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Spring boot Jpa与部分嵌入Id的关系

spring-boot-jpa spring-data-jpa spring-boot java

Lolly · 技术社区 · 3 年前

我正在Spring Boot数据JPA中创建实体关系。由于这些表是遗留的,我无法修改或添加列。问题是我得到了一个错误,如果点部分嵌入的Id。

我的实体类如下所示:

Class Customer {
   @EmbededId
   private CustomerPk id;
   
   @Column("NAME")
   private String name; 
   
   @OneToMany(fetch=FetchType.LAZY, cascade=CascadeType.ALL, mappedBy="customerDetails")
   private List<Purchase> purchaseDetails;
   ...
}

@Embeddable
Class CustomerPk {
   @Column("CUSTOMER_ID")
   private String customerId
       
   @Column("PURCHASE_ID")
   private String productId; 

   @Column("PURCHASE_DATE")
   private String date; 

   
 }

采购实体如下所示:

Class Purchase {
       @EmbededId
       private PurchasePK id;
       
       @Column("TRANSACTION_NAME")
       private String transactionName; 
       
       @ManyToOne(fetch=FetchType.LAZY, cascade=CascadeType.ALL)
       @JoinColumns({
         @JoinColumn(name="CUSTOMER_ID" referencedColumnName="CUSTOMER_ID")
         @JoinColumn(name="PURCHASE_ID" referencedColumnName="PURCHASE_ID")
       )}
       private Customer customerDetails;
       ...
    }

    @Embeddable
    Class PurchasePK {
       @Column("CUSTOMER_ID")
       private String customerId
           
       @Column("PURCHASE_ID")
       private String productId; 

       @Column("TRANSACTION_DATE")
       private String date; 
     }

有了上面的结构,我得到了org.hibernate。AnnotationException:PURCHASE.CUSTOMER的referencedColumnNames(CUSTOMER_ID,PURCHASE_ID)引用未映射到单个属性的CUSTOMER的详细信息。

如果我从CustomerPK中删除日期属性,我就可以启动服务器。但根据当前的要求,我需要日期作为CustomerPK类的一部分。

我想如果我使用复合键的一部分作为Join Columns,我会得到这个错误。

0 回复 | 直到 3 年前

Ibrahim AlTamimi 3 年前

工作版本:

@Entity
public class Customer {
    @EmbeddedId
    private CustomerPk id;

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

    @OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, mappedBy = "customerDetails")
    private List<Purchase> purchaseDetails;
}

@Embeddable
public class CustomerPk implements Serializable {
    @Column(name = "CUSTOMER_ID")
    private String customerId;

    @Column(name = "PURCHASE_ID")
    private String productId;

    @Column(name = "PURCHASE_DATE")
    private String date;
}

@Entity
public class Purchase {
    @EmbeddedId
    private PurchasePK id;

    @Column(name = "TRANSACTION_NAME")
    private String transactionName;

    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
    @JoinColumns({
            @JoinColumn(name = "CUSTOMER_ID", referencedColumnName = "CUSTOMER_ID", insertable = false, updatable = false),
            @JoinColumn(name = "PURCHASE_ID", referencedColumnName = "PURCHASE_ID", insertable = false, updatable = false),
            @JoinColumn(name = "PURCHASE_DATE", referencedColumnName = "PURCHASE_DATE", insertable = false, updatable = false)
    })
    private Customer customerDetails;
}

@Embeddable
public class PurchasePK implements Serializable {
    @Column(name = "CUSTOMER_ID")
    private String customerId;

    @Column(name = "PURCHASE_ID")
    private String productId;

    @Column(name = "TRANSACTION_DATE")
    private String date;
}

结论

提供的信息来自射线有效,您错过了添加所需的连接列以表示完整的实体关系,关于您的注释射线重点,是的,你是对的,两列的用法不同,但两列都有自己的名称,在运行时不会覆盖任何行值。

以上表格和表示的结果如下:


create table customer
(
    customer_id   varchar(255) not null,
    purchase_date varchar(255) not null,
    purchase_id   varchar(255) not null,
    name          varchar(255),
    primary key (customer_id, purchase_date, purchase_id)
);
create table purchase
(
    customer_id      varchar(255) not null,
    transaction_date varchar(255) not null,
    purchase_id      varchar(255) not null,
    transaction_name varchar(255),
    purchase_date    varchar(255),
    primary key (customer_id, transaction_date, purchase_id)
);
alter table purchase
    add constraint FK6rkrb8rq8x56kai7g5gm32d1y foreign key (customer_id, purchase_date, purchase_id) references customer;