对象引用了一个未保存的瞬态实例-在刷新之前保存瞬态实例：Spring Data JPA

Question

I have Below 3 models :

Model 1: Reservation

    @Entity
    public class Reservation  {
    
        public static final long NOT_FOUND = -1L;
    
        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        public Long id;
        
        @OneToMany(mappedBy = "reservation", cascade = CascadeType.ALL, orphanRemoval = true)
        public List<RoomReservation> roomReservations = new ArrayList<>();
}

模型2: 房间预订:

 public class RoomReservation extends{
    
        @Id
        @GeneratedValue(strategy = GenerationType.IDENTITY)
        public Long id;
    
        @JsonIgnore
        @ManyToOne(fetch = FetchType.LAZY)
        @JoinColumn(name = "RESERVATION_ID")
        public Reservation reservation;
    
        @OneToMany(mappedBy = "roomReservation", cascade = CascadeType.ALL, orphanRemoval = true)
        public List<GuestDetails> guestDetails = new ArrayList<>();
    }

Model 3：客人详细信息：

public class GuestDetails {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Long id;

    public Long guestId;

    @JsonIgnore
    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "ROOM_RESERVATION_ID")
    public RoomReservation roomReservation;

    public Boolean isPrimary;

    @Transient
    public Guest guest;

}

这三者之间的关系为：

预订 --基于 RESERVATION_ID 一对多--> 房间预订 --基于 ROOM_RESERVATION_ID 一对多--> 客人详细信息

我正在获取预订对象并尝试更新客人详细信息，但遇到以下错误：

org.hibernate.TransientPropertyValueException: object references an unsaved transient instance - save the transient instance before flushing : com.model.GuestDetails.roomReservation -> com.model.RoomReservation
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1760)
    at org.hibernate.jpa.spi.AbstractEntityManagerImpl.convert(AbstractEntityManagerImpl.java:1677)
    at org.hibernate.jpa.internal.TransactionImpl.commit(TransactionImpl.java:82)
    at org.springframework.orm.jpa.JpaTransactionManager.doCommit(JpaTransactionManager.java:517)
... 73 common frames omitted

我已经按照常见问题中建议的将cascadeType更改为ALL，但仍然遇到相同的错误。请不要重复提出已经问过的这种问题的解决方案，因为我已经尝试了所有相关的解决方案。

请告诉我我做错了什么。谢谢。

修改GuestDetails后保存Reservation对象的代码：

Reservation existingReservation = reservationRepository.findOne(reservationId);
Reservation reservation = reservationParser.createFromJson(reservationNode);
existingReservation.roomReservations.forEach(roomReservation -> {
                    RoomReservation updatedRoomReservation = reservation.roomReservations.stream().filter(newRoomReservation -> Objects.equals(roomReservation.id, newRoomReservation.savedReservationId)).findFirst().orElse(null);
                    if(updatedRoomReservation != null){
                        roomReservation.guestDetails = updatedRoomReservation.guestDetails;
                    }
                });
reservationRepository.save(existingReservation);

- utsav anand

---

请发布实际执行保存操作的代码 - 您要保存哪种类类型？ - PaulNUK

{btsdaf} - LenglBoy

{btsdaf} - soorapadman

{btsdaf} - utsav anand

谢谢大家，问题已经解决了，空的子对象导致了这个问题。 - utsav anand

Answer 1

这可能是由于事务语义不正确引起的。

如果所引用的实例未在当前事务中获取，则计为瞬态。

最简单的解决方法是将@Transactional添加到该方法中：

@Transactional
public void saveReservation(...) {
    Reservation existingReservation = reservationRepository.findOne(reservationId);
    Reservation reservation = reservationParser.createFromJson(reservationNode);
    // ...
    reservationRepository.save(existingReservation);
}

- OrangeDog

Answer 2

... save the transient instance before flushing : 
    com.model.GuestDetails.roomReservation -> com.model.RoomReservation

这个异常清楚地说明了在数据库中不存在包含在GuestDetails中的RoomReservation（很可能是它的id为null）。

通常情况下，您可以通过以下方式解决此异常：

• 在保存GuestDetails之前保存RoomReservation实体。

• 或者将@ManyToOne GuestDetail-->RoomReservation的cascade=CascadeType.ALL（或至少是{CascadeType.MERGE, CascadeType.PERSIST}）。

但首先，我有几点需要注意：

• 不要在类中使用公共字段，这违反了封装概念。

• 在具有双向关联的情况下，您可以在Setter方法中设置关联的另一侧。

对于您的情况，您应该更改RoomReservation类：

public class RoomReservation{

    //..... other lines of code

    @OneToMany(mappedBy = "roomReservation", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<GuestDetails> guestDetails = new ArrayList<>();

    public void setGuestDetails(List<GuestDetails> guestDetails) {

           this.guestDetails.clear();

           // Assuming that by passing null or empty arrays, means that you want to delete
           // all GuestDetails from this RoomReservation entity
           if (guestDetails == null || guestDetails.isEmpty()){
               return;
           }

           guestDetails.forEach(g -> g.setRoomReservation(this));
           this.guestDetails.addAll(guestDetails);
    }

    public List<GuestDetails> getGuestDetails() {
        // Expose immutable collection to outside world  
        return Collections.unmodifiableList(guestDetails);
    }

    // You may add more methods to add/remove from [guestDetails] collection
}

保存预订：

Reservation existingReservation = reservationRepository.findOne(reservationId);
Reservation reservation = reservationParser.createFromJson(reservationNode);
existingReservation.roomReservations.forEach(roomReservation -> {
                    Optional<RoomReservation> updatedRoomReservation = reservation.roomReservations.stream().filter(newRoomReservation -> Objects.equals(roomReservation.id, newRoomReservation.savedReservationId)).findFirst();
                    if(updatedRoomReservation.isPresent()){
                        // roomReservation already exists in the database, so we don't need to save it or use `Cascade` property
                        roomReservation.setGuestDetails( updatedRoomReservation.get().getGuestDetails());
                    }
                });
reservationRepository.save(existingReservation);

希望它有所帮助！

- O.Badr

Answer 3

如果您使用的是 JPA 2.0，则 OneToMany 的默认获取类型为 LAZY。如果在 lambda 后，您的 updatedRoomReservation 是 null（如您在 orElse 中设置的那样），则 existingReservation.roomReservation.guestDetails 将永远不会被加载并将为 null。

因此，当您保存 existingReservation 时，会出现错误。

- user8826113

Answer 4

GuestDetails - 添加所需的级联类型：

@ManyToOne(fetch = FetchType.LAZY, cascade=CascadeType.ALL)
@JoinColumn(name = "ROOM_RESERVATION_ID")
public RoomReservation roomReservation;

RoomReservation - 添加所需的级联类型：

@JsonIgnore
@ManyToOne(fetch = FetchType.LAZY, cascade=CascadeType.AL)
@JoinColumn(name = "RESERVATION_ID")
public Reservation reservation;

然后在使用for-each循环之前/之后，您需要持久化数据。这取决于您的safe()方法。

Reservation reservation = reservationParser.createFromJson(reservationNode);
entityManager.persist(reservation);

然后保存它。告诉我你的结果。也许可以直接工作，而不需要改变/添加级联类型。

- LenglBoy

---

我已经尝试将cascadeType.ALL添加到Guest Details，并在循环后持久化对象，但仍然得到相同的错误。 - utsav anand

Answer 5

您可以保存从Json获取的预订。 JPA将更新与相同ID的行。

您收到错误是因为guestDetails仍然引用了updatedRoomReservation。 如果您不想保存来自json的整个预订，您必须设置正确的RoomReservation。

e.g.:

if(updatedRoomReservation != null){
    roomReservation.guestDetails = updatedRoomReservation.guestDetails;
    guestDetails.forEach(guestDetail -> guestDetail.roomReservation = roomReservation);
}

- Tom