首页 > 解决方案 > Spring boot JPA - 双向关系:返回外键而不是相关对象

问题描述

我有两个实体:CustomerAddress

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)。

为客户返回执行 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"
        }
    }
]

当使用 POST 请求添加与 DB 中存在的地址属性完全相同的新客户时,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
    }
]

所以我期望的是,每当添加一个新客户,其地址已经存在于数据库中,它应该返回地址对象,而不是来自 json 响应的外键。

代码:

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;
    [...]

客户控制器.java

//...
@PostMapping
    public void createCustomer(@RequestBody Customer customer) {
        customerService.createCustomer(customer);
    }

[...]

以及将客户保存到数据库的服务,它还确保如果数据库中已经存在地址,则不会创建任何地址(它检查参数中的每个属性是否相等):

//...
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);
    }

[...]

标签: javaspring-bootrestjpaforeign-keys

解决方案


由于 JsonIdentityInfo,您可能会遇到这种行为,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 具有类似 Jpa 的注释,但这会使这成为一个不同的问题)并且数据被正确保存。

请参阅javadocs

在实践中,这是通过将第一个实例序列化为完整的对象和对象标识,以及对对象的其他引用作为参考值来完成的


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