我有两个实体:客户和地址。
这是一种***双向关系***-一个地址可以有多个客户(oneToMany),而一个客户只能有一个地址(manyToOne)。
执行客户退货的GET请求:
[
{
"id": 1,
"name": "Foo",
"contact": "5512312",
"email": "Foo@gmail.com",
"address": {
"id": 1,
"street": "X",
"postalCode": 123,
"houseNo": "10",
"city": "New York"
}
}
]
当使用POST请求添加一个新客户(其地址属性与DB中存在的客户完全相同)时,json响应返回与DB中现有对象相关的外键,而不是对象本身:
[
{
"id": 1,
"name": "Foo",
"contact": "5512312",
"email": "Foo@gmail.com",
"address": {
"id": 1,
"street": "X",
"postalCode": 123,
"houseNo": "10",
"city": "New York"
}
},
{
"id": 2,
"name": "Bar",
"contact": "5512312",
"email": "Bar@gmail.com",
"address": 1 <----------- it returns the foreign key instead of the object
}
]
因此,我期望的是,每当添加一个新客户时,如果该客户的地址 * 已经存在于数据库中 *,那么它应该返回 *address对象 *,而不是json响应中的外键。
验证码:
客户.java
@JsonIdentityInfo(
generator = ObjectIdGenerators.PropertyGenerator.class,
property = "id")
@Entity
@Table
public class Customer {
@Id
@SequenceGenerator(
name = "customer_sequence",
sequenceName = "customer_sequence",
allocationSize = 1
)
@GeneratedValue(
strategy = GenerationType.SEQUENCE,
generator = "customer_sequence"
)
private Long id;
private String name;
private String contact;
private String email;
@ManyToOne(cascade = CascadeType.ALL)
@JoinColumn(name = "address_id", nullable = false)
private Address address;
[...]
地址.java
@JsonIdentityInfo(
generator = ObjectIdGenerators.PropertyGenerator.class,
property = "id")
@Entity
@Table
public class Address {
@Id
@SequenceGenerator(
name = "address_sequence",
sequenceName = "address_sequence",
allocationSize = 1
)
@GeneratedValue(
strategy = GenerationType.SEQUENCE,
generator = "address_sequence"
)
private Long id;
private String street;
private int postalCode;
private String houseNo;
private String city;
@JsonIgnore
@OneToMany(mappedBy = "address")
private Set<Customer> customers;
[...]
CustomerController.java
//...
@PostMapping
public void createCustomer(@RequestBody Customer customer) {
customerService.createCustomer(customer);
}
[...]
而***service***将客户保存到DB中,这也确保如果数据库中已经存在地址,则不会创建地址(它检查每个属性是否与参数相等):
//...
public void createCustomer(Customer customer) {
Optional<Customer> customerWithExistingAddress = customerRepository.findAll()
.stream()
.filter(x -> x.getAddress().equals(customer.getAddress()))
.findFirst();
customerWithExistingAddress.ifPresent(c -> customer.setAddress(c.getAddress()));
customerRepository.save(customer);
}
[...]
2条答案
按热度按时间hkmswyz61#
你可能会因为JsonIdentityInfo而得到这种行为,所以这是一个序列化问题,而不是持久化问题。我假设你使用的是关系数据库(Hibernate for NoSql有类似Jpa的注解,但这会使这成为一个不同的问题),并且数据被正确持久化。
参见the javadocs:
在实践中,这是通过将第一个示例序列化为完整对象和对象标识,并将对该对象的其他引用序列化为引用值来实现的
q5iwbnjs2#
删除@JsonIdentityInfo