JPA：持續存在兩個具有OneToMany關系的實體的問題

Question

我有2個實體Customer和Address，關系是一個地址可以屬於多個客戶。

下面是客戶類，您可以看到它具有對地址對象的引用，在基礎客戶表中，它是地址的ID。 我省略了getter和setter以及一些簡單的變量。

@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;
........
}

下面是地址類。

//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;
 ......
}

我試圖在一個電話中保留新客戶和新地址，以像下面這樣保留。 我省略了我設置的一些變量。

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)    

但是我收到一條錯誤消息，說客戶表中的address_id為null。 我以為JPA會插入新地址，然后插入新客戶，並將地址ID列設置為新地址ID？ 我的想法錯了嗎？ 還是我在映射中犯了一個錯誤，或者我是如何持久化實體的？

我可以執行此操作的另一種方法是先持久保存地址，然后持久保存客戶，但如果可能的話，希望在單個持久保存中執行。

下面是基礎表。

//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;

謝謝。

Answer 1

如果要與Customer保存新Address ，則需要添加CascadeType.ALL

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

並通過這種方式保存所有內容（您無需將客戶添加到地址列表中，因為客戶僅通過外鍵address_id引用地址即可）

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

但這不是一種非常方便的方法，因為地址類似於引用。 因此，通過保存每個客戶來更新參考地址是不尋常的。