如何在 POST 主体中发送数据而不在 Spring 引导中设置自动生成的主键值？

Question

我有两个 classes/tables--- 客户和地址具有双向一对一关系。 Address_id 是外键。

这是实体图

我正在尝试通过 postman 发送数据，但我想发送值而不在 post 正文中设置主键属性。如果我只为客户省略 id 属性，它就可以工作。 但是如果我对地址做同样的事情，它就不起作用了。

这是成功插入数据的 post 正文。

<Customer>
<firstName>Dave</firstName>
<lastName>Bautista</lastName>
<gender>M</gender>
<date>2012-01-26T09:00:00.000+0000</date>
<addressdto>
<id>7</id>
<city>BANKURA</city>
<country>WEST BENGAL</country>
</addressdto>
</Customer>

如果我在 addressdto 中省略 <id></id> 那么我会在 postman--

中收到此错误

Failed to add customer due to could not execute statement; SQL [n/a]; constraint [null]; nested exception is org.hibernate.exception.ConstraintViolationException: could not execute statement

控制台出错--

2020-04-23 23:10:12.093 ERROR 3824 --- [io-8080-exec-10] o.h.engine.jdbc.spi.SqlExceptionHelper   : 
Cannot add or update a child row
: a foreign key constraint fails (`liqbtest`.`customers`, CONSTRAINT `FK_DETAIL` FOREIGN KEY 
(`address_id`) REFERENCES `address` (`id`))

CustomerDto

package com.spring.liquibase.demo.dto;

import java.util.Date;

import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlRootElement;
import com.sun.xml.txw2.annotation.XmlElement;


@JacksonXmlRootElement(localName = "Customer")
public class CustomerDto {

    private int id;
    private String firstName;
    private String lastName;
    private String gender;
    private Date date;
    private AddressDto addressdto;


    public CustomerDto() {
        super();
    }
..getters and setters

addressDto

public class AddressDto {

    private int id;
    private String city;
    private String country;


    public AddressDto() {
        super();
    }

EntityToDtoMapper

public Customer mapToEntity(CustomerDto customerDto) {
        Address address=new Address();
        address.setCity(customerDto.getAddressdto().getCity());
        address.setCountry(customerDto.getAddressdto().getCountry());
        address.setId(customerDto.getAddressdto().getId());

        Customer customer=new Customer();
        customer.setId(customerDto.getId());
        customer.setFirstName(customerDto.getFirstName());
        customer.setLastName(customerDto.getLastName());
        customer.setGender(customerDto.getGender());
        customer.setDate(customerDto.getDate());
        customer.setAddress(address);
        return customer;    
    }

家庭控制器

@PostMapping("/customer")
     public ResponseEntity<String> addCustomer(@RequestBody CustomerDto customerDto){
        String message="";
        ResponseEntity<String> finalMessage=null;
        try {

        Customer customer=mapper.mapToEntity(customerDto);
        customerService.addCustomer(customer);
        message="Customer with "+customer.getId()+" sucessfully added";
        finalMessage= new ResponseEntity<>(message, HttpStatus.OK);

    }catch(Exception e) {
        message="Failed to add customer due to "+e.getMessage();
        finalMessage= new ResponseEntity<>(message, HttpStatus.NOT_ACCEPTABLE);
    }
        return finalMessage;
    }

请告诉我这样做的正确方法是什么，我认为我们不需要提供 id 字段。我该如何处理？在 EntityToDtoMapper mapToEntity() 方法中，如果我从 addressDto 中省略 setId() 那么它根本不起作用。

地址实体

@Entity
public class Address {
    @Id
    private int id;
    private String city;
    private String country; 
    @OneToOne(mappedBy="address",cascade=CascadeType.ALL)
    private Customer customer;
    public Address() {
        super();
    }
    ...getters and setters

客户实体

@Entity
@Table(name="customers")
public class Customer {

    @Id
    private int id;
    @Column(name="first_name")
    private String firstName;
    @Column(name="last_name")
    private String lastName;
    private String gender;

    @Temporal(TemporalType.TIMESTAMP)
    private Date date;

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

    public Customer() {
        super();
    }
...getters an setters

Answer 1

将 @GeneratedValue 添加到您的 ID 字段以正确自动生成实体 ID。请更改

@Id
private int id;

到

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id", unique = true, nullable = false)

在您的两个实体上。

然后在您的 mapToEntity 方法中添加行：

address.setCustomer(customer);

Answer 2

通过将新字段变量添加为 addressId

来更改您的 Customer 实体

@Entity
@Table(name="customers")
public class Customer {

    @Id
    private int id;

    @Column(name="first_name")
    private String firstName;

    @Column(name="last_name")
    private String lastName;

    private String gender;

    @Column(name="address_id")
    private int addressId;

    @Temporal(TemporalType.TIMESTAMP)
    private Date date;

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

    public Customer() {
        super();
    }
}

并如下更改 EntityDTOMapper

public Customer mapToEntity(CustomerDto customerDto) {
    Address address=new Address();
    address.setCity(customerDto.getAddressdto().getCity());
    address.setCountry(customerDto.getAddressdto().getCountry());
    address.setId(customerDto.getAddressdto().getId());

    Customer customer=new Customer();
    customer.setId(customerDto.getId());
    customer.setFirstName(customerDto.getFirstName());
    customer.setLastName(customerDto.getLastName());
    customer.setGender(customerDto.getGender());
    customer.setDate(customerDto.getDate());

    customer.setAddressId(customerDto.getAddressdto().getId());
    //customer.setAddress(address);

    return customer;    
}

如何在 POST 主体中发送数据而不在 Spring 引导中设置自动生成的主键值？

How can I send data in POST body without setting the auto generated primary key values in Spring Boot?

post

hibernate

hibernate-mapping

spring-boot

postman