春靴 - 使用额外列进行多对多关系的分页

Question

在对多对多关系使用额外列时，我找不到一种干净简单的方法来进行分页。

我的模型看起来像这样：

我有一个用户和一个产品模型。每个用户可以消费n种产品。每个消耗将存储在一个额外的表中，因为我想存储额外的信息，如日期等。我已经实现了如下模型并且它的工作原理，但我想将用户的消费作为可寻呼的，而不是检索整个集合。实现它的最佳方法是什么？

@Entity

public class User {

    @Id

    @GeneratedValue(strategy = GenerationType.AUTO)

    private Long id;

    @OneToMany(

            mappedBy = "user",

            cascade = CascadeType.ALL,

            orphanRemoval = true

    )

    private List<Consumption> consumptionList = new ArrayList<>(); // never set this attribute

    public List<Consumption> getConsumptionList() {

        return consumptionList;

    }

    public void addConsumption(Product product) {

        Consumption consumption = new Consumption(this, product);

        consumptionList.add(consumption);

        product.getConsumptionList().add(consumption);

    }

    public void removeConsumption(Consumption consumption) {

        consumption.getProduct().getConsumptionList().remove(consumption);

        consumptionList.remove(consumption);

        consumption.setUser(null);

        consumption.setProduct(null);

    }

}

--

@Entity

@NaturalIdCache

@org.hibernate.annotations.Cache(usage = CacheConcurrencyStrategy.READ_WRITE)

public class Product {

    @Id

    @GeneratedValue(strategy = GenerationType.AUTO)

    private Long id;

    @OneToMany(

            mappedBy = "product",

            cascade = CascadeType.ALL,

            orphanRemoval = true

    )

    private List<Consumption> consumptionList = new ArrayList<>();

    public List<Consumption> getConsumptionList() {

        return consumptionList;

    }

}

这是我存储消耗的类。

@Entity

public class Consumption {

    @EmbeddedId

    private UserProductId id;

    @ManyToOne(fetch = FetchType.LAZY)

    @MapsId("userId")

    private User user;

}

我希望能够调用一个方法，例如“getConsumptionList（Page page）”，然后返回一个Pageable。

我希望你能帮助我！

Answer 1

好吧，如果使用Spring Boot，你可以使用存储库：@Repositorypublic interface ConsumptionRepo extends JpaRepository<Consumption, Long>{&nbsp; &nbsp; List<Consumption> findByUser(User user, Pageable pageable);}然后你可以简单地调用它ConsumptionRepo.findByUser(user, PageRequest.of(page, size);

Answer 2

由于@mtshaikh想法，我终于为我的问题找到了解决方案：只需实现一个分页服务：public Page<Consumption> getConsumptionListPaginated(Pageable pageable) {&nbsp; &nbsp; &nbsp; &nbsp; int pageSize = pageable.getPageSize();&nbsp; &nbsp; &nbsp; &nbsp; int currentPage = pageable.getPageNumber();&nbsp; &nbsp; &nbsp; &nbsp; int startItem = currentPage * pageSize;&nbsp; &nbsp; &nbsp; &nbsp; List<Consumption> list;&nbsp; &nbsp; &nbsp; &nbsp; if (consumptionList.size() < startItem) {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; list = Collections.emptyList();&nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; int toIndex = Math.min(startItem + pageSize, consumptionList.size());&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; list = consumptionList.subList(startItem, toIndex);&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; return new PageImpl<>(list, PageRequest.of(currentPage, pageSize), consumptionList.size());&nbsp; &nbsp; }