java - Hibernate JPA ManyToOne 组合键

Question

我正在尝试设置我的实体以允许进行 pk。我的数据库包含两个字段，

dealer_detail_id pk user_detail_id pk

两者都在相应表中加入 id。

到目前为止，我已经试过了，但没有成功。

@Embeddable
public class DealerUserPk implements Serializable {

    private Integer dealerDetail;
    private Integer userDetail;

经销商用户

@Embeddable
@Table(name = "dealer_user", schema = "account") 
public class DealerUser implements Serializable {

    @EmbeddedId
    private DealerUserPk id;

    @Id
    @ManyToOne
    @JoinColumn(name = "dealer_detail_id", referencedColumnName = "id")
    private DealerDetail dealerDetail;

    @Id
    @ManyToOne
    @JoinColumn(name = "user_detail_id", referencedColumnName = "id")
    private UserDetail userDetail;

经销商详情

@Entity
@Table(name = "dealer_detail", schema = "account") 
public class DealerDetail implements Serializable {

    @Id
    private Integer id;

用户详细信息

@Entity
@Table(name = "user_detail", schema = "account") 
public class UserDetail implements Serializable {

    @Id
    private Integer id;

任何人都可以发现我做错了什么吗？

Answer 1

这是正确的:

@Embeddable
public class DealerUserPk implements Serializable {

    private Integer dealerDetail;
    private Integer userDetail;

1. 但您的 DealerUser 带有可嵌入注释，它应该是 @Entity 因为您正在使用 @Table 注释。

2. 需要添加MapsId如下

    @Entity
    @Table(name = "dealer_user", schema = "account") 
     public class DealerUser implements Serializable {
   
   @EmbeddedId
   private DealerUserPk id;
   
   @MapsId("dealerDetail")
   @ManyToOne
   @JoinColumn(name = "dealer_detail_id", referencedColumnName = "id")
   private DealerDetail dealerDetail;
   
   @Id
   @MapsId("userDetail")
   @JoinColumn(name = "user_detail_id", referencedColumnName = "id")
   private UserDetail userDetail;
   

尝试一下。