Question

我有一个用户类

@Entity(name = "users")
@Table(name = "users")
public class User implements UserDetails {

static final long serialVersionUID = 1L;

@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "user_id", nullable = false)
private Long id;

@Column(name = "username", nullable = false, unique = true)
private String username;

@Column(name = "password", nullable = false)
private String password;
}

绑定到一个简单的存储库

public interface UserRepository extends PagingAndSortingRepository<User, Long> {
}

我有一个具有嵌套用户对象的教师类

@Entity
@Table(name = "instructors")
public class Instructor {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
@Column(name = "instructor_id", nullable = false, updatable = false)
private Long id;

@OneToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "user_id")
private User user;

@OneToMany(fetch = FetchType.EAGER)
@JoinColumn(name = "course_id")
private List<Course> courses;

}

它与以下存储库一起保存

public interface InstructorRepository extends PagingAndSortingRepository<Instructor, Long> {

}

我发布的JSON

{
"user": {
    "id": 1
}
}

当我尝试POST到/instructors时。用户无效。是否有一些我缺少的东西让JPA将两者结合在一起？我尝试将CascadeType.ALL添加到字段上，并且只抛出一个分离的持久异常。

Answer 1

将CascadeType.ALL保留为Instructor，就像您已经尝试过的那样：

@ManyToOne(fetch = FetchType.EAGER, cascade=CascadeType.ALL)
@JoinColumn(name = "user_id")
private User user;

另外，将以下内容添加到User。似乎和我一起工作。它提供了映射信息，并使JPA处理User托管

@OneToMany(mappedBy="user")//, cascade=CascadeType.ALL)
private List<Instructor> instructors = new ArrayList<>();

我已在上面评论了cascadeType，但如果您希望User保留其Instructors的全部内容，则可能会有用。

坚持嵌套实体Spring Rest数据

1 个答案: