Question

我在数据库端有两个表（oracle）

create table GROUPS
(
  ID   NUMBER not null,
  GROUP_NAME VARCHAR2(30)
)alter table GROUPS
  add constraint ID primary key (ID)

和

create table ITEM_GROUP
(
  ITEM_ID VARCHAR2(30) not null,
  GROUP_ID    NUMBER not null
)

alter table ITEM_GROUP
  add constraint ITEM_GROUPD_ID primary key (ITEM_ID, GROUP_ID)

alter table ITEM_GROUP
  add constraint ITEM_GROUP_FK01 foreign key (GROUP_ID)
  references GROUPS (ID);

比我在Java端有映射类。我想做点什么，当我选择组来取他所有的项目时，我想用hibernate保存项目就是这样。

@Entity
@Table(name = "GROUPS")
public class Group {

    @Id
    @Column(name = "ID", nullable = false)
    @javax.persistence.SequenceGenerator(name = "groupIdGenerator", sequenceName = "GROUP_SEQ", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "groupIdGenerator")
    private int id;

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

    @JsonManagedReference
    @OneToMany(fetch = FetchType.EAGER, mappedBy="group",cascade = CascadeType.ALL)
    private List<GroupItems> groupItems = new ArrayList<>();
    // setters and getters 

}


@SuppressWarnings("serial")
@Embeddable
public class GroupItemPK implements Serializable {

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

    @Column(name = "GROUP_ID")
    private int id;
    // getters , setters , constructors , equals  hashcode methods

}



@Entity
@Table(name = "ITEM_GROUP")
public class GroupITEM {

    @EmbeddedId
    private GroupITEMtPK id;

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "ID")
    @JsonBackReference
    private Group group;
}

我感兴趣的是我在构建关系中犯了什么错误吗？如果我做了我的错误，因为我无法做出选择和保存查询而无例外。

我正在尝试使用我的代码

List<Group> list = sessionFactory.getCurrentSession().createQuery("from Group a").list();

这是我的Exception

org.hibernate.engine.jdbc.spi.SqlExceptionHelper could not extract ResultSet [n/a]
java.sql.SQLSyntaxErrorException: ORA-00904: "GROUPITE0_"."ID": invalid identifier

在冬眠和春天，很多人和一个人都在玩

0 个答案: