JPA @ManyToOne 只有 JSON swagger body 中的 ID？

Question

我创建了一个 JPA object，它与另外 2 个 object 有 @ManyToOne 关系。但是我在让它工作时遇到了问题。特别是关于 JSON body。如果它们看起来不重要，我已经删除了一些注释

public class Booking implements Serializable {

    @Id
    @ApiModelProperty(readOnly = true)
    @GeneratedValue(strategy = GenerationType.TABLE)
    private Long id;

    @NotNull
    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.DETACH)
    @JoinColumn(name = "customer_id")
    private Customer customer;

    @NotNull
    @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.DETACH)
    @JoinColumn(name = "taxi_id")
    private Taxi taxi;

    @NotNull
    @Future(message = "Bookings can not be in the past. Please choose one from the future")
    @Column(name = "booking_date")
    @Temporal(TemporalType.DATE)
    private Date bookingDate;
}

public class Customer implements Serializable {
    @Id
    @ApiModelProperty(readOnly = true)
    @GeneratedValue(strategy = GenerationType.TABLE)
    private Long id;
    private String firstName;
    private String lastName;
    private String email;
    private String phoneNumber;
    @OneToMany(mappedBy = "taxi", cascade = CascadeType.ALL)
    @ApiModelProperty(hidden = true)
    private List<Booking> bookings;
}

public class Taxi implements Serializable {
    @Id
    @ApiModelProperty(readOnly = true)
    @GeneratedValue(strategy = GenerationType.TABLE)
    private Long id;
    private String registration;
    private String numberOfSeats;
    @OneToMany(mappedBy = "taxi", cascade = CascadeType.ALL)
    @ApiModelProperty(hidden = true)
    private List<Booking> bookings;
}

休息服务

public Response createBooking(
            @ApiParam(value = "Customer ID, Taxi ID and a booking date required to create a booking", required = true)
                    org.jboss.quickstarts.wfk.booking.Booking booking)
    {
        if (booking == null) {
            throw new RestServiceException("Bad Request", Response.Status.BAD_REQUEST);
        }

        Response.ResponseBuilder builder;

        try {
            // Go add the new Booking.
            service.create(booking);
            builder = Response.status(Response.Status.CREATED).entity(booking);


        } catch (ConstraintViolationException ce) {
            //Handle bean validation issues
            Map<String, String> responseObj = new HashMap<>();

            for (ConstraintViolation<?> violation : ce.getConstraintViolations()) {
                responseObj.put(violation.getPropertyPath().toString(), violation.getMessage());
            }
            throw new RestServiceException("Bad Request", responseObj, Response.Status.BAD_REQUEST, ce);

        } catch (Exception e) {
            // Handle generic exceptions
            throw new RestServiceException(e);
        }

        log.info("createBooking completed. Booking = " + booking.toString());
        return builder.build();
    }

服务 class 其中 crud 是存储库

@Inject
private BookingRepository crud;

Booking create(Booking booking) throws ConstraintViolationException, ValidationException, Exception {
        log.info("BookingService Testing: " + booking.toString());
        log.info("BookingService.create() - Creating " + booking.getCustomer().getId() + " " + booking.getTaxi().getId());
        
        // Check to make sure the data fits with the parameters in the Booking model and passes validation.
        //validator.validateBooking(booking);

        // Write the booking to the database.
        return crud.create(booking);
    }

存储库class

@Inject
private EntityManager em;
Booking create(Booking booking) throws ConstraintViolationException, ValidationException, Exception {
        log.info("BookingRepository.create() - Creating " + booking.getCustomer().getId() + " " + booking.getTaxi().getId());

        // Write the booking to the database.
        em.persist(booking);

        return booking;
    }

但是当我查看我的 swagger 文档时，它显示了这一点。但我不想创建新的 object。我想访问 ID 为

的 objects

{
  "customer": {
    "firstName": "string",
    "lastName": "string",
    "email": "string",
    "phoneNumber": "string"
  },
  "taxi": {
    "registration": "string",
    "numberOfSeats": 0
  },
  "bookingDate": "2021-11-11T22:20:34.952Z"
}

所以我想将其发送为 body

{
  "customer_id": 0,
  "taxi_id": 0,
  "bookingDate": "2021-11-11T22:20:34.952Z"
}

有人body对此有什么解决办法吗？如果您需要更多代码，请告诉我。

Answer 1

您需要更改 RestService 以接受以下实例 class 而不是 Booking:

public class BookingCreationRequest {
    @JsonProperty("customer_id")
    private Long customerId;
    
    @JsonProperty("taxi_id")
    private Long taxiId;
    
    private Date bookingDate;
}

所以你的 RestService 中的内容类似于以下内容：

public Response createBooking(BookingCreationRequest bookingCreationRequest) {
     (...)
     
     service.create(bookingCreationRequest);

     (...)
}

现在在您的服务 class 中，您需要获取与请求关联的 Customer 和 Taxi 实体，构建 Booking 对象并将其持久化在数据库中：

@Inject
private BookingRepository crud;

@Inject
private TaxiService taxiService;

@Inject
private CustomerService customerService;

Booking create(BookingCreationRequest bookingCreationRequest) throws ConstraintViolationException, ValidationException, Exception {
    (...) 

    Taxi taxi = taxiService.getTaxiById(bookingCreationRequest.getTaxiId());
    Customer customer = customerService.getCustomerById(bookingCreationRequest.getCustomerId());
    Booking booking = new Booking();
    booking.setTaxi(taxi);
    booking.setCustomer(customer);
    booking.setBookingDate(bookingCreationRequest.getBookingDate());

    return crud.create(booking);
}

JPA @ManyToOne 只有 JSON swagger body 中的 ID？

JPA @ManyToOne Only having the ID in the JSON swagger body?

java

jpa

swagger

java-ee-7