Spring 是否可以创建具有两个主键值的实体？

我在一个

实体

中遇到了一个问题，它包含一个

唯一的\u键

，所以我可能需要用两个

主键创建该
实体
，这可能吗

这是我的
实体

@Entity
public class UserAnswerQuestion extends DateAudit {

    @Id
    @Column(name = "useranswerquestion_id")
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "useranswerquestion_seq")
    @SequenceGenerator(name = "useranswerquestion_seq", allocationSize = 1)
    private Long id;

    @ManyToOne
    private User user;
    @ElementCollection
    private List<Answer> answerList;
    @ManyToOne
    private Question question;

    private Boolean passed;

    private Boolean shown;

    public UserAnswerQuestion(){
    }
....

然后在
UserAnswerQuestion
实体中，我将其更改为：

 @EmbeddedId
 private UserAnswerQuestionId userAnswerQuestionId;

但现在的错误是：

org.hibernate.id.IdentifierGenerationException: null id generated for:class com.pew.model.useranswers.UserAnswerQuestion
    at org.hibernate.event.internal.AbstractSaveEventListener.saveWithGeneratedId(AbstractSaveEventListener.java:125) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
    at org.hibernate.event.internal.DefaultPersistEventListener.entityIsTransient(DefaultPersistEventListener.java:192) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
    at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:135) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
    at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:62) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
    at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:800) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
    at org.hibernate.internal.SessionImpl.persist(SessionImpl.java:785) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]

编辑2
我正在仔细阅读错误，看起来问题出在

@ElementCollection
private List<Answer> answerList;

主键始终仅为1且唯一。它可以是由多列元组组成的复合键。
使用
@IdClass
包括您的
UserAnswerQuestionId
应该可以：

@Entity
@IdClass(UserAnswerQuestionId.class)
public class UserAnswerQuestion extends DateAudit {

    @Id
    private Long id;

    @Id
    private Long user_id;

    @ElementCollection
    private List<Answer> answerList;
    @ManyToOne
    private Question question;

    private Boolean passed;

    private Boolean shown;

    public UserAnswerQuestion(){
    }

@实体
@IdClass（UserAnswerQuestionId.class）
公共类UserAnswerQuestion扩展了DateAudit{
@身份证
私人长id；
@身份证
专用长用户标识；
@元素集合
私人名单；
@许多酮
私人问题；
私有布尔传递；
显示私有布尔值；
公共用户回答问题（）{
}
也许您需要更好地解释您的用例。从上下文信息来看，您似乎希望检索用户以及他们为各种问题提供的答案。在这种情况下，您可能只需要如下所示的关联表


public class User extensed DateAudit{//这假设您在域模型中检索用户及其答案时有经验。
@身份证
私有长id；//用户id。
@许多
@JoinTable（name=“ANSWERS\u FOR\u QUESTIONS”，//更合适的名称可能是：ANSWERS\u BY\u USERS
joinColumns={@JoinColumn（name=“user_id”）}，//fk
inverseJoinColumns={@JoinColumn（name=“question_id”）}）//fk
私有集answers=newhashset（）；

@Entity(name = "answer")
public class Answer extends DateAudit {

@Id
@Column(name = "answer_id")
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "answer_seq")
@SequenceGenerator(name = "answer_seq", allocationSize = 1)
private Long id;

@Column(name = "answerToQuestion")
private String answerToQuestion;

@Entity
@IdClass(UserAnswerQuestionId.class)
public class UserAnswerQuestion extends DateAudit {

    @Id
    private Long id;

    @Id
    private Long user_id;

    @ElementCollection
    private List<Answer> answerList;
    @ManyToOne
    private Question question;

    private Boolean passed;

    private Boolean shown;

    public UserAnswerQuestion(){
    }

public class User extends DateAudit { //This assumes you are intested in retrieving User and their answer(s) in the domain model.

    @Id
    private Long id; //userid.



    @ManyToMany
    @JoinTable(name = “ANSWERS_FOR_QUESTIONS”, //More appropriate name may be : ANSWERS_BY_USERS
           joinColumns = { @JoinColumn(name = “user_id”) }, //fk
           inverseJoinColumns = { @JoinColumn(name = “question_id”) }) //fk
    private Set<Answers> answers = new HashSet<Answers>();