Hibernate一对多关系java.sql.SQLIntegrityConstraintViolationException:列'person_id‘不能为空。
Hibernate一对多关系java.sql.SQLIntegrityConstraintViolationException:列'person_id‘不能为空。
提问于 2018-11-23 15:48:31
我是新来的hibernate，学习doc保存持久对象
遵循冬眠病，这是人与电话的一对多关系。
@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
}我是一个执着的人，加上一部手机，错误被抛出
@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Dao持久化
public class PersonDaoImpl implements PersonDao {
@PersistenceContext
private EntityManager entityManager;
@Override
public void save(Person person) {
entityManager.persist(person);
}更新服务代码，服务只需保存人员
@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
回答 5
Stack Overflow用户
回答已采纳
发布于 2018-11-23 16:21:58
添加@GeneratedValue注释，以指定两个实体的主键将在代码之外填充。
@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，您需要持久化Person对象而不是Phone对象，因为没有从Phone配置到Person的级联。如果无法做到这一点，请将CascadeType on Person切换到none，并将级联设置在Phone上，如上面所示。
您还应该在@JoinColumn实体上添加一个Phone注释，这样hibernate就知道了外键列。
Stack Overflow用户
发布于 2018-11-23 16:42:47
你漏掉了什么。你可以试试这个。
人实体
@Entity
public class Person {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private long id;
private String username;
@OneToMany(mappedBy = "person")
private List<Phone> phones = new ArrayList<>();
public String getUsername() {
return username;
}
public void setUsername(String username) {
this.username = username;
}
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public List<Phone> getPhones() {
return phones;
}
public void setPhones(List<Phone> phones) {
this.phones = phones;
}
//omit getter and setter
}电话实体
@Entity
@Table(name = "phone")
public class Phone {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private long id;
@Column(name = "number")
private String number;
@ManyToOne(cascade = CascadeType.PERSIST)
private Person person;
public String getNumber() {
return number;
}
public void setNumber(String number) {
this.number = number;
}
public Person getPerson() {
return person;
}
public void setPerson(Person person) {
this.person = person;
}
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
//ommit setter and getter
}电话刀
public interface PhoneDao {
public Phone save(Phone phone);
}PhoneDaoImpl
@Repository
public class PhoneDaoImpl implements PhoneDao {
@PersistenceContext
private EntityManager entityManager;
@Override
public Phone save(Phone phone) {
return entityManager.merge(phone);
}
}PersonDaoImpl
@Repository
public class PersonDaoImpl implements PersonDao{
@PersistenceContext
private EntityManager entityManager;
@Override
public Person save(Person person) {
return entityManager.merge(person);
}
}测试方法
@Test
@Transactional
@Commit
public void say()
{
Phone phone = new Phone();
phone.setNumber("jghjkhk");
Person person = new Person();
person.setUsername("7576");
phone.setPerson(person);
Phone pers = phoneDao.save(phone);
Assert.assertNotNull(pers);
}现在试一试。看起来不错。
Stack Overflow用户
发布于 2018-11-23 16:14:52
我认为您需要设置person->id的值，然后使用getter方法将id传递给您的phone对象，而不是传递person对象。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/53449592
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