JPA@ManyToOne只有JSONswagger正文中的ID？

我創建了一個@ManyToOne與其他 2 個物件有關系的 JPA物件。但是我在讓它作業時遇到了問題。特別是關于 JSON 正文。如果它們看起來不重要，我已經洗掉了一些注釋

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;
}

休息服務

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();
    }

crud 是存盤庫的服務類

@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);
    }

存盤庫類

@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;
    }

但是當我查看我的 swagger 檔案時，它顯示了這一點。但我不想創建新物件。我想訪問帶有 id 的物件

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

所以我想把它作為正文發送

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

有沒有人對此有任何解決方案？如果您需要更多代碼，請告訴我。

uj5u.com熱心網友回復：

You need to change your RestService to accept an instance of the following class instead of Booking:

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

So something like the following in your RestService:

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

     (...)
}

Now in your service class, you need to get both the Customer and the Taxi entities associated with the request, build the Booking object and persist it in the database:

@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);
}

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