在對(duì)多對(duì)多關(guān)系使用額外列時(shí)，我找不到一種干凈簡(jiǎn)單的方法來進(jìn)行分頁(yè)。我的模型看起來像這樣：我有一個(gè)用戶和一個(gè)產(chǎn)品模型。每個(gè)用戶可以消費(fèi)n種產(chǎn)品。每個(gè)消耗將存儲(chǔ)在一個(gè)額外的表中，因?yàn)槲蚁氪鎯?chǔ)額外的信息，如日期等。我已經(jīng)實(shí)現(xiàn)了如下模型并且它的工作原理，但我想將用戶的消費(fèi)作為可尋呼的，而不是檢索整個(gè)集合。實(shí)現(xiàn)它的最佳方法是什么？@Entitypublic 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; }}這是我存儲(chǔ)消耗的類。@Entitypublic class Consumption { @EmbeddedId private UserProductId id; @ManyToOne(fetch = FetchType.LAZY) @MapsId("userId") private User user;}我希望能夠調(diào)用一個(gè)方法，例如“getConsumptionList（Page page）”，然后返回一個(gè)Pageable。我希望你能幫助我！
2 回答
ibeautiful
TA貢獻(xiàn)1993條經(jīng)驗(yàn) 獲得超6個(gè)贊
好吧，如果使用Spring Boot，你可以使用存儲(chǔ)庫(kù)：
@Repository
public interface ConsumptionRepo extends JpaRepository<Consumption, Long>{
List<Consumption> findByUser(User user, Pageable pageable);
}
然后你可以簡(jiǎn)單地調(diào)用它
ConsumptionRepo.findByUser(user, PageRequest.of(page, size);
MMTTMM
TA貢獻(xiàn)1869條經(jīng)驗(yàn) 獲得超4個(gè)贊
由于@mtshaikh想法，我終于為我的問題找到了解決方案：
只需實(shí)現(xiàn)一個(gè)分頁(yè)服務(wù)：
public Page<Consumption> getConsumptionListPaginated(Pageable pageable) {
int pageSize = pageable.getPageSize();
int currentPage = pageable.getPageNumber();
int startItem = currentPage * pageSize;
List<Consumption> list;
if (consumptionList.size() < startItem) {
list = Collections.emptyList();
} else {
int toIndex = Math.min(startItem + pageSize, consumptionList.size());
list = consumptionList.subList(startItem, toIndex);
}
return new PageImpl<>(list, PageRequest.of(currentPage, pageSize), consumptionList.size());
}
添加回答
舉報(bào)
0/150
提交
取消