Question

Hibernate需要我的实体ID：

（throws org.hibernate.AnnotationException：没有为实体指定标识符）

@Entity
@Table(name = "CHILDREN")
public class ChildDb {

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "INNER_ID", nullable = false)
    private ParentDb parent;

    @Column(name = "NAME")
    private String name;

    @Column(name = "VALUE")
    private double value;
}

这里是父实体：

@Entity
@Table(name = "PARENT")
public class ParentDb {
    @Id
    @Column(name = "INNER_ID")
    @GeneratedValue(strategy = GenerationType.AUTO, generator = "G1")
    @SequenceGenerator(name = "G1", sequenceName = "SOME_SEQ")
    private long id;

    @Column(name = "TIMESTAMP")
    private long timestamp;

    @OneToMany(fetch = FetchType.EAGER, mappedBy = "parent", cascade = {CascadeType.ALL}, orphanRemoval=true)
    private List<ChildDb> children;
}

但是，id应该是复合的 - 父ID和名称。

如何定义？

Answer 1

我相信这样做会有效：

@Entity
@Table(name = "CHILDREN")
public class ChildDb implements Serializable{

@Id
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "INNER_ID", nullable = false)
private ParentDb parent;

@Id
@Column(name = "NAME")
private String name;

@Column(name = "VALUE")
private double value;

}

Answer 2

在我看来，你做的奇怪的事情可能就是这样：

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "INNER_ID", nullable = false)
private ParentDb parent;

您的ChildDB中是否还有“INNER_ID”。如果是OneToMany关系，@ ManyToOne方必须是外键的所有者。

使用ManyToOne关系定义创建复合ID

2 个答案: