4

Spring Data JPA 是否有可能做到这一点？像这样：

public interface UserDao extends CrudRepository<User, Long> {

@EntityGraph(value = :graphName, type = EntityGraph.EntityGraphType.LOAD)
    @Query(value = "SELECT DISTINCT u FROM User u")
    List<User> findAllWithDetailsByGraphName(@Param(value="graphName") String graphName);

}

能够在运行时将 graphName 传递给方法并使用一组需要的集合调用加载？此构造不起作用，产生编译时错误。围绕它的任何游戏也失败了......

因此，我在 User 类中有多个集合，我想按条件加载它们；

@Entity
@Table(name="user")
@NamedEntityGraphs({
        @NamedEntityGraph(name = "User.details", attributeNodes = {
                @NamedAttributeNode("phones"), @NamedAttributeNode("emails"), @NamedAttributeNode("pets")}),
        @NamedEntityGraph(name = "User.phones", attributeNodes =
                {@NamedAttributeNode("phones")}),
        @NamedEntityGraph(name = "User.emails", attributeNodes =
                {@NamedAttributeNode("emails")}),
        @NamedEntityGraph(name = "User.pets", attributeNodes =
                {@NamedAttributeNode("pets")})
})
public class User {
    @Id
    @GeneratedValue(strategy= GenerationType.AUTO, generator="native")
    @GenericGenerator(name = "native", strategy = "native")
    @Column(name="user_id")
    private Long userId;

    @Column(name="name")
    private String name;

// more fields omitted

    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private Set<Phone> phones;

    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private Set<Email> emails;

    @OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
    private Set<Pet> pets;
}

现在，我只需隐式声明所有方法，如下所示：

@EntityGraph(value = "User.phones", type = EntityGraph.EntityGraphType.LOAD)
@Query(value = "SELECT DISTINCT u FROM User u")
List<User> findAllWithPhones();

谢谢你的建议！

4

1 回答 1

6

您可以定义一个接受实体图作为参数的基本 JPA 存储库。Specification这在与s结合使用时特别有用。因此，下面是一个基于Specifications 的示例。也可以使用不同类型的参数构造其他查询。

@NoRepositoryBean
public interface MyBaseRepository<T, ID extends Serializable> extends PagingAndSortingRepository<T, ID> {

    T findOne(Specification<T> spec, EntityGraphType entityGraphType, String entityGraphName);

    List<T> findAll(Specification<T> spec, Sort sort, EntityGraphType entityGraphType, String entityGraphName);

}

实现基础存储库：

@NoRepositoryBean
public class MyBaseRepositoryImpl<T, ID extends Serializable> extends SimpleJpaRepository<T, ID> implements MyBaseRepository<T, ID> {

    private EntityManager em;

    public MyBaseRepositoryImpl(JpaEntityInformation<T, ?> entityInformation, EntityManager entityManager) {
        super(entityInformation, entityManager);
        this.em = entityManager;
    }

    public MyBaseRepositoryImpl(Class<T> domainClass, EntityManager em) {
        super(domainClass, em);
        this.em = em;
    }

    @Override
    public T findOne(Specification<T> spec, EntityGraph.EntityGraphType entityGraphType, String entityGraphName) {
        TypedQuery<T> query = getQuery(spec, (Sort) null);
        query.setHint(entityGraphType.getKey(), em.getEntityGraph(entityGraphName));
        return query.getSingleResult();
    }

    @Override
    public List<T> findAll(Specification<T> spec, Sort sort, EntityGraph.EntityGraphType entityGraphType, String entityGraphName) {
        TypedQuery<T> query = getQuery(spec, sort);
        query.setHint(entityGraphType.getKey(), em.getEntityGraph(entityGraphName));
        return query.getResultList();
    }

}

指定自定义基础存储库：

<jpa:repositories base-package="my.domain" base-class="my.repository.MyBaseRepositoryImpl" />

从自定义基础存储库扩展：

public interface UserRepository extends JpaRepository<User, Long>, MyBaseRepository<User, Long>, JpaSpecificationExecutor<User> {
}

使用自定义存储库的方法：

Specification mySpecs = ...
List<User> user = picklistRepository.findAll(mySpecs, EntityGraphType.LOAD, "User.phones");
于 2018-06-26T13:02:06.047 回答