我在Hibernate中有一个一对多的Map，当我试图通过spring JPARepository从数据库中获取它时，总是给我一个空集合。
我有一个User类，看起来像这样。

@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 + "]";
}

}

字符串
我的JumsList实体（实际上是一个带有额外列的连接表）看起来像这样：

@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;
}

}

型
我的用户存储库就是这样：

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

型
每当我得到一个User并试图通过getUserList（）访问它的UserList时，它总是给我一个空集。我试过FetchType.EAGER，和Hibernate一样。初始化和初始化的老方法是使用@ translate和getFreindsList（）.size（）。我相信getFreindsList（）的结果是一个HashSet，而不是某种Hibernate代理集，这表明Hibernate甚至没有尝试懒惰-初始化它？
Hibernate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=?;

型
正如你所看到的，hibernate确实在尝试查询friendslist表，但是数据并没有进入集合。我看到的唯一不寻常的事情是，我们在friendslist中有两个相同的select语句，而我们通常期望的是1。也许第二个有不同的userid，那么是不是第一个？但是我不知道第二个select语句来自哪里。