hibernate中从父级到子级的ID

时间:2013-06-14 15:21:29

标签: java sql hibernate jpa mule

这是我的代码。我想基于父类生成一个自动ID。我正在使用一种方法来创建机场,所以我的ID就是空值。将生成AirportModel中的ID,但我不知道如何在子类中创建它。

@Entity(name = "Airport")
@Table(name = "ai_airport")
public class AirportModel {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id", nullable = false)
private Long id;
    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "airport_id")
    private List<AirportTranslatedModel> translations;

第二课(孩子):

    @Entity(name = "AirportTranslated")
    @IdClass(AirportTranslatedModelKey.class)
    @Table(name = "ai_translated_airport")
    public class AirportTranslatedModel 
        @Id
        @Column(name="airport_id")
        private Long airportId;

        @Id
        @Column(name="language_code", length=2)
        private String languageCode;

第三个(键):

@Embeddable
public class AirportTranslatedModelKey implements Serializable {

    @Column(name="airport_id")
    private Long airportId;

    @Column(name="language_code", length=2)
    private String languageCode;

我仍然有同样的错误;日志:

Hibernate: insert into ai_airport (active, airport_code, city_code, country_code, externa
l_id, is_default, latitude, longitude, market_code, min_connection_time_DD, min_connection_time_DI, min_connection_time_id, min_connection_time_II, time_diff, VERSION) values (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)

Hibernate: insert into ai_translated_airport (airport_long_name, airport_short_name, airp
ort_id, language_code) values (?, ?, ?, ?)
ERROR org.hibernate.engine.jdbc.spi.SqlExceptionHelper - Column 'airport_id' cannot be null

1 个答案:

答案 0 :(得分:1)

您当前的设置具有通过Long映射的AirportTranslatedModel airport_id字段 - 您需要手动设置airportId以使其在数据库中设置ID。这可能需要您持久保存AirportModel并可能在建立AirportModel-&gt; AirportTranslatedModel关联之前刷新其PK并且可用,以便您可以设置AirportTranslatedModel.airportId。

JPA 2虽然允许派生的ID。如果您希望AirportTranslatedModel从AirportModel分配其ID,则需要与它建立关系。 http://wiki.eclipse.org/EclipseLink/Examples/JPA/2.0/DerivedIdentifiers

有一个简单的例子

如果您要以类似的方式为类建模,它可能看起来像:

public class AirportModel {
  ..
  @OneToMany(mappedby="airportModel", cascade = CascadeType.ALL, orphanRemoval = true)
  private List<AirportTranslatedModel> translations;
  ..
}

public class AirportTranslatedModel {
  @Id
  @JoinColumn(name="airport_id")
  private AirportModel airportModel;

  @Id
  @Column(name="language_code", length=2)
  private String languageCode;
  ..
}
public class AirportTranslatedModelKey implements Serializable {
    private Long airportModel;

    private String languageCode;
}

请注意,如果您只是将它用作pk类,则无需制作AirportTranslatedModelKey并可嵌入。另请注意,AirportTranslatedModelKey包含Long airportModel - 这必须与AirportModel中的pk类型以及AirportTranslatedModel中的relationship属性的名称相匹配。

这将允许AirportTranslatedModel从AirportModel中提取airport_id值并将其用作PK,即使它们可能尚未生成,但两个实体仍然是新的。