Question

我的实体出了问题。我进行交易以在这两个实体之间建立多对一的连接。我是这样做的：

User user = new User();
    user.setName("a");
    user.setLastName("b");
    Set<Adress> a = new HashSet<Adress>();
    Adress a1 = new Adress();
    Adress a2 = new Adress();
    a1.setCity("a1");
    a2.setCity("a2");
    a.add(a1);
    a.add(a2);
    user.setAdress(a);
    userProxy.save(user);

我的邀请：

@Entity
public class User {

@Id
@GeneratedValue(strategy = GenerationType.TABLE)
private Long id;

@NotNull
private String name;

@NotNull
private String lastName;

@OneToMany(mappedBy="user", cascade=CascadeType.ALL)
private Set<Adress> adress = new HashSet<Adress>();

public User(String name, String lastName) {
    this.name = name;
    this.lastName = lastName;
}

public User() {

}

public Set<Adress> getAdress() {
    return adress;
}

public void setAdress(Set<Adress> adress) {
    this.adress = adress;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public String getLastName() {
    return lastName;
}

public void setLastName(String lastName) {
    this.lastName = lastName;
}

public Long getId() {
    return id;
}
}

第二实体

@Entity
public class Adress {
@Id
@GeneratedValue(strategy=GenerationType.TABLE)
private Long id;

private String city;

@ManyToOne
@JoinColumn(name="user_id")
private User user;

public String getCity() {
    return city;
}

public void setCity(String city) {
    this.city = city;
}

public Long getId() {
    return id;
}

}

表格中的数据用户保存正常，但在表格地址字段中，user_id为＆＃34; NULL＆＃34;任何人都可以向我解释为什么会这样？我和@ManyToOne尝试了很多组合，但对我没用。

有关详细信息，请参阅UserProxy：

@Service
public class UserProxyDao {

private UserDao userDao;

@Autowired
public UserProxyDao(UserDao userDao) {
    this.userDao = userDao;
}

public void save(User user) {
    userDao.save(user);
}
}

然而，如果我将@NotNull放在Adress中的字段用户实体验证失败...我真的不知道为什么会这样

Caused by: javax.validation.ConstraintViolationException: validation failed for classes [pl.rd.j2ee.api.domain.Adress] during persist time for groups [javax.validation.groups.Default, ]

Answer 1

只要您先执行此操作，就可以在一个操作中执行此操作。

a1.setUser(user);
a2.setUser(user);

您可以随时将User添加到Address构造函数中并在那里执行。

public Address (User user) {
    this.user = user;
}

Spring Data Jpa多对一给出了NULL

1 个答案: