java - OneToMany 关系不加载集合

Question

我有一个与 TravellerVote 存在关系的类 Traveller。我希望检索旅行者发送的选票和收到的选票。

等级旅行者

@Entity
@PrimaryKeyJoinColumn(name="user_id")
public class Traveller extends User{

    @OneToMany(mappedBy="voter", cascade={CascadeType.PERSIST,  CascadeType.REMOVE})
    protected List<TravellerVote> myVotes;

    @OneToMany(mappedBy="partner", cascade={CascadeType.PERSIST,  CascadeType.REMOVE})
    protected List<TravellerVote> receivedVotes;

    public Traveller(){
        super();
        myVotes = new ArrayList<TravellerVote>();
        receivedVotes = new ArrayList<TravellerVote>();
    }

    public void addVote(TravellerVote vote){
        myVotes.add(vote);
    }
}

旅行者投票类别

@Entity
public class TravellerVote {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(length=11)
    protected long id;

    @ManyToOne
    @JoinColumn(name="voter_id")
    protected Traveller voter;

    @ManyToOne
    @JoinColumn(name="partner_id")
    protected Traveller partner;

    @Column(nullable=false)
    protected Date date;

    @Column(length=300, nullable=true)
    protected String comment;

    @Column(nullable=false)

    protected boolean positive;
        public TravellerVote(Traveller voter, Traveller partner, String comment, boolean positive){
        this.date = new Date();
        this.voter = voter;
        this.partner = partner;
        this.comment = comment;
        this.positive = positive;   
    }
}

实体正确地保存在数据库中，如下:

TravellerDao dao = new TravellerDao();
Traveller t1 = dao.findByUsername("ramiromd");
Traveller t2 = dao.findByUsername("danimd");
Traveller t3 = dao.findByUsername("famartinezd");

t1.addVote(new TravellerVote(t1, t2, "Test comment", true));
t2.addVote(new TravellerVote(t2, t1, "Test comment", true));
t3.addVote(new TravellerVote(t3, t1, "Test comment", false));

dao.save(t1);
dao.save(t2);
dao.save(t3);

但是，当在其他 Controller 中尝试获取投票时，集合有 0 个元素。

TravellerDao dao = new TravellerDao();
Traveller t1 = dao.findByUsername("ramiromd");
t1.getReceivedVotes().size(); // Must be 2 not 0.
t1.getMyVotes().size(); // Must be 1 not 0.

TravelerDao 保存方法:

public void save(Traveller traveller) {
    EntityManager em = PersistenceHelper.getEm();
    em.getTransaction().begin();
    em.persist(traveller);
    //em.flush();
    em.getTransaction().commit();
}

这是findByUsername方法:

public Traveller findByUsername(String username){

    EntityManager em = PersistenceHelper.getEm();
    String qstring = "SELECT t FROM Traveller  t WHERE t.username LIKE :username";
    TypedQuery<Traveller> query = em.createQuery(qstring, Traveller.class);
    query.setParameter("username", username);


    return query.getSingleResult();
}

有什么想法吗？

Answer 1

在这些场景中，启用实现框架的调试会很有帮助。例如，如果您使用 Hibernate，请将 org.hibernate 包的调试级别设置为“DEBUG”。这将显示所有被触发的 SQL。在您的问题中，如果您将 fetch 设置为 LAZY ，您将看到这些语句触发了精确的 SQL:-

t1.getReceivedVotes().size(); // Must be 2 not 0.
t1.getMyVotes().size();

希望这能给您足够的提示来解决这个问题。