Hibernate / Spring数据未在OneToMany映射中获取集合

Question

我在Hibernate中有一个一对多的映射，当我尝试通过Spring JPARepository将其从数据库中移出时，总是给我一个空集合。

我有一个看起来像这样的User类。

@Indexed
@Entity
@Table(name="usr", indexes = {@Index(columnList = "email", unique = true)})
public class User{
    private Long userId;
    private String email;
    private String firstname;
    private String lastname;
    private String phone;
    private String passwordHash;

    private Set<PollOption> votes;
    private Set<FriendsList> friendsList; 

//no args constructor
    public User() {
        votes = new HashSet<>();
        friendsList = new HashSet<>();
    }

/**
 * Gets the value of id
 *
 * @return the value of id
 */
@Id
@GenericGenerator(name = "userautoinc", strategy = "org.hibernate.id.enhanced.SequenceStyleGenerator",
          parameters = {
          @Parameter(name = "sequence_name", value = "userautoinc"),
          @Parameter(name = "optimizer", value = "hilo"),
          @Parameter(name = "initial_value", value = "1"),
          @Parameter(name = "increment_size", value = "1") }
)
@GeneratedValue(generator = "userautoinc")
@Column(name="userid")
    public Long getUserId() {
    return this.userId;
}

/**
 * Sets the value of id
 *
 * @param argId Value to assign to this.id
 */
public void setuserId(final Long argId) {
    this.userId = argId;
}

/**
 * Gets the value of email
 *
 * @return the value of email
 */
@Field
@NaturalId
@Column(name="email", nullable = false)
    public String getEmail() {
    return this.email;
}

/**
 * Sets the value of email
 *
 * @param argEmail Value to assign to this.email
 */
public void setEmail(final String argEmail) {
    this.email = argEmail;
}

/**
 * Gets the value of firstname
 *
 * @return the value of firstname
 */
@Field
@Column(name="firstname", nullable = false)
public String getFirstname() {
    return this.firstname;
}

/**
 * Sets the value of firstname
 *
 * @param argFirstname Value to assign to this.firstname
 */
public void setFirstname(final String argFirstname) {
    this.firstname = argFirstname;
}

/**
 * Gets the value of lastname
 *
 * @return the value of lastname
 */
@Field
@Column(name="lastname", nullable = false)
    public String getLastname() {
    return this.lastname;
}

/**
 * Sets the value of lastname
 *
 * @param argLastname Value to assign to this.lastname
 */
public void setLastname(final String argLastname) {
    this.lastname = argLastname;
}

/**
 * Gets the value of phone
 *
 * @return the value of phone
 */
@Column(name="phone", nullable = true)
public String getPhone() {
    return this.phone;
}

/**
 * Sets the value of phone
 *
 * @param argPhone Value to assign to this.phone
 */
public void setPhone(final String argPhone) {
    this.phone = argPhone;
}

/**
 * Gets the value of passwordHash
 *
 * @return the value of passwordHash
 */
@Column(name="passwordhash", nullable = false)
public String getPasswordHash() {
    return this.passwordHash;
}

/**
 * Sets the value of passwordHash
 *
 * @param argPasswordHash Value to assign to this.passwordHash
 */
public void setPasswordHash(final String argPasswordHash) {
    this.passwordHash = argPasswordHash;
}


@ManyToMany(mappedBy = "voters", fetch = FetchType.LAZY, cascade = CascadeType.ALL)
public Set<PollOption> getVotes(){
    return this.votes;
}

public void setVotes(Set<PollOption> votes){
    this.votes = votes;
}

@OneToMany(mappedBy = "user", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
public Set<FriendsList> getFriendsList() {
    return this.friendsList;
}

public void setFriendsList(Set<FriendsList> friendList) {
    this.friendsList = friendsList;
}

//To String Method
@Override
public String toString() {
    return "User [userId=" + userId + ", email=" + email + ", firstname=" + firstname + ", lastname=" + lastname
    + ", phone=" + phone + ", passwordHash=" + passwordHash + "]";
}

}

我的FriendsList实体（实际上是具有额外列的联结表）如下所示：

@Entity
@Table(name = "friendslist")  
public class FriendsList{


    private FriendsListPK friendsListPK;

    private User user;
    private User friend;
    private FriendsListStatus status;

    @Embeddable
    public static class FriendsListPK implements Serializable{

        private Long userId;
        private Long friendId;

        private FriendsListPK(){
        }

        private FriendsListPK(Long userId, Long friendId){
            this.userId = userId;
            this.friendId = friendId;
        }

        @JoinColumn(name = "userid", referencedColumnName = "userid")
        public Long getUserId(){
             return this.userId;
        }

        public void setUserId(Long userId){
            this.userId = userId;
        }

        @JoinColumn(name = "friendid", referencedColumnName = "userid")
        public Long getFriendId(){
             return this.friendId;
        }

        public void setFriendId(Long friendId){
            this.friendId = friendId;
        }

        @Override
        public boolean equals(Object o){
            boolean ret = true;
            if (o == null || this.getClass() != o.getClass()){ 
                ret = false;
            } else{
            FriendsListPK that = (FriendsListPK) o;
            ret = this.userId.equals(that.getUserId()) &&
            this.friendId.equals(that.getFriendId());
            }
            return ret;
        }

        @Override
        public int hashCode(){
            return Objects.hash(this.userId, this.friendId);
        }

    }

/**
 * Gets the value of friendsListPK
 *
 * @return the value of friendsListPK
 */
@EmbeddedId
public FriendsListPK getFriendsListPK() {
   return this.friendsListPK;
}

/**
 * Sets the value of friendsListPK
 *
 * @param argFriendsListPK Value to assign to this.friendsListPK
 */
public void setFriendsListPK(FriendsListPK argFriendsListPK) {
    this.friendsListPK = argFriendsListPK;
}

/**
 * Gets the value of status
 *
 * @return the value of status
 */
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "statusid", referencedColumnName = "statusid", nullable = false)
public FriendsListStatus getStatus() {
    return this.status;
}

/**
 * Sets the value of status
 *
 * @param argStatus Value to assign to this.status
 */
public void setStatus(FriendsListStatus argStatus) {
    this.status = argStatus;
}


/**
 * Gets the value of user
 *
 * @return the value of user
 */
@ManyToOne
@MapsId("userId")
public User getUser() {
    return this.user;
}

/**
 * Sets the value of user
 *
 * @param argUser Value to assign to this.user
 */
public void setUser(User argUser) {
    this.user = argUser;
}

/**
 * Gets the value of friend
 *
 * @return the value of friend
 */
@ManyToOne
@MapsId("friendId")
public User getFriend() {
    return this.friend;
}

/**
 * Sets the value of friend
 *
 * @param argFriend Value to assign to this.friend
 */
public void setFriend(User argFriend) {
    this.friend = argFriend;
}

}

而我的用户存储库就是这样的：

@Repository
public interface UserDao extends JpaRepository<User, Long>{
    public User findByEmail(String email);
}

每当我获得一个用户并尝试通过getFriendsList（）访问其FriendsList时，它总是给我一个空集。 我已经尝试过FetchType.EAGER以及Hibernate.Initialize并使用@Transactional和getFriendsList（）。size（）初始化旧方法。 我相信getFreindsList（）的结果是HashSet而不是某种冬眠代理集的事实表明，冬眠甚至没有尝试延迟初始化它？

休眠生成的SQL如下：

select user0_.userid as userid1_11_, user0_.email as email2_11_,  user0_.firstname as firstname3_11_, user0_.lastname as lastname4_11_, user0_.passwordhash as passwordhash5_11_, user0_.phone as phone6_11_ from usr user0_ where user0_.email =?;
select friendslis0_.user_userid as user_userid2_4_0_, friendslis0_.friend_userid as friend_userid1_4_0_, friendslis0_.friend_userid as friend_userid1_4_1_, friendslis0_.user_userid as user_userid2_4_1_, friendslis0_.statusid as statusid3_4_1_, user1_.userid as userid1_11_2_, user1_.email as email2_11_2_, user1_.firstname as firstname3_11_2_, user1_.lastname as lastname4_11_2_, user1_.passwordhash as passwordhash5_11_2_, user1_.phone as phone6_11_2_, friendslis2_.statusid as statusid1_5_3_, friendslis2_.statusname as statusname2_5_3_ from friendslist friendslis0_ inner join usr user1_ on friendslis0_.friend_userid=user1_.userid inner join friendsliststatus friendslis2_ on friendslis0_.statusid=friendslis2_.statusid where friendslis0_.user_userid=?;
select friendslis0_.user_userid as user_userid2_4_0_, friendslis0_.friend_userid as friend_userid1_4_0_, friendslis0_.friend_userid as friend_userid1_4_1_, friendslis0_.user_userid as user_userid2_4_1_, friendslis0_.statusid as statusid3_4_1_, user1_.userid as userid1_11_2_, user1_.email as email2_11_2_, user1_.firstname as firstname3_11_2_, user1_.lastname as lastname4_11_2_, user1_.passwordhash as passwordhash5_11_2_, user1_.phone as phone6_11_2_, friendslis2_.statusid as statusid1_5_3_, friendslis2_.statusname as statusname2_5_3_ from friendslist friendslis0_ inner join usr user1_ on friendslis0_.friend_userid=user1_.userid inner join friendsliststatus friendslis2_ on friendslis0_.statusid=friendslis2_.statusid where friendslis0_.user_userid=?;

如您所见，休眠IS确实是在尝试查询friendslist表，但是数据并没有进入集合。 我与众不同的唯一发现是，我们有2个相同的select语句进入了我们通常期望的1个好友列表中。也许第二个语句具有不同的用户名，并且正在覆盖第一个？ 但是我不知道第二选择来自何处。

Answer 1

这似乎是一个棘手的问题，我的解释如下：

FiendsList实体具有一个复合主键，该主键由userId和friendId列组成。 它们都引用User.userId列：

    @JoinColumn(name = "userid", referencedColumnName = "userid")
    public Long getUserId(){
         return this.userId;
    }

    @JoinColumn(name = "friendid", referencedColumnName = "userid")
    public Long getFriendId(){
         return this.friendId;
    }

但是，在User实体上，您仅将映射放在FiendList.user字段中：

@OneToMany(mappedBy = "user", fetch = FetchType.EAGER, cascade = CascadeType.ALL)
public Set<FriendsList> getFriendsList() {
    return this.friendsList;
}

问题是休眠不足以让休眠状态知道要做什么，因为它需要两列才能正确获取FriendsList实体（因为它具有复合pk）。

您可能需要你一旦为了得到让用户实体执行附加查询FriendList通过将检索到的User.userId值作为的一部分FriendListPk ：

@Repository
public interface FriendListDao extends JpaRepository<FriendList, FriendListPK>{
    ... // use one of the find methods to get by pk.
}

Answer 2

糟糕，我很傻，原来我在二传手中只是想念一个“ s”。

public void setFriendsList(Set<FriendsList> friendList) {
    this.friendsList = friendsList;
}

实际上，“朋友列表”应该是“朋友列表”。 Hibernate正在获取数据，但是当它试图将其分配给字段时，现有的实例变量将被默默地分配给自己而不是分配器的参数。