我有这两个简单的实体Something and Property.
Something实体与Property具有多对一关系,因此当我创建一个新的Something行时,我会分配一个现有的Property.
Something实体与Property具有多对一关系,因此当我创建一个新的Something行时,我会分配一个现有的Property.
东西：
@Entity
@Table(name = "something")
public class Something implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Column(name = "name")
private String name;
@Column(name = "owner")
private String owner;
@ManyToOne
private Property property;
// getters and setters
@Override
public String toString() {
return "Something{" +
"id=" + getId() +
",name='" + getName() + "'" +
",owner='" + getOwner() + "'" +
",property=" + getProperty() +
"}";
}
属性：
@Entity
@Table(name = "property")
public class Property implements Serializable {
private static final long serialVersionUID = 1L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Column(name = "shape")
private String shape;
@Column(name = "color")
private String color;
@Column(name = "dimension")
private Integer dimension;
// getters and setters
@Override
public String toString() {
return "Property{" +
"id=" + getId() +
",shape='" + getShape() + "'" +
",color='" + getColor() + "'" +
",dimension='" + getDimension() + "'" +
"}";
}
}
这是SomethingRepository(Spring)：
@SuppressWarnings("unused")
@Repository
public interface SomethingRepository extends JpaRepository<Something,Long> {
}
通过REST控制器和JSON,我想创建一个新的东西：
@RestController
@RequestMapping("/api")
public class SomethingResource {
private final SomethingRepository somethingRepository;
public SomethingResource(SomethingRepository somethingRepository) {
this.somethingRepository = somethingRepository;
}
@PostMapping("/somethings")
public Something createSomething(@RequestBody Something something) throws URISyntaxException {
Something result = somethingRepository.save(something);
return result;
}
}
{
"name": "MyName","owner": "MySelf","property": {
"id": 1
}
}
问题是：在方法.save(something)之后,变量result包含持久化实体,但没有field属性字段,验证(它们为null)：
输出JSON：
{
"id": 1,"name": "MyName","property": {
"id": 1,"shape": null,"color": null,"dimension": null
}
}
我希望在保存操作后验证/返回它们.
要解决这个问题,我必须在REST控制器中注入/声明EntityManager,并调用方法EntityManager.refresh(something)(或者我必须调用.findOne(something.getId())方法来获得完整的持久化实体)：
@RestController
@RequestMapping("/api")
@Transactional
public class SomethingResource {
private final SomethingRepository somethingRepository;
private final EntityManager em;
public SomethingResource(SomethingRepository somethingRepository,EntityManager em) {
this.somethingRepository = somethingRepository;
this.em = em;
}
@PostMapping("/somethings")
public Something createSomething(@RequestBody Something something) throws URISyntaxException {
Something result = somethingRepository.save(something);
em.refresh(result);
return result;
}
}
通过这种解决方法,我得到了预期的保存entith(使用正确的JSON)：
{
"id": 4,"shape": "Rectangle","color": "Red","dimension": 50
}
}
是否有自动方法/注释,使用JPA或Spring或Hibernate,以获得“完整”持久化实体？
我想避免在每个REST或Service类中声明EntityManager,或者我希望每次我想要新刷新的持久化实体时都避免调用.findOne(Long)方法.
非常感谢,
安德里亚
解决方法
这还不够：
Something result = somethingRepository.save(something);
您需要手动合并传入的实体：
Something dbSomething = somethingRepository.findOne(
Something.class,something.getId()
);
dbSomething.setName(something.getName());
dbSomething.setOwner(something.getOwner());
somethingRepository.save(dbSomething);
由于属性属性使用默认值FetchType.EAGER,因此实体应初始化属性属性.
但是,从REST控制器调用存储库两次是很奇怪的.你应该有一个服务层来完成@Transactional服务方法中的所有操作.这样,您不需要重新保存实体,因为它已经被管理.
@Transactional
public Something mergeSomething(Something something) {
Something dbSomething = somethingRepository.findOne(
Something.class,something.getId()
);
dbSomething.setName(something.getName());
dbSomething.setOwner(something.getOwner());
return dbSomething;
}
现在,您需要仔细合并您发送的每个属性.在您的情况下,如果您为属性发送null,您应该决定是否应该取消@ManyToOne引用.因此,它取决于您当前的应用程序业务逻辑要求.
更新
如果您确定始终发送回先前提取的同一实体,则可以使用合并.
em.refresh(result);
