保存后刷新并获取实体（JPA/Spring Data/Hibernate）

我有这两个简单的实体Something and Property. The Something实体具有多对一关系Property，所以当我创建一个新的Something行，我分配一个现有的Property.

某物：

@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() +
            "}";
    }

财产：

@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 id1 是数据库中的现有行）：

{
  "name": "MyName",
  "owner": "MySelf",
  "property": {
    "id": 1
  }

}

问题是：方法之后.save(something)，变量result包含持久化实体，但不包含 field 的字段property，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
  }
}

我希望它们在保存操作后得到验证/返回。

为了解决这个问题，我必须注入/声明EntityManager在 REST 控制器中，并调用该方法EntityManager.refresh(something)（或者我必须打电话给.findOne(something.getId())拥有完整持久实体的方法）：

@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;
    }
}

通过此解决方法，我已经保存了预期的实体（具有正确的 JSON）：

{
  "id": 4,
  "name": "MyName",
  "owner": "MySelf",
  "property": {
    "id": 1,
    "shape": "Rectangle",
    "color": "Red",
    "dimension": 50
  }
}

是否有使用 JPA 或 Spring 或 Hibernate 的自动方法/注释，以便拥有“完整”的持久实体？

我想避免声明EntityManager在每个 REST 或服务类中，或者我想避免调用.findOne(Long)每当我想要新的刷新的持久实体时。

而不是定义EntityManager在你的每一个resource，您可以通过创建一个来定义它一次CustomJpa存储库。参考 https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.custom-implementations

然后使用refresh你的EntityManager直接在每个存储库中。

请参考下面的例子：

自定义存储库接口

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);
}

自定义存储库实施

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);
  }
}

在 Spring Boot 应用程序类中启用自定义 JPARepository

@SpringBootApplication
@EnableJpaRepositories (repositoryBaseClass = CustomRepositoryImpl.class)
public class Application {
  public static void main(String[] args) {
    SpringApplication.run(Application.class, args);
  }
}

你的东西存储库

public interface SomethingRepository extends CustomRepository<Something, Long> {

}

直接在 SomethingResource 中使用 Refresh（假设某物是一个实体）

@RestController
@RequestMapping("/api")
@Transactional
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);
        somethingRepository.refresh(result);
        return result;
    }
}