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

我有以下3种型号：

模式一：预订

    @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<>();
}

模式2：客房预订：

 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<>();
    }

型号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;

}

这三个之间的关系如下：

预订-一对一的RESERVATION_ID->房间预订-一对多的ROOM_RESERVATION_ID->客人详细信息

我正在获取预订对象并尝试更新访客详细信息，但出现以下错误：

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的建议将cascadeType更改为ALL，但仍然出现相同的错误。请不要重复，因为我已经尝试了针对已经问过的这种问题的所有解决方案

请让我知道我在做什么错。 谢谢

通过更改GuestDetails保存预订对象的代码：

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);

Answer 1

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

该异常清楚地表明， RoomReservation包含的GuestDetails在数据库中不存在（并且很可能其id为null ）。

通常 ，您可以通过以下方法解决此异常：

• 保存GuestDetails之前先保存RoomReservation实体

• 或为@ManyToOne GuestDetail-->RoomReservation cascade = CascadeType.ALL （或至少{CascadeType.MERGE, CascadeType.PERSIST} ）

但是首先，我要说明两点：

• 不要在您的课程中使用公共字段，这违反了封装概念 。

• 当您具有双向关联时，可以在Setter方法中设置关联的另一端。

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);

希望能帮助到你！

Answer 2

GuestDetails添加所需的CasadeType：

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

RoomReservation-添加所需的CascadeType：

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

然后，您需要在使用for-each循环之前/之后保留数据。 取决于您safe() Method。

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

然后将其安全。 告诉我你的结果。 也许可以直接工作而无需更改/添加级联类型。

Answer 3

您可以保存从Json获得的预订。 JPA将更新具有相同ID的行。

您收到的错误是因为guestDetails仍具有对updatedRoomReservation的引用。 如果您不想从json保存整个预订，则必须设置正确的RoomReservation。

例如：

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

Answer 4

如果您使用的是JPA 2.0，则OneToMany的默认获取类型为LAZY。 如果你的拉姆达后，您updatedRoomReservation为空 （如您在设置否则容易），那么existingReservation.roomReservation.guestDetails将永远不会被加载，并会为空。

因此，当您保存existingReservation ，您会收到错误消息。

Answer 5

这可能是由错误的事务语义引起的。

如果在当前事务中未获取引用的实例，则将其视为瞬态。

最简单的解决方案是将@Transactional添加到方法中：

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