从@MappedSuperClass创建具有@id复合主键的JPA实体

Question

我有一个JPA实体的类层次结构，基类是定义了一个ID的MappedSuperclass。我试图在子类中使用复合键，但这似乎不起作用

@MappedSuperclass
public class BaseEntity implements Serializable {
    @Id  
    @GeneratedValue(strategy = GenerationType.IDENTITY)  
    @Basic(optional = false)  
    @Column(name = "id")  
    protected Long id;  

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }
}


@Entity
@EntityListeners(EntityBaseListener.class)
@Inheritance(strategy=InheritanceType.JOINED)  
@Table(name = "catalog_entity")  
public class BaseCatalogEntity extends BaseEntity {

    @Column(name = "created_at", nullable = false)
    @Temporal(TemporalType.TIMESTAMP)
    private Date createdAt;

    @Column(name = "updated_at", nullable = false)
    @Temporal(TemporalType.TIMESTAMP)
    private Date updatedAt;


    public void setCreatedAt(Date date)
    {
        createdAt = date;
    }

    public void setUpdatedAt(Date date)
    {
        updatedAt = date;
    }

    public Date getCreatedAt() {
        return createdAt;
    }

    public Date getUpdatedAt() {
        return updatedAt;
    }

}


@Entity
@Table(schema = "student_catalog")
@IdClass(value = StudentCatalog.StudentCatalogPK.class)
public class StudentCatalog extends BaseCatalogEntity {

    @Id
    @Column(name = "name", nullable = false, length = 100)
    private String name;

    @Id
    @Column(name = "version", nullable = false)
    private Integer version;

    @Column(name = "description" , length = 255)
    private String description; 

    @Column(name = "vendor" , length = 50)
    private String vendor; 

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public Integer getVersion() {
        return version;
    }

    public void setVersion(Integer version) {
        this.version = version;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

    public String getVendor() {
        return vendor;
    }

    public void setVendor(String vendor) {
        this.vendor = vendor;
    }

    public static class StudentCatalogPK implements Serializable {

        private Long id;
        private String name;
        private Integer version;

        public Long getId() {
            return id;
        }

        public void setId(Long id) {
            this.id = id;
        }

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }

        public Integer getVersion() {
            return version;
        }

        public void setVersion(Integer version) {
            this.version = version;
        }

        @Override
        public boolean equals(Object obj) {
            boolean result = false;
            if(obj != null && (obj instanceof StudentCatalogPK)) {
                StudentCatalogPK other = (StudentCatalogPK)obj;
                result = (Objects.equals(this.id, other.id) && Objects.equals(this.name, other.name) &&
                        Objects.equals(this.version, other.version));
            }
            return result;
        }

        @Override
        public int hashCode() {
            return (27780 + (this.id != null ? this.id.hashCode() : 0) +
                           (this.version != null ? this.version.hashCode() : 0) +
                           (this.name != null ? this.name.hashCode() : 0));
        }
    }
}

Answer 1

在JPA中重新定义子类中的id是不合法的。这将导致表映射和多态查询中的歧义。

当业务密钥用于DB标识时，扩展超类中定义的密钥是一个常见的问题。我建议只使用代理键（如UUID）作为DB标识，使用业务键作为实例标识。