Question

我有2个实体Customer和Address，这个关系是一个可以属于多个客户的地址。

以下是客户类，因为您可以看到它具有对地址对象的引用，在基础客户表中它是地址的ID。我省略了getter和setter以及一些简单的变量。

@Entity
@Table(name = "customer")
public class Customer implements Serializable  {

@Id
@GeneratedValue
@Column(name = "customer_id")
private int customerId;
@ManyToOne
@JoinColumn(name = "store_id")
private Store store;
@ManyToOne
@JoinColumn(name = "address_id")
private Address address;
........
}

以下是地址类。

//Address Class
@Entity
@Table(name = "address")
public class Address implements Serializable {

@Id
@GeneratedValue
@Column(name = "address_id")
private int addressId;
@JoinColumn(name = "city_id")
@ManyToOne
private City city;
@OneToMany(cascade = CascadeType.ALL, mappedBy = "address")
@JsonIgnore
List<Customer> customers;
 ......
}

我尝试在一次通话中坚持新客户和新地址，如下所示。我省略了一些我设定的变量。

Customer cus = new Customer();
Address addr= new Address();
........
cus.setAddress(addr)
List<Customer> cusList= new ArrayList<>();
cusList.add(cus);
addr.setCustomers(cusList);
entityManager.persist(cus)

但是我得到一个错误，说客户表中的address_id为空。我原本以为JPA会插入新地址然后插入新客户，地址id列设置为新地址ID？我的想法在这里错了吗？或者我在映射中犯了错误或者我如何坚持实体？

我能做到这一点的另一种方法是先坚持地址然后坚持客户，但如果可能的话，我宁愿在一个坚持下做。

以下是基础表格。

//Customer Table
CREATE TABLE `customer` (
   `customer_id` smallint(5) unsigned NOT NULL AUTO_INCREMENT,
   `store_id` tinyint(3) unsigned NOT NULL,
   `first_name` varchar(45) NOT NULL,
   `last_name` varchar(45) NOT NULL,
   `email` varchar(50) DEFAULT NULL,
   `address_id` smallint(5) unsigned NOT NULL,
   `active` tinyint(1) NOT NULL DEFAULT '1',
   `create_date` datetime NOT NULL,
   `last_update` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE      CURRENT_TIMESTAMP,
   PRIMARY KEY (`customer_id`),
   KEY `idx_fk_store_id` (`store_id`),
   KEY `idx_fk_address_id` (`address_id`),
   KEY `idx_last_name` (`last_name`),
   CONSTRAINT `fk_customer_address` FOREIGN KEY (`address_id`) REFERENCES    `address` (`address_id`) ON UPDATE CASCADE,
   CONSTRAINT `fk_customer_store` FOREIGN KEY (`store_id`) REFERENCES  `store` (`store_id`) ON UPDATE CASCADE
  ) ENGINE=InnoDB AUTO_INCREMENT=608 DEFAULT CHARSET=utf8;

/Address Table

CREATE TABLE `address` (
  `address_id` smallint(5) unsigned NOT NULL AUTO_INCREMENT,
  `address` varchar(50) NOT NULL,
  `address2` varchar(50) DEFAULT NULL,
  `district` varchar(20) NOT NULL,
  `city_id` smallint(5) unsigned NOT NULL,
  `postal_code` varchar(10) DEFAULT NULL,
  `phone` varchar(20) NOT NULL,
  `last_update` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE    CURRENT_TIMESTAMP,
  PRIMARY KEY (`address_id`),
  KEY `idx_fk_city_id` (`city_id`),
  CONSTRAINT `fk_address_city` FOREIGN KEY (`city_id`) REFERENCES `city` (`city_id`) ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=619 DEFAULT CHARSET=utf8;

感谢。

Answer 1

如果您要使用Address保存新的Customer，则需要添加CascadeType.ALL

@ManyToOne(cascade = CascadeType.ALL)
@JoinColumn(name = "address_id")
private Address address;

以这种方式保存所有内容（您不需要将客户添加到地址列表中，因为客户仅通过外键address_id来引用该地址）

Customer cus = new Customer();
Address addr = new Address();
cus.setAddress(addr)
entityManager.persist(cus)

但这不是一种非常方便的方式，因为地址就像引用一样。因此，通过保存每个客户更新参考地址是不常见的。

JPA：持续存在2个具有OneToMany关系的实体的问题

1 个答案: