Question

我有这堂课，我想坚持上这堂课，但是我遇到错误。

@Entity
@Table(name = "Participant")
public class Participant {

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

    @OneToOne(cascade = {CascadeType.ALL}, fetch = FetchType.EAGER)
    @JoinColumn(name="user_id")
    User user;

    @OneToOne(cascade = {CascadeType.ALL}, fetch = FetchType.EAGER)
    @JoinColumn(name="meet_id")
    Meet meet;

    @Enumerated(EnumType.STRING)
    @Column(name = "rol")
    Rol rol;
}

当我尝试保存时，出现此错误：

2019-05-20 22:03:29.637 ERROR 17416 --- [nio-8080-exec-9] o.a.c.c.C.[.[.[/].[dispatcherServlet]    : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exceptio
n [Request processing failed; nested exception is org.springframework.dao.InvalidDataAccessApiUsageException: detached entity passed to persist: com.example.demo.model.User; nested exceptio
n is org.hibernate.PersistentObjectException: detached entity passed to persist: com.example.demo.model.User] with root cause

org.hibernate.PersistentObjectException: detached entity passed to persist: com.example.demo.model.User
        at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:124) ~[hibernate-core-5.0.12.Final.jar!/:5.0.12.Final]
        at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:765) ~[hibernate-core-5.0.12.Final.jar!/:5.0.12.Final]
        at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:758) ~[hibernate-core-5.0.12.Final.jar!/:5.0.12.Final]

如何解决我的问题？

在我的服务中，我做到了

public class MeetServiceImpl {
@Autowired
MeetRepository meetRepository;

@Autowired
UserRepository userRepository;

@Autowired
ParticipantRepository participantRepository;

static Logger log = Logger.getLogger(MeetServiceImpl.class.getName());

public Meet createMeet(Long idUser, Meet meet) {
    Meet currentMeet = meetRepository.findByName(meet.getName());
    User user = userRepository.findById(idUser);
    Participant participant = new Participant();

        participant.setRol(Rol.OWNER);
        Meet meet1 = new Meet();
        meet1.setName("nuevo");
        meet1.setState(State.start);

        participant.setMeet(meet1);
        participant.setUser(user);
        participantRepository.save(participant);
        return meetRepository.save(meet);
}

我不明白问题是什么，我做错了什么，这似乎没有逻辑

额外：这是型号，下面我展示了有关我使用的类的模型

@Entity
@Table(name = "Meet")
public class Meet implements Serializable {

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

    @Column(unique = true, name = "name")
    private String name;

    @Enumerated(EnumType.STRING)
    @Column(name = "state")
    private State state;

和班见面

@Entity
@Table(name = "User")
public class User implements Serializable {

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

    @Column(name = "name")
    private String name;

    @Column(name = "password")
    private String password;

    @Column(unique = true, name = "email")
    private String email;

    @Enumerated(EnumType.STRING)
    @Column(name = "life")
    Life life;

当我评论ID时

// @ GeneratedValue（strategy = GenerationType.AUTO）

我收到此错误

com.mysql.jdbc.exceptions.jdbc4.MySQLIntegrityConstraintViolationException: Duplicate entry '0' for key 'PRIMARY'

Answer 1

您的创建方法似乎不是事务性的。您可以在类或方法的顶部添加@Transactional吗？

Answer 2

您已经在对象Meet上进行了两次插入：

participantRepository.save(participant);
return meetRepository.save(meet);

问题是： 1-第一行（1）保存参与者和会议，因为您已经在参加者类别：

@OneToOne(cascade = {CascadeType.ALL}, fetch = FetchType.EAGER)
 @JoinColumn(name="meet_id")
 Meet meet;

因此，只要您保存与会议相关的参与者，会议便会保存自动

2-当您评论此行时：// @ GeneratedValue（strategy = GenerationType.AUTO） id不会生成，所以我们有两个相遇，在插入状态mnt由jpa生成，id为0。所以数据库说：键'PRIMARY'的条目'0'重复

Answer 3

如果将Spring JPA与Entity Manager结合使用，则可以在服务层顶部添加@Transactional。

 @Transactional
    public class MeetServiceImpl {

如何保存复合类

3 个答案: