java - Hibernate一对多关系java.sql.SQLIntegrityConstraintViolationException : Column 'person_id' cannot be null

Question

我是 hibernate 新手，学习 doc 保存持久对象

已关注 hibernate doc这是人和电话的一对多关系

@Entity
@Table(name = "phone")
public class Phone {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

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

    @ManyToOne(fetch = FetchType.LAZY)
    private Person person;

  //omit setter and getter
}

@Entity
@Table(name = "person")
public class Person {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;
    private String username;

    @OneToMany(cascade = CascadeType.ALL, mappedBy = "person")
    private List<Phone> phones = new ArrayList<>();
   //omit getter and setter 
}

我是个执着的人，加一个电话就报错

@Test
public void say() {
    Person person = new Person();
    person.setUsername("aaaa");

    Phone phone = new Phone();
    phone.setNumber("111");

    person.getPhones().add(phone);

    personService.save(person);
}

这是道的执着

public class PersonDaoImpl implements PersonDao {
    @PersistenceContext
    private EntityManager entityManager;

    @Override
    public void save(Person person) {
        entityManager.persist(person);
}

更新服务代码，服务只是救人

@Service(value = "personService")
public class PersonServiceImpl implements PersonService {
    @Autowired
    private PersonDao personDao;

    @Transactional
    @Override
    public void save(Person person) {
        personDao.save(person);
    }
}

错误信息:

    23:35:47.059 [main] DEBUG org.hibernate.engine.spi.ActionQueue - Executing identity-insert immediately
23:35:47.062 [main] DEBUG org.hibernate.SQL - 
    insert 
    into
        phone
        (number, person_id) 
    values
        (?, ?)
23:35:47.297 [main] DEBUG org.hibernate.engine.jdbc.spi.SqlExceptionHelper - could not execute statement [n/a]
java.sql.SQLIntegrityConstraintViolationException: Column 'person_id' cannot be null

Answer 1

添加 @GeneratedValue 注释以指定两个实体的主键将在您的代码之外填充。

@Entity
@Table(name = "phone")
public class Phone {

    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private long id;

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

    @JoinColumn("person_id")
    @ManyToOne(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
    private Person person;

  //omit setter and getter
}


public class Person {
    @Id
    @GeneratedValue(strategy=GenerationType.AUTO)
    private long id;
    private String username;

    @OneToMany(mappedBy = "person")
    private List<Phone> phones = new ArrayList<>();
   //omit getter and setter 
}

此外，您需要保留Person 对象而不是Phone 对象，因为没有配置从Phone 到 的级联人。如果您不能这样做，请将 Person 上的 CascadeType 切换为 none，然后将级联放在 Phone 上，如上所示。

您还应该在 Phone 实体上添加一个 @JoinColumn 注释，以便 hibernate 知道外键列。