Hibernate-一对一关系单表继承

Question

我有3个实体BaseFoo ， FooAbc ， FooAbcDetail 。 FooAbc扩展了基本实体BaseFoo 。 我正在尝试在FooAbc和FooAbcDetail之间FooAbc一对一的关系。

@Data
@EqualsAndHashCode(callSuper = true)
@Entity
@Table(name = "foo")
@Audited
@AuditOverride(forClass = Auditable.class)
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "type_id", discriminatorType = DiscriminatorType.INTEGER)
public abstract class BaseFoo extends Auditable implements Serializable {

    @Id
    @Column(name = "id")
    private Long id;

    //other fields
}

@Data
@EqualsAndHashCode(callSuper = true)
@Entity
@Audited
@AuditOverride(forClass = BaseFoo.class)
@DiscriminatorValue("2")
public class FooAbc extends BaseFoo implements Serializable {

    @EqualsAndHashCode.Exclude
    @ToString.Exclude
    @NotAudited
    @OneToOne(mappedBy = "fooAbc",
              fetch = FetchType.LAZY,
              cascade = CascadeType.ALL,
              orphanRemoval = true,
              optional = false)
    private FooAbcDetail fooAbcDetail;

    //other fields
}

@Data
@Entity
@Table(name = "foo_abc_detail")
public class FooAbcDetail implements Serializable {

    @Id
    @Column(name = "foo_id"/* foo_abc_id (I tried both) */)
    private Long id;

    @ToString.Exclude
    @EqualsAndHashCode.Exclude
    @MapsId //also tried @MapsId("id")
    @OneToOne(fetch = FetchType.LAZY)
    private FooAbc fooAbc; //also tried BaseFoo

    //other fields
}

当项目启动时，Hibernate抛出：

org.hibernate.MappingException：无法在org.hibernate.mapping.Table（public.foo_abc_detail）及其相关的超表和辅助表中找到逻辑名称为ID的列

这里有什么问题？

环境

• 休眠5.3.10.Final
• Spring Boot 2.1.7。发布

Answer 1

该错误告诉您foo_abc_detail表上没有名为foo_id列foo_abc_detail上的列仅称为id 。

@Data
@EqualsAndHashCode(callSuper = true)
@Entity
@Table(name = "foo")
@Audited
@AuditOverride(forClass = Auditable.class)
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@DiscriminatorColumn(name = "type_id", discriminatorType = DiscriminatorType.INTEGER)
public abstract class BaseFoo extends Auditable implements Serializable {

    @Id
    @Column(name = "id")
    private Long id;

    //other fields
}

@Data
@EqualsAndHashCode(callSuper = true)
@Entity
@Audited
@AuditOverride(forClass = BaseFoo.class)
@DiscriminatorValue("2")
public class FooAbc extends BaseFoo implements Serializable {

    @EqualsAndHashCode.Exclude
    @ToString.Exclude
    @NotAudited
    @OneToOne(mappedBy = "fooAbc",
              fetch = FetchType.LAZY,
              cascade = CascadeType.ALL,
              orphanRemoval = true,
              optional = false)
    private FooAbcDetail fooAbcDetail;

    //other fields
}

@Data
@Entity
@Table(name = "foo_abc_detail")
public class FooAbcDetail implements Serializable {

    @Id
    @Column(name = "id")
    private Long id;

    @ToString.Exclude
    @EqualsAndHashCode.Exclude
    @MapsId //also tried @MapsId("id")
    @OneToOne(fetch = FetchType.LAZY)
    private FooAbc fooAbc; //also tried BaseFoo

    //other fields
}

Answer 2

我解决了。 我不知道这是否是最好的方法，但是可以！ 我通过@JoinColumn替换了@MapsId注释。 FooAbcDetail类的最终结果如下。

@Data
@Entity
@Table(name = "foo_abc_detail")
public class FooAbcDetail implements Serializable {

    @Id
    @Column(name = "id")
    private Long id;

    @ToString.Exclude
    @EqualsAndHashCode.Exclude
    @OneToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "id", referencedColumnName = "id", nullable = false)
    private FooAbc fooAbc;

}