java - Spring JPA插入连接表实体时不允许NULL

Question

我有表结构

卖家

| ID |

客户

| ID |

交易

| ID | SELLER_ID | CUSTOMER_ID |

我用实体来表示这些:

@Entity
@Table(name = "SELLER")
public class Seller {

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

    @OneToMany(mappedBy = "seller", fetch = FetchType.LAZY, orphanRemoval = true)
    private List<Transaction> transactions;

}

<小时/>

@Entity
@Table(name = "CUSTOMER")
public class Customer {

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

    @OneToMany(mappedBy = "customer", fetch = FetchType.LAZY, orphanRemoval = true)
    private List<Transaction> transactions;
}

<小时/>

@Entity
@Table(name = "TRANSACTION")
public class Transaction {

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

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(
            name = "SELLER_ID",
            insertable = false,
            updatable = false,
            nullable = false,
            foreignKey = @ForeignKey(name = "TRA_SLR_FK")
    )
    private Seller seller;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(
            name = "CUSTOMER_ID",
            insertable = false,
            updatable = false,
            nullable = false,
            foreignKey = @ForeignKey(name = "TRA_CMR_FK")
    )
    private Customer customer;

}

<小时/>

当我尝试创建事务并使用 CrudRepository 保存它时:

public interface TransactionRepository extends CrudRepository<Transaction,Long> {}

使用:

Customer customer = customerRepository.findOne(1L); // Similar CrudRepository    
Seller seller = sellerRepository.findOne(1L); // Similar CrudRepository    
Transaction transaction = new Transaction(seller, customer);
transaction = transactionRepository.save(transaction);

我收到错误:

NULL not allowed for column "CUSTOMER_ID"; SQL statement: insert into transaction (id) values (null) [23502-195]

当我调用save时，为什么不使用customer来填充CUSTOMER_ID？

Answer 1

根据doc ，insertable、updateable等属性指的是SQL生成，而不是插入或更新相关实体。