FK 在一对多关系中引用的 JPA 组合 PK答案

【问题标题】：JPA Composite PK referenced by FKs in One to Many RelationshipFK 在一对多关系中引用的 JPA 组合 PK
【发布时间】：2020-02-07 14:45:56
【问题描述】：

我正在尝试将下表映射到 JPA。 user_tax 和 tax 与 user_tax 和 user 之间的关系是一对多。我有一个复合主键，我需要将外键映射到这两个键，这让我感到困惑。

错误信息：org.hibernate.AnnotationException: mappedBy reference an unknown target entity property: entity.Tax.user_tax in entity.UserTax.taxs

             tax           user_tax         user
           --------        --------        ------
         PK|t_id  |--------| t_id |PK-FK  |u_name|
           |t_name|   PK-FK| u_id |-------|u_id  | PK
           |      |        | name |       |      |

这是我的实体：

 @Entity
 @Table(name = "user")
 @Inheritance(strategy = InheritanceType.SINGLE_TABLE)
    public class User implements Serializable {

       @Id
       @GeneratedValue(strategy = GenerationType.IDENTITY)
         private Long id;

       @Column(name="u_name")
         private String uname;

          getters + setters
        }

 @Entity
 @Table(name = "tax")
 @Inheritance(strategy = InheritanceType.SINGLE_TABLE)
    public class Tax implements Serializable {

        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
          private Long id;

        @Column(name = "t_name")
          private String tname;

   @Embeddable
    public class UserTaxId implements Serializable {

      @Column(name="u_id")
        private Long uId;

        @Column(name="t_id")
        private Long tId;

  @Entity
  @Table(name = "user_tax")
  @Inheritance(strategy = InheritanceType.SINGLE_TABLE)
    public class UserTax implements Serializable {

      @EmbeddedId
        private UserTaxId userTaxId;

      @OneToMany(fetch = FetchType.LAZY, mappedBy = "user_tax")
        private List<User> users;

      @OneToMany(fetch = FetchType.LAZY, mappedBy = "user_tax")
        private List<Tax> taxs;

【问题讨论】：

标签： spring jpa foreign-keys one-to-many composite-key

【解决方案1】：

您的 1:n 映射是向后的（即 UserTax 只能有一个 User 和一个 Tax）并且您使用的是派生标识。尝试像这样映射UserTax：

@Entity
@Table(name = "user_tax")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class UserTax implements Serializable {

    @EmbeddedId
    private UserTaxId userTaxId;

    @ManyToOne(fetch = FetchType.LAZY)
    @MapsId("uId") // maps uId attribute of embedded id
    private User user;

    @ManyToOne(fetch = FetchType.LAZY)
    @MapsId("tId") // maps tId attribute of embedded id
    private Tax tax;

    ...
}

在第 2.4.1 节的JPA 2.2 spec 中讨论了派生的身份（带有示例）。

【讨论】：

【解决方案2】：

经过 3 天的研究，我将在这里发布对我有用的东西。

Brian Vosburgh 正确发布了 UserTax 类：

 @Entity
@Table(name = "user_tax")
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class UserTax implements Serializable {

    @EmbeddedId
    private UserTaxId userTaxId;

    @ManyToOne(fetch = FetchType.LAZY)
    @MapsId("uId") // maps uId attribute of embedded id
    private User user;

    @ManyToOne(fetch = FetchType.LAZY)
    @MapsId("tId") // maps tId attribute of embedded id
    private Tax tax;

    ...
}

但是，我收到错误消息并且我的代码没有编译。然后我还必须编辑 User 和 Tax 类：

 @Entity
     @Table(name = "user")
     @Inheritance(strategy = InheritanceType.SINGLE_TABLE)
        public class User implements Serializable {

           @Id
           @GeneratedValue(strategy = GenerationType.IDENTITY)
             private Long id;

           @OneToMany(
            mappedBy = "tid",
            cascade = CascadeType.ALL,
            orphanRemoval = true
        )
        private List<UserTax> tax = new ArrayList<>();

           @Column(name="u_name")
             private String uname;

              getters + setters
            }

    @Entity
     @Table(name = "tax")
     @Inheritance(strategy = InheritanceType.SINGLE_TABLE)
        public class Tax implements Serializable {

            @Id
            @GeneratedValue(strategy = GenerationType.IDENTITY)
              private Long id;

           @OneToMany(
            mappedBy = "uid",
            cascade = CascadeType.ALL,
            orphanRemoval = true
        )
        private List<UserTax> taxs = new ArrayList<>();

            @Column(name = "t_name")
              private String tname;

    setters+getters

    }

这是我找到问题解决方案的链接：https://vladmihalcea.com/the-best-way-to-map-a-many-to-many-association-with-extra-columns-when-using-jpa-and-hibernate/

【讨论】：