为什么我在实体的映射中获得了这个"重复列"进入一个Hibernate实体?处理这种情况的最佳方法是什么?

时间:2016-12-11 11:38:24

标签: hibernate jpa hibernate-mapping

我不是很喜欢Hibernate而且我有以下问题\怀疑。

我遇到这种情况:

1)我有这个 Room 实体类,代表住宿的房间:

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

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private Long id;


    @ManyToOne
    @JoinColumn(name = "id_accomodation_fk", nullable = false)
    private Accomodation accomodation;

    @ManyToOne
    @JoinColumn(name = "id_room_tipology_fk", nullable = false)
    private RoomTipology roomTipology;

    @OneToMany(mappedBy = "room")
    private List<RoomMedia> roomMediaList;

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

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

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

    @Column(name = "is_enabled")
    private Boolean isEnabled;

    ........................................................................
    ........................................................................
    GETTER AND SETTER METHODS
    ........................................................................
    ........................................................................
}

2)然后我有这个 RoomMedia 实体类,它代表与特定房间实体相关的照片:

@Entity
@Table(name = "room_media")
public class RoomMedia {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id")
    private Long id;

    @Column(name = "id_room")
    private Long idRoom;

    @ManyToOne
    @JoinColumn(name = "id_room", nullable = false)     // da rinominare anche sul DB in room_fk
    private Room room;

    @Lob
    @Column(name = "media")
    private byte[] media;

    private String description;

    ........................................................................
    ........................................................................
    GETTER AND SETTER METHODS
    ........................................................................
    ........................................................................

}

正如您在前面的代码中所看到的,这两个实体类以这种方式链接到togheter:

会议室实例代表特定会议室,并以这种方式包含其照片列表:

@OneToMany(mappedBy = "room")
private List<RoomMedia> roomMediaList;

这意味着单个会议室与许多 RoomMedia 实例相关联。 mappedBy =&#34;房间&#34;表示要在房间 RoomMedia 之间建立链接,请使用 RoomMedia 实体的会议室字段(代表 RoomMedia 实例的特定房间。

然后在 RoomMedia 实体类中,我有这个链接:

@ManyToOne
@JoinColumn(name = "id_room", nullable = false)    
private Room room;

表示许多 RoomMedia 实例与单个 Room intsance相关。加入操作是使用 room_media 表上的 id_room 字段完成的(它是在表级而非实体级别完成的。)

如果所有这些推理都有问题,请更正。

我的问题是,在应用程序启动时,我收到此错误消息:

Caused by: javax.persistence.PersistenceException: [PersistenceUnit: default] Unable to build Hibernate SessionFactory
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.persistenceException(EntityManagerFactoryBuilderImpl.java:1249)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.access$600(EntityManagerFactoryBuilderImpl.java:120)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:860)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:850)
    at org.hibernate.boot.registry.classloading.internal.ClassLoaderServiceImpl.withTccl(ClassLoaderServiceImpl.java:425)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl.build(EntityManagerFactoryBuilderImpl.java:849)
    at org.springframework.orm.jpa.vendor.SpringHibernateJpaPersistenceProvider.createContainerEntityManagerFactory(SpringHibernateJpaPersistenceProvider.java:60)
    at org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean.createNativeEntityManagerFactory(LocalContainerEntityManagerFactoryBean.java:353)
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.buildNativeEntityManagerFactory(AbstractEntityManagerFactoryBean.java:373)
    at org.springframework.orm.jpa.AbstractEntityManagerFactoryBean.afterPropertiesSet(AbstractEntityManagerFactoryBean.java:362)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.invokeInitMethods(AbstractAutowireCapableBeanFactory.java:1642)
    at org.springframework.beans.factory.support.AbstractAutowireCapableBeanFactory.initializeBean(AbstractAutowireCapableBeanFactory.java:1579)
    ... 39 more
Caused by: org.hibernate.MappingException: Repeated column in mapping for entity: com.betrivius.domain.RoomMedia column: id_room (should be mapped with insert="false" update="false")
    at org.hibernate.mapping.PersistentClass.checkColumnDuplication(PersistentClass.java:709)
    at org.hibernate.mapping.PersistentClass.checkPropertyColumnDuplication(PersistentClass.java:731)
    at org.hibernate.mapping.PersistentClass.checkColumnDuplication(PersistentClass.java:753)
    at org.hibernate.mapping.PersistentClass.validate(PersistentClass.java:506)
    at org.hibernate.mapping.RootClass.validate(RootClass.java:270)
    at org.hibernate.cfg.Configuration.validate(Configuration.java:1360)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1851)
    at org.hibernate.jpa.boot.internal.EntityManagerFactoryBuilderImpl$4.perform(EntityManagerFactoryBuilderImpl.java:857)
    ... 48 more

为什么Hibernate无法同时处理这两个问题?问题是什么?我该如何解决?

我认为也许这可能取决于在 RoomMedia 实例中我仍然不再需要 id_room 字段,因为在此实体中我也有把:

@ManyToOne
@JoinColumn(name = "id_room", nullable = false)    
private Room room;

本身包含此ID。

所以我认为删除这个字段:

@Column(name = "id_room")
private Long idRoom;

我会解决我的问题,是吗?

但现在我可能会遇到另一个问题:在我的DAO中,我可能会遇到一些涉及 idRoom 的复杂查询。因此,为了方便起见,我将这些私人房间; 私人Long idRoom; 字段放入我的 RoomMedia 实体类。

我怎样才能避免这个问题呢?

1 个答案:

答案 0 :(得分:2)

您绝对应该从映射中删除此列

@Column(name = "id_room")
private Long idRoom;

和Hibernate告诉你一样。

  

但现在我可能会遇到另一个问题:在我的DAO中,我可能会遇到一些涉及idRoom的复杂查询。因此,为了方便起见,我将拥有这两间私人房间;和私人Long idRoom;字段进入我的RoomMedia实体类。

正确的映射是

@ManyToOne
@JoinColumn(name = "id_room", nullable = false)    
private Room room;

它可以为您提供所需的一切

查询示例

Query query1 = em.createQuery("select rm from RoomMedia rm where rm.room.id = :id");

Query query2 = em.createQuery("select rm from RoomMedia rm join rm.room r where r.id = :id");
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