我有两个简单的实体Something
和Property
。
Something
实体与Property
具有多对一关系,因此当我创建新的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
(春天):
@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Something
:
@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;
}
}
这是输入中的JSON(property
id
1是数据库中的现有行):
{
"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,但没有字段property
的字段,经过验证(它们是null
) :
输出JSON:
{
"id": 1,
"name": "MyName",
"owner": "MySelf",
"property": {
"id": 1,
"shape": null,
"color": null,
"dimension": null
}
}
我希望在保存操作后验证/返回它们。
要解决此问题,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())
方法才能完成坚持实体):
@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的entith(使用正确的JSON):
{
"id": 4,
"name": "MyName",
"owner": "MySelf",
"property": {
"id": 1,
"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, 安德烈
答案 0 :(得分:11)
您可以通过创建自定义 JpaRepository来定义一次,而不是在每个资源中定义EntityManager
。 Reference
然后直接在您的每个存储库中使用refresh
的{{1}}。
参考以下示例:
CustomRepository接口
EntityManager
CustomRepository实施
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.repository.NoRepositoryBean;
import java.io.Serializable;
@NoRepositoryBean
public interface CustomRepository<T, ID extends Serializable> extends JpaRepository<T, ID> {
void refresh(T t);
}
在Spring Boot Application Class中启用自定义JPARepository
import org.springframework.data.jpa.repository.support.JpaEntityInformation;
import org.springframework.data.jpa.repository.support.SimpleJpaRepository;
import org.springframework.transaction.annotation.Transactional;
import javax.persistence.EntityManager;
import java.io.Serializable;
public class CustomRepositoryImpl<T, ID extends Serializable> extends SimpleJpaRepository<T, ID>
implements CustomRepository<T, ID> {
private final EntityManager entityManager;
public CustomRepositoryImpl(JpaEntityInformation entityInformation, EntityManager entityManager) {
super(entityInformation, entityManager);
this.entityManager = entityManager;
}
@Override
@Transactional
public void refresh(T t) {
entityManager.refresh(t);
}
}
您的东西存储库
@SpringBootApplication
@EnableJpaRepositories (repositoryBaseClass = CustomRepositoryImpl.class)
public class Application {
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}
直接在SomethingResource中使用Refresh (假设某事是实体)
public interface SomethingRepository extends CustomRepository<Something, Long> {
}
答案 1 :(得分:2)
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property
属性使用的是默认FetchType.EAGER
,因此该实体应初始化property
属性。
但是,从REST控制器调用存储库两次是很奇怪的。您应该有一个服务层,可以在@Transactional
服务方法中执行所有操作。这样,您就不需要重新保存实体,因为它已经被管理了。
@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null
发送property
,则应决定是否应取消@ManyToOne
引用。因此,它取决于您当前的应用程序业务逻辑要求。
如果您确定始终发送回先前提取的同一实体,则可以使用merge
。
em.refresh(result);
但是你的property
属性只是一个id,而不是一个真正的子实体,所以你必须自己在服务层解决这个问题。
答案 2 :(得分:0)
当您持久保存实体时,它将处于托管状态,因此如果您只是调用something.getProperty();
,它将从数据库加载并填充property
实体的something
值
public Something save(Something something) {
em.persist(something);
something.getProperty();
return something;
}
通常情况下,当你有多对一的关系时,应该自动获取。如果没有调用实体中对象的getter,也会通过触发新的DB Find请求来填充它们。
答案 3 :(得分:0)
在Spring Boot JpaRepository中:
如果我们的修改查询更改了持久性上下文中包含的实体,则该上下文将过时。
为了从数据库中获取具有最新记录的实体。
使用@Modifying(clearAutomatically = true)
@Modifying批注具有clearAutomatically属性,该属性定义执行修改查询后是否应清除基础持久性上下文。
示例:
@Modifying(clearAutomatically = true)
@Query("UPDATE NetworkEntity n SET n.network_status = :network_status WHERE n.network_id = :network_id")
int expireNetwork(@Param("network_id") Integer network_id, @Param("network_status") String network_status);