Java “双向一通多回传”;“不允许为空”列;保存
这是实体的简化版本,我只显示相关部分Java “双向一通多回传”;“不允许为空”列;保存,java,spring,hibernate,spring-data,jpa-2.0,Java,Spring,Hibernate,Spring Data,Jpa 2.0,这是实体的简化版本,我只显示相关部分 @Entity @Data public class Wrapper { @Id @GeneratedValue(strategy = GenerationType.IDENTITY) private Integer id @OneToOne(mappedBy = "wrapper", cascade = CascadeType.ALL, fetch = FetchTy
@Entity
@Data
public class Wrapper {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id
@OneToOne(mappedBy = "wrapper", cascade = CascadeType.ALL, fetch = FetchType.EAGER, orphanRemoval = true)
private Application application;
public Wrapper(Application application) {
this.application = application;
application.setWrapper(this);
}
}
@Data
@Entity
@EqualsAndHashCode(exclude = "wrapper")
public class Application {
@Id
private Integer id;
@JsonIgnore
@OneToOne
@JoinColumn(name = "id")
@MapsId
private Wrapper wrapper;
@OneToMany(mappedBy = "application", cascade = CascadeType.ALL, orphanRemoval = true, fetch = FetchType.EAGER)
@SortNatural
private SortedSet<Apartement> ownedApartements = new TreeSet<>();
}
@Entity
@Data
public class Apartement {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
@ManyToOne(fetch = FetchType.LAZY, optional = false)
@JoinColumn(name = "application_id", insertable = false, updatable = false)
private Application application;
}
@Repository
public interface WrapperRepository extends JpaRepository<Wrapper, Integer> {
}
给定以下实体和以下代码:
Apartement apartement1 = new Apartement()
Apartement apartement2 = new Apartement()
Wrapper wrapper = new Wrapper(new Application());
Application application = wrapper.getApplication();
application.getOwnedApartements().addAll(Arrays.asList(apartement1, apartement2));
apartement1.setApplication(application);
apartement2.setApplication(application);
WrapperRepository.saveAndFlush(wrapper);
我在日志中看到三个插页。
首先是包装器,然后是应用程序,最后是Apartment。但由于某些原因,应用程序_id在第一次保存时为空。但我知道这是一种双向关系
我得到的错误是:
Caused by: org.h2.jdbc.JdbcSQLException: NULL not allowed for column "APPLICATION_ID"; SQL statement:
insert into Apartement (id) values (null) [23502-197]
为什么会发生这种情况?我需要按正确的顺序存储所有东西吗?我是否需要首先存储包装器和应用程序,然后在获得应用程序ID后最终存储Apartment?
无法休眠一次存储所有三个?或者自己解决这个问题?对不起,我修好了
问题是
@ManyToOne(fetch = FetchType.LAZY, optional = false)
@JoinColumn(name = "application_id", insertable = false, updatable = false)
private Application application;
我删除了insertable=false、updateable=false,并添加了optional=false
成功了
@JoinColumn(name = "application_id", optional = false)
试试这个:
Apartement apartement1 = new Apartement()
Apartement apartement2 = new Apartement()
Wrapper wrapper = new Wrapper(new Application());
Application application = wrapper.getApplication();
application.getOwnedApartements().addAll(Arrays.asList(apartement1, apartement2));
apartement1.setApplicationId(application.getId());
apartement2.setApplicationId(application.getId());
WrapperRepository.saveAndFlush(wrapper);
这和我的一模一样。有什么区别吗?我不认为你的意思是假的?
Apartement apartement1 = new Apartement()
Apartement apartement2 = new Apartement()
Wrapper wrapper = new Wrapper(new Application());
Application application = wrapper.getApplication();
application.getOwnedApartements().addAll(Arrays.asList(apartement1, apartement2));
apartement1.setApplicationId(application.getId());
apartement2.setApplicationId(application.getId());
WrapperRepository.saveAndFlush(wrapper);