Hibernate JPA IdentifierGenerationException：為帶有@embeddedid的類生成的null id

Question

我無法將我的數據庫域模型映射到程序實體，在一種情況下，實體本質上是一個連接表（一個句點），它結合了另外兩個實體（時間段和一天）。 然后，另一個實體（課程）引用該期間實體，確定它何時發生。

當我嘗試使用saveOrUpdate(lesson)以新的時段保存課程時，hibernate會拋出一個IdentifierGenerationException

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

數據庫如下所示（不是真正的數據庫，只是關鍵表和列）

在java hibernate模型中，我使用了一個嵌入式id作為句點類的主鍵，然后課程類引用了一個句點。

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
}

嵌入式ID只有主鍵列：

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
}

然后是使用定義為的時段的課程類：

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實體類都是非常基本的pojos，它們的id使用GenerationType.AUTO 。 所以我的問題是：

1. 導致此IdentifierGenerationException的原因
2. 如何在保持相同數據庫模型的同時避免它

提前致謝

Answer 1

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;

在PeriodId類中，拋棄那些整數。 我已經用這種方式完成了與你類似的映射，並且它可以工作。

Answer 2

我能夠為我的案例（scala代碼）創建以下映射，並且可以完全拋棄@Embeddable類：

@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 = _
}

那么以下內容應該對你有用

@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，您在嵌入式id類中缺少生成的值注釋。

@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
}