保存后刷新并获取实体(JPA/Spring Data/Hibernate) [英] Refresh and fetch an entity after save (JPA/Spring Data/Hibernate)

问题描述

我有这两个简单的实体Something 和Property.Something 实体与 Property 具有多对一的关系，因此当我创建一个新的 Something 行时，我分配了一个现有的 属性.

I've these two simple entities Something and Property. The Something entity has a many-to-one relationship to Property, so when I create a new Something row, I assign an existing Property.

事情:

@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(Spring):

This is the 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 Something:

Through a REST controller and a JSON, I want to create a new 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):

This is the JSON in input (the property id 1 is an existing row in the database):

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

}

问题是:在方法.save(something)之后，变量result包含持久化的实体，但没有字段property，已验证(它们是 null):

The problem is: after the method .save(something), the variable result contains the persisted entity, but without the fields of field property, validated (they are 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.

I expect that they are validated/returned after the save operation.

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)(或者我必须调用一个 .findOne(something.getId()) 方法来拥有完整的持久化实体):

To workaround this, I have to inject/declare the EntityManager in the REST controller, and call the method EntityManager.refresh(something) (or I have to call a .findOne(something.getId()) method to have the complete persisted entity):

@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内容(使用正确的 JSON):

With this workaround, I've the expected saved entith (with a correct 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，以便完整"?持久化实体?

Is there an automatic method/annotation, with JPA or Spring or Hibernate, in order to have the "complete" persisted entity?

我想避免在每个 REST 或服务类中声明 EntityManager，或者我想避免每次我想要调用 .findOne(Long) 方法新刷新的持久化实体.

I would like to avoid to declare the EntityManager in every REST or Service class, or I want avoid to call the .findOne(Long) method everytime I want the new refreshed persisted entity.

推荐答案

不是在每个资源中定义EntityManager，你可以通过创建一个<强>自定义 JpaRepository.参考

Instead of defining EntityManager in each of your resource, you can define it once by creating a Custom JpaRepository. Reference

然后直接在每个存储库中使用 EntityManager 的 refresh.

Then use the refresh of your EntityManager in each of your repository directly.

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);
}

自定义存储库实现

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Something是一个实体)

Use Refresh directly in SomethingResource (Assuming Something is an Entity)

@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;
    }
}

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