JPA：问题持续存在一个OneToMany关系的2个实体

Question

我有2个实体Customer和Address，关系是一个地址可以属于多个客户。

以下是客户类，您可以看到它具有对地址对象的引用，在底层客户表中它是地址的标识。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; 
...... 
} 

我试图在一次调用中坚持一个新客户和一个新地址，像下面一样保持。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)  

但我得到一个错误，说客户表中的address_id为空。我原以为JPA会插入新的地址，然后插入新的地址标识列设置为新的地址标识的客户？我的思想错了吗？或者我在映射中犯了一个错误，或者我如何坚持这些实体？

另一种方法，我可以做到这一点是先坚持住地址，然后坚持客户，但宁愿如果可能的话在一个坚持做。

下面是基础表。

//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; 

谢谢。

Answer 1

如果要保存新Address你需要添加CascadeType.ALL

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

而且通过这种方式，所有保存Customer（你不需要给客户添加到地址列表，因为客户请参阅地址只是一个外键address_id）

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

但是，这不是一个非常方便的方法，因为地址是类似的参考。因此，通过节省每个客户来更新地址参考是很不寻常的。