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Spring boot JPA - 双向关系:返回外键而不是相关对象

如何解决Spring boot JPA - 双向关系:返回外键而不是相关对象

我有两个实体:客户地址

这是一种双向关系 - 一个地址可以有多个客户(onetoMany),一个客户只能有一个地址(manyToOne)。

为客户退货执行 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"
        }
    }
]

一个新客户的地址属性数据库中存在的完全相同时 - 正在通过 POST 请求添加,json 响应返回与数据库中现有对象相关的外键,而不是对象本身:>

[
    {
        "id": 1,"city": "New York"
        }
    },{
        "id": 2,"name": "Bar","email": "Bar@gmail.com","address": 1 <----------- it returns the foreign key instead of the object
    }
]

所以我期望是,每当一个新客户的地址已经存在数据库中时,被添加 - 它应该返回地址对象而不是来自json响应的外键。

代码

Customer.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;
    [...]

Address.java

@JsonIdentityInfo(
        generator = ObjectIdGenerators.PropertyGenerator.class,property = "id")
@Entity
@Table
public class Address {
    @Id
    @SequenceGenerator(
            name = "address_sequence",sequenceName = "address_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);
    }

[...]

服务将客户保存到数据库中,这也确保如果数据库中已经存在地址,则不会创建任何地址(它检查每个属性等于参数):

//...
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);
    }

[...]

解决方法

您可能因为 JsonIdentityInfo 而得到这种行为,所以这是一个序列化问题,您没有持久性问题。我假设您使用的是关系数据库(NoSql 的 Hibernate 具有 Jpa 之类的注释,但这会使这成为一个不同的问题)并且数据被正确持久化。

the javadocs

在实践中,这是通过将第一个实例序列化为完整对象和对象标识,并将对对象的其他引用作为引用值来完成的

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