Another entity (a lesson) then has a reference to this period entity, determining when it occurs.

在一种情况下，我无法将我的数据库域模型映射到程序实体，其中实体本质上是一个连接表（一个周期），它结合了两个其他实体（一个时间段和一天）。另一个实体（一个课程）然后引用了这个时期实体，确定它何时发生。

When I try to save a lesson with a new period using saveOrUpdate(lesson)hibernate throws an IdentifierGenerationException

当我尝试使用saveOrUpdate(lesson)hibernate以新的时间段保存课程时会抛出 IdentifierGenerationException

org.hibernate.id.IdentifierGenerationException: null id generated for:class com.trials.domain.Period

org.hibernate.id.IdentifierGenerationException：为：class com.trials.domain.Period 生成了空 ID

The database looks like below (not the real database, just the key tables and columns)

数据库如下所示（不是真正的数据库，只是关键的表和列）

In the java hibernate model, I have used an embedded id for the primary key of the period class and the lesson class then has a reference to a period.

在 java 休眠模型中，我使用了一个嵌入的 id 作为周期类的主键，然后课程类引用了一个周期。

Period.java

期间.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
}

And the embedded id just has the primary key columns:

嵌入的 id 只有主键列：

PeriodId.java

周期ID.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
}

Then there is the lesson class that uses the period defined as:

然后是使用定义为的时间段的课程类：

Lesson.java

课程.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
}

The Timeslot and Day entity classes are both very basic pojos, and their ids use GenerationType.AUTO. So my problems are:

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. What causes this IdentifierGenerationException
2. How to avoid it while keeping the same database model

1. 是什么导致了这个 IdentifierGenerationException
2. 如何在保持相同数据库模型的同时避免它

Thanks in advance

提前致谢

采纳答案by Dalton

Put those guys

把那些家伙

@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;

inside the PeriodId class and throw away those ints. I have done a mapping similar to yours this way and it works.

在 PeriodId 类中并丢弃这些整数。我已经以这种方式完成了类似于您的映射，并且它有效。

回答by Rowanto

On a first glance, You're missing the generated value annotation in the embedded id class.

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
}

回答by Petr Osipov

I was able to create the following mapping for my case (scala code) and could totally throw away the @Embeddable class:

我能够为我的案例（scala 代码）创建以下映射，并且可以完全丢弃 @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 = _
}

So the following should work for you then

所以以下应该对你有用

@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
}