如何获得jpa的参与者?

时间:2019-05-21 00:53:29

标签: java spring jpa

我有这个实体:

@Entity
@Table(name = "Participant")
public class Participant {

    @Column(name = "user")
    User user;

    @Column(name = "meet")
    Meet meet;

    @Enumerated(EnumType.STRING)
    @Column(name = "rol")
    Rol rol;

与我见面的班级是:

@Entity
@Table(name = "Meet")
public class Meet implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @Column(unique = true, name = "name")
    private String name;

    @Enumerated(EnumType.STRING)
    @Column(name = "state")
    private State state;

我想参加见面会

@Query("SELECT p FROM Participant p where p.meet= :id")
    Participant findByMeet(@Param("id") Long id);

我如何获取聚会中的参与者列表

2 个答案:

答案 0 :(得分:0)

假设@IdLongParticipant的{​​{1}}应该就是这样。

@Repository
public interface ParticipantRepository implements CrudRepository<Participant, Long> {
  List<Participant> findAllByMeet(Meet meet);
}

Spring Data将为您动态提供实现。

答案 1 :(得分:0)

首先,将文件添加到域模型中。

@Entity 
@Table(name = "Participant")
public class Participant {

    @Column(name = "user")
    User user;

    @ManyToOne
    @Column(name = "meet_id") // It's a foreign key of Participant 
    Meet meet;

    @Enumerated(EnumType.STRING)
    @Column(name = "rol")
     Rol rol;
}

@Entity
@Table(name = "Meet")
public class Meet implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    @Column(unique = true, name = "name")
    private String name;

    @Enumerated(EnumType.STRING)
    @Column(name = "state")
    private State state;

    @OneToMany(mappedBy = "Meet", cascade = CascadeType.ALL, orphanRemoval = true)
    private Set<Participant> participants = new HashSet<>();
}