JPA 2抽象コレクションに参加する基準

Question

Criteria APIを使用して、クラスPlayerを階層にManagerと結合する方法はありますか？ 私はDAOで与えられたIDのマネージャーを持つすべてのプレイヤーを見つけたいと、このような何かしようとしたと思います：

問合せ：

CriteriaBuilder cb = em.getCriteriaBuilder(); 
CriteriaQuery<Player> query = cb.createQuery(Player.class); 
Root<Player> root = query.from(Player.class); 

Join<Player, PlayerTeam> playerTeamJoin = root.join(Player_.playerTeam); 
Join<PlayerTeam, PlayerTeamAssignment> playerTeamAssignmentJoin = playerTeamJoin.join(PlayerTeam_.playerTeamAssignments); 

// Desired statement: 
Predicate predicate = cb.equal(playerTeamAssignmentJoin.get(ManagerAssignment_.manager), manager); 

クラス階層：

@Entity 
class Player { 
    @OneToOne(mappedBy = "player") 
    PlayerTeam playerTeam; 
} 

@Entity 
class PlayerTeam { 
    @JoinColumn(name = "PLAYER_ID", referencedColumnName = "ID") 
    Player player; 
    @OneToMany(mappedBy = "playerTeam") 
    Collection<PlayerTeamAssignment> playerTeamAssignments; 
} 

@Entity 
@DiscriminatorColumn(name = "ASSIGNMENT_TYPE", discriminatorType = DiscriminatorType.STRING) 
public abstract class PlayerTeamAssignment { // It has three implementations 
    @ManyToOne 
    @JoinColumn(name = "PLAYER_TEAM_ID", referencedColumnName = "ID") 
    PlayerTeam playerTeam; 
} 

@Entity 
@DiscriminatorValue("INDIVIDUAL") 
public class ManagerAssignment extends PlayerTeamAssignment { 
    @ManyToOne 
    @JoinColumn(name = "MANAGER_ID", referencedColumnName = "ID") 
    Manager manager; // id inside 
} 

Answer 1

を私はすでに私の問題の解決策を見つけました。 は私が

Join<Player, PlayerTeam> playerTeamJoin = root.join(Player_.playerTeam); 
Join<PlayerTeam, PlayerTeamAssignment> playerTeamAssignments = root.join(PlayerTeam_.playerTeamAssignments); 
Path managerPath = ((Path) playerTeamAssignments.as(ManagerAssignment.class)).get(ManagerAssignment_.manager)); 

パスを使用していた、あなたは、述語の例のためにそれを使用することができます。

Predicate predicate = cb.equal(managerPath.get(Manager_.globalId), managerId);