سؤال

UML Diagram

I need association OneToMany from Country to superclass Place (@MappedSuperclass). It could be bidirectional. I wolud need something like @OneToAny...

@MappedSuperclass
public class Place {

private String name;
private Country country;

@Column
public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

@ManyToOne
@JoinColumn(name="country_id")
public Country getCountry() {
    return country;
}

public void setCountry(Country country) {
    this.country = country;
}
}

Country:

@Entity
   public class Country {
   private long id;
   private String name;
   private List<Place> places;

   @Any(metaColumn = @Column(name = "place_type"), fetch = FetchType.EAGER)
   @AnyMetaDef(idType = "integer", metaType = "string", metaValues = {
         @MetaValue(value = "C", targetEntity = City.class),
         @MetaValue(value = "R", targetEntity = Region.class) })
   @Cascade({ org.hibernate.annotations.CascadeType.ALL })
   //@JoinColumn(name="unnecessary") 
   //@OneToMany(mappedBy="country")  // if this, NullPointerException...
   public List<Place> getPlaces() {
      return places;
   }
//and rest of class

Without @JoinColunm there is exception

Caused by: org.hibernate.AnnotationException: @Any requires an explicit @JoinColumn(s): tour.spring.bc.model.vo.Country.places

In table City and Region is foreign key to table Country (Region.country_id, City.country_id) which is ok. But I do NOT need foreign key in table Country to tables Region and City so I don't need @JoinColum.

I was looking for solution a lot but it seems there is no good solution.

هل كانت مفيدة؟

المحلول

@Any doesn't make sense here since foreign key is at the Places side and therefore it doesn't need additional meta column.

I'm not sure if it's possible to create a polymorphic relationship to @MappedSuperclass. However, you can try to declare Place as @Entity @Inheritance(InheritanceType.TABLE_PER_CLASS), it should produce the same database schema and allow polymorpic relationship.

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