从 JPA 查询返回页面对象

Question

我想创建带有分页的 JPA。我尝试实现这个：

@Override
    public Page<PaymentTransactions> findAll(Specification<PaymentTransactions> spec, Pageable pageable) {

        int pageNumber = pageable.getPageNumber();
        int pageSize = pageable.getPageSize();      

        String hql = "select e from " + PaymentTransactions.class.getName() + " e LIMIT :limit OFFSET :offset";
        TypedQuery<PaymentTransactions> query = entityManager.createQuery(hql, PaymentTransactions.class).setParameter("limit", pageSize).setParameter("offset", pageNumber);
        Page<PaymentTransactions> paymentTransactions = (Page<PaymentTransactions>) query.getResultList();
        return paymentTransactions;
    }

如何在不使用转换的情况下返回页面对象？

编辑我也尝试过这个：

终点：

@Autowired
private PaymentTransactionRepository transactionRepository;

@GetMapping
public Page<PaymentTransactionsDTO> page(@PathVariable int page, @PathVariable int size) {
        return transactionRepository
                .findAll(page, size)
                .map(mapper::toDTO);
    }

存储库：

public interface PaymentTransactionRepository extends CrudRepository<PaymentTransactions, Integer>, JpaSpecificationExecutor<PaymentTransactions> {

    @Query(nativeQuery=true, 
            value="SELECT * FROM payment_transactions \n-- #pageable\n",
            countQuery="SELECT count(*) FROM payment_transactions")
    Page<PaymentTransactions> findAll(Pageable page);
}

编辑2。我也尝试过这个：

@GetMapping("page")
    public Page<PaymentTransactionsDTO> page(@PathVariable int page, @PathVariable int size) {
        PageRequest pageRequest = PageRequest.of(page, size);
        return transactionRepository.findAll(pageRequest).map(mapper::toDTO);
    }

界面：

public interface PaymentTransactionRepository extends CrudRepository<PaymentTransactions, Integer>, JpaSpecificationExecutor<PaymentTransactions> {

    Page<PaymentTransactions> findAll(Pageable page);
}

执行：

@Override
    public Page<PaymentTransactions> findAll(Pageable page) {
        int pageNumber = page.getPageNumber();
        int pageSize = page.getPageSize();

        String hql = "select e from " + PaymentTransactions.class.getName() + " e LIMIT :limit OFFSET :offset";
        TypedQuery<PaymentTransactions> query = entityManager.createQuery(hql, PaymentTransactions.class)
                .setParameter("limit", pageSize).setParameter("offset", pageNumber);
        Page<PaymentTransactions> paymentTransactions = (Page<PaymentTransactions>) query.getResultList();
        return paymentTransactions;
    }

Answer 1

我已经完成了 2 个 SQL 查询。也许它会有用。

1. 为列表对象创建SQL并添加页面参数：

   query.setMaxResults(pageable.getPageSize());
   query.setFirstResult(pageable.getPageNumber() * pageable.getPageSize());
   

2. 使用包装的 Select count(*) 创建相同的 SQL

   countQuery.getSingleResult();
   

然后将其设为 paga 对象

Page<ViewContract> page = (Page<ViewContract>) new PageImpl(
              selectQueryResultsList, 
              pageable, 
              countQueryResult);

Answer 2

为了获得分页和排序的功能，我们必须扩展PagingAndSortingRepositoryorJpaRepository接口。

对于您的情况，我们可以通过以下代码实现相同的目的：

终点：

@Autowired
private PaymentTransactionRepository transactionRepository;

@GetMapping
public Page<PaymentTransactionsDTO> page(@PathVariable int page, @PathVariable int size) {
    PageRequest pageRequest = PageRequest.of(page, size);
    return transactionRepository
            .findAll(pageRequest)
            .map(mapper::toDTO);
}

存储库：

public interface PaymentTransactionRepository extends JpaRepository<PaymentTransactions, Integer> {
    Page<PaymentTransactions> findAll(Pageable page);
}

请尝试使用这个，它应该有效。