如何在oneToMany關系中正確創建新子項以避免托管重繪和分離物體錯誤？【Spring PostgreSQL】

預訂系統涉及物體：用戶、預訂、課程。
問題： ReservationService 不會將預訂添加到資料庫或用戶。Reservation 預訂的自動生成的 long id = new Reservation(); 是 0 [我試過序列生成型別，沒用]。然后，reservationService.addReservation(reservation); 導致錯誤：分離的物體傳遞給持久化。關系：

用戶 -- oneToMany --> 預訂 -- oneToOne --> 課程

預訂控制器：

@Controller
public class ReservationController {

    private final AppUserService appUserService;
    private final ReservationService reservationService;
    private final CourseService courseService;

    public ReservationController(AppUserService appUserService, ReservationService reservationService, CourseService courseService) {
        this.appUserService = appUserService;
        this.reservationService = reservationService;
        this.courseService = courseService;
    }

    @RequestMapping(value = "bookCourse/{courseId}")
    public String bookCourse(@PathVariable("courseId") long id, Principal principal) {

        AppUser user = appUserService.getAppUser(principal.getName());
        Course course = courseService.getCourse(id);

        Reservation reservation = new Reservation();

        reservation.setAppUser(user);
        reservation.setCourse(course);

        reservationService.addReservation(reservation);

        user.getReservations().add(reservation); //this line causes the error:       
        //org.hibernate.PersistentObjectException: detached entity passed to persist: application.domain.Course

        appUserService.editAppUser(user);

        return "redirect:/reservations.html";
    }

預訂服務實施：

@Service("reservationService")
@Transactional
public class ReservationServiceImpl implements ReservationService {

    private final ReservationRepository reservationRepository;

    @Autowired
    public ReservationServiceImpl(ReservationRepository reservationRepository) {
        this.reservationRepository = reservationRepository;
    }

    @Transactional
    public void addReservation(Reservation reservation) {
        reservationRepository.save(reservation);
    };

用戶.java：

@Entity
@Table(name="appuser")
public class AppUser {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    long id;

    @OneToMany(mappedBy="appUser", cascade = CascadeType.ALL)
    private Set<Reservation> reservations;

    //getters and setters
}

預訂.java：

@Entity
@Table(name="reservation")
public class Reservation {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @OneToOne(cascade = CascadeType.ALL)
    @JsonManagedReference
    private Course course;

    @ManyToOne(fetch = FetchType.EAGER)//(mappedBy="reservation")
    private AppUser appUser;

    //getters and setters

課程.java：

@Entity
@Table(name="course")
public class Course {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @NotEmpty
    private String name;

    @NotEmpty
    private String price;

    @JsonBackReference
    @OneToOne(mappedBy="course")
    private Reservation reservation;
   
    //getters and setters

In the database, an appuser_reservation junction table has been created to link the ids of a user and a reservation.
In the database, reservation has appuser_id and course_id "automatically" created, and of bigint type.

The ReservationController method is referenced by clicking in course list html:

<a href="bookCourse/${course.id}">Book!</a> 

[course already is in the Database]

Framework is Spring, database is PostgreSQL.

If there are more information needed, please let me know!

Since I don't know if the issue is that new Reservation() creates a reservation with a null id, so then it can't be properly saved to DB by the service, or there is some other, more general issue, I don't know if the title of this question is appropriate.

我知道錯誤本身已經被討論過，但我現在好幾天都無法解決這個問題，我將非常感謝所有的洞察力。也許我在理解關系或 new Entity() 的作業方式方面犯了一個錯誤。

在 bookCourse{id} 中創建一個新的 Reservation() 有什么問題；并將其寫入資料庫？

uj5u.com熱心網友回復：

好的，我們親自見面并解決了問題。

與 JPA 和 Hibernate 的關系是錯誤的。
您需要將 Reservation 中的關系更改為：

@Entity
@Table(name="reservation")
public class Reservation { 
...
@ManyToOne
private Course course;
...
}

你在這里有一個現場繪圖，所以你可以記住什么是什么以及關系是什么。

預訂預訂的自動生成的長 ID = new Reservation(); 是 0

首先，您需要將創建的物體添加到資料庫并獲取它以獲取自動生成的 id，如下所示：

Reservation reservation = new Reservation();  
// fill the reservation

// I assume there is repository.save(reservation); 
Reservation reservationWithID = reservationService.add(reservation);

// assign reservationWithID to User Reservation List to create connection

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