Запрос на соединение критериев для составного первичного ключа в спящем режиме

Question

Требуется запрос на присоединение критериев для составного первичного ключа.Объекты:

1. ArtWork

   @Entity
   @Table(name = "artwork")
   public class ArtWork implements io.malevich.web.entity.Entity {
   
   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   private Long id;
   @Id
   @Column(name = "language", columnDefinition = "CHAR(2)")
   private String language;
   @Column(name = "art_name", nullable = false)
   private String artName;
   @Column(name = "creation_date", nullable = false)
   private Date creationDate;
   @Column(name = "edition_flag", nullable = false, columnDefinition = "tinyint(1)")
   private boolean editionFlag;
   @Column(name = "replica_flag", nullable = false, columnDefinition = "tinyint(1)")
   private boolean replicaFlag;
   @Column(name = "number_of_editions")
   private Long numberOfEditions;
   @Column(name = "original_id")
   private Long originalId;
   
   @ManyToOne
   @JoinColumns({
           @JoinColumn(
                   name = "category_id",
                   referencedColumnName = "id", insertable = false, updatable = false),
           @JoinColumn(
                   name = "language",
                   referencedColumnName = "language", insertable = false, updatable = false)
   })
   private Category category;
   
   @ManyToOne
   @JoinColumns({
           @JoinColumn(
                   name = "gallery_id",
                   referencedColumnName = "id", insertable = false, updatable = false),
           @JoinColumn(
                   name = "language",
                   referencedColumnName = "language", insertable = false, updatable = false)
   })
   private Gallery gallery;
   
   @ManyToOne
   private Specialization specialization;
   
   @ManyToOne
   @JoinColumns({
           @JoinColumn(
                   name = "author_id",
                   referencedColumnName = "id", insertable = false, updatable = false),
           @JoinColumn(
                   name = "language",
                   referencedColumnName = "language", insertable = false, updatable = false)
   })
   private Author author;
   
   @Column
   private String description;
   @Column
   private Double price;
   
   //getter setter
   }
   

2. Пользователь:

   @javax.persistence.Entity
   @Table(name = "user")
   public class User implements Entity, UserDetails {
       @Id
       @GeneratedValue(strategy = GenerationType.IDENTITY)
       private Long id;
   
       @Column(unique = true, length = 255, nullable = false)
       private String name;
   
       @Column(length = 255, nullable = false)
       private String password;
   
       @ElementCollection(fetch = FetchType.EAGER)
       private Set<Role> roles = new HashSet<>();
   
       @Column(name = "user_type_id")
       private Long userTypeId;
   
       @ManyToOne
       @JoinColumn(name = "person_id", referencedColumnName = "id")
       private Person person;
   
       @ManyToOne
       @JoinColumn(name = "organization_id", referencedColumnName = "id")
       private Organization organization;
   
       @ManyToOne
       @JoinColumn(name = "file_id", referencedColumnName = "id")
       private File file;
   
   
       @Column(name = "activity_flag")
       private boolean activityFlag;
   //gettter and setter
   }
   

3. Состояния счета

   @javax.persistence.Entity
   @Table(name = "account_states")
   public class AccountStates implements Entity {
   
   @Id
   @GeneratedValue(strategy = GenerationType.IDENTITY)
   private Long id;
   
   @Column(insertable = false, updatable = false)
   private String language;
   
   @ManyToOne
   @JoinColumns({ @JoinColumn(name = "artwork_id", referencedColumnName = "id"),
           @JoinColumn(name = "language", referencedColumnName = "language") })
   private ArtWork artwork;
   
   @ManyToOne
   @JoinColumn(name = "art_owner_id", referencedColumnName = "id")
   private User artOwner;
   
   @Column(name = "quantity")
   private Long quantity;
   
   @Temporal(TemporalType.DATE)
   @Column(name = "buy_date")
   private Date buyDate;
   }
   

4. Состояние учетной записи Дао: открытый класс JpaAccountStatesDao расширяет JpaDao реализует AccountStatesDao {

   public JpaAccountStatesDao() {
   super(AccountStates.class);
   }
   
   @Override
   public AccountStates find(Long artOwnerId, Long artworkId, String language) {
   final CriteriaBuilder builder = this.getEntityManager().getCriteriaBuilder();
   final CriteriaQuery<AccountStates> criteriaQuery = builder.createQuery(AccountStates.class);
   
   Root<AccountStates> root = criteriaQuery.from(AccountStates.class);
   Predicate p1 = builder.and(builder.equal(root.get("artwork"), artworkId),
           builder.equal(root.get("artwork"), language), builder.equal(root.get("artOwner"), artOwnerId));
   
   criteriaQuery.where(p1);
   
   TypedQuery<AccountStates> typedQuery = this.getEntityManager().createQuery(criteriaQuery);
   
   return typedQuery.getSingleResult();
   }
   }
   

Я хочу найти состояния учетной записи, где указан идентификатор artOwner= 1 и language = en и artwork id = 1.

Может кто-нибудь предложить правильный запрос для того же?

Answer 1

Я нашел решение для того же, я попытался передать весь объект вместо идентификатора объекта.

Итак, окончательный запрос:

@Override
public AccountStates find(User artOwner, Artwork artwork) {


 final CriteriaBuilder builder = this.getEntityManager().getCriteriaBuilder();
    final CriteriaQuery<AccountStates> criteriaQuery = builder.createQuery(AccountStates.class);

    Root<AccountStates> root = criteriaQuery.from(AccountStates.class);
    Predicate p1 = builder.and(builder.equal(root.get("artwork"), artwork),
            builder.equal(root.get("artOwner"), artOwner));

    criteriaQuery.where(p1);

    TypedQuery<AccountStates> typedQuery = this.getEntityManager().createQuery(criteriaQuery);

    return typedQuery.getSingleResult();
    }
}

Теперь он работает успешно ..Спасибо