java - 如何解决 org.hibernate.PropertyValueException

Question

我有一个 BaseEntity 作为超类：

@MappedSuperclass
public abstract class BaseEntity<ID> {

    public static final String IS_DELETED = "is_deleted";

    @Id
    @GeneratedValue
    private ID id;

    @Column(name = IS_DELETED, columnDefinition = "TINYINT(1)")
    private Boolean isDeleted;
}

还有另外两个实体：

1. 扩展 BaseEntity 的用户

   @Entity
   @Table(name = User.TABLE_NAME)
   public class User extends BaseEntity<Long> {
   
       public static final String TABLE_NAME = "user";
       public static final String USER_NAME = "user_name";
       public static final String PASS_WORD = "pass_word";
       public static final String IS_ACTIVE = "is_active";
   
       @Column(name = USER_NAME)
       private String userName;
   
       @Column(name = PASS_WORD)
       private String passWord;
   
       @Column(name = IS_ACTIVE, columnDefinition = "TINYINT(1)")
       private Boolean isActive;
   
       @OneToOne(mappedBy = TABLE_NAME, cascade = CascadeType.ALL)
       private UserProfile userProfile;
   
       public User(){
           userName = null;
           passWord = null;
           isActive = null;
           userProfile = new UserProfile();
       }
   }
   

2. 用户资料

   @Entity
   @Table (name = UserProfile.TABLE_NAME)
   public class UserProfile {
   
       public static final String TABLE_NAME = "user_profile";
       public static final String FIRST_NAME = "first_name";
       public static final String LAST_NAME = "last_name";
       public static final String PHONE_NUMBER = "phone_number";
       public static final String EMAIL = "email";
       public static final String AGE = "age";
       public static final String USER_ID = "user_id";
   
       @Id
       @GeneratedValue(strategy = GenerationType.AUTO)
       private Long id;
   
       @Column(name = FIRST_NAME)
       private String firstName;
   
       @Column(name = LAST_NAME)
       private String lastName;
   
       @Column(name = PHONE_NUMBER)
       private String phoneNumber;
   
       @Column(name = EMAIL)
       private String email;
   
       @Column(name = AGE)
       private Integer age;
   
       @OneToOne
       @JoinColumn(name = USER_ID, nullable = false)
       private User user;
   }
   

在我的 UserService 中，当用户想要注册时，我设置了 UserProfile 字段，如 firstName、lastName 等。然后将 User 对象传递到存储库以保存它，并且当我将 cascadeType 设置为 all 时，不应该有问题，因为 UserProfile 应该被插入到数据库中，或者我错过了什么：

public class UserServiceImpl {

    public void signUp() {
        User user = new User();
        user.getUserProfile().setId(user.getId());

        System.out.println("Enter firstname:");
        user.getUserProfile().setFirstName(ApplicationContext.getStringScanner().nextLine());

        System.out.println("Enter lastname:");
        user.getUserProfile().setLastName(ApplicationContext.getStringScanner().nextLine());

        System.out.println("Enter username:");
        String userName = ApplicationContext.getStringScanner().next();

        while (repository.getUserByUserName(userName) != null) {
            System.out.println("\nUsername not available!");
            System.out.println("Enter username:");
            userName = ApplicationContext.getStringScanner().next();
        }
        user.setUserName(userName);

        System.out.println("Enter Password:");
        String passWord = ApplicationContext.getStringScanner().next();
        user.setPassWord(passWord);

        System.out.println("Enter Your Phone Number:");
        String phoneNumber = ApplicationContext.getStringScanner().next();
        user.getUserProfile().setPhoneNumber(phoneNumber);

        System.out.println("Enter Your Email:");
        String email = ApplicationContext.getStringScanner().next();
        user.getUserProfile().setEmail(email);

        System.out.println("Enter Your age:");
        Integer age = ApplicationContext.getStringScanner().nextInt();
        user.getUserProfile().setAge(age);

        repository.save(user);
    }
}

保存在 UserRepository 中：

@Override
public void save(User user) {
    EntityManager entityManager = entityManagerFactory.createEntityManager();
    entityManager.getTransaction().begin();
    entityManager.persist(user);
    entityManager.getTransaction().commit();
    entityManager.close();
}

但是，在调用 UserService signUp 方法后，我得到了这个异常：

org.hibernate.PropertyValueException：非空属性引用空值或瞬态值：model.UserProfile.user

标签： javahibernate