Spring Data JPA：使用连接表进行排序和分页

Question

我有一个场景，我想对3个表参与的结果进行过滤，排序和分页。

目前我使用Spring Data JPA规范功能在单个实体上执行：qazxsw poi。

这很好用，但现在我有另一个场景，其中sort / filter属性分布在3个表中，这些表通过一对多关系连接。

这是我的场景：

repository.findAll(specification, pageRequest)

有没有办法使用

@Entity
public class CustomerEntity ... {
  ...

  @Column(nullable = false)
  public String                                 customerNumber;

  @OneToMany(mappedBy = "customer", cascade = CascadeType.ALL, orphanRemoval = true)
  public List<CustomerItemEntity> items;
}


@Entity
public class CustomerItemEntity ... {
  ...

  @Column(nullable = false)
  public String                                 itemNumber;

  @ManyToOne(optional = false)
  @JoinColumn(name = "customerId")
  public CustomerEntity customer;

  @OneToMany(mappedBy = "item", cascade = CascadeType.ALL, orphanRemoval = true)
  public List<DocumentEntity> documents;
}


@Entity
public class DocumentEntity ... {
  ...

  @Column(nullable = false)
  public LocalDate                                 validDate;

  @ManyToOne(optional = false)
  @JoinColumn(name = "itemId")
  public CustomerItemEntity item;
}

和PageRequest Specification，customerNumber和itemNumber同时用于过滤，排序和分页？

Answer 1

尝试这样的事情：

validDate

但我不知道为什么你需要

Specification<CustomerEntity> joins = (customer, query, cb) ->  {
    // from CustomerEntity c
    // join c.items i
    Join<CustomerEntity, CustomerItemEntity> items = customer.join("items");

    // join i.documents d
    Join<CustomerItemEntity, DocumentEntity> documents = items.join("documents");

    // // where c.customerNumber = ?1 and i.itemNumber = ?2 and d.validDate = ?3 
    return cb.and( 
            customer.equal(customer.get("customerNumber", customerNumber)),
            items.equal(items.get("itemNumber", itemNumber)), 
            documents.equal(documents.get("validDate", validDate))
    );
};

// sort by c.customerNumber asc
PageRequest pageRequest = new PageRequest(0, 2, new Sort(Sort.Direction.ASC, "customerNumber"));

Page<CustomerEntity> customerPage = CustomerRepo.findAll(joins, pageRequest);

？

你可以使同样更简单：

Specification

但是这一切都没有意义，因为你的三个实体具有顺序的一对多关联。在这种情况下，您可以只使用最后一个条件而不是三个条件：

@Query("select c from CustomerEntity c join c.items i join i.documents d where c.customerNumber = ?1 and i.itemNumber = ?2 and d.validDate = ?3")
Page<CustomerEntity> getCustomers(String customerNumber, String itemNumber, LocaleDate validDate, Pageable pageable);  

。然后查询方法变得更加容易：

where d.validDate = ?1

UPDATE

要通过连接实体的字段添加排序，我们可以使用

@Query("select c from CustomerEntity c join c.items i join i.documents d where d.validDate = ?1")
Page<CustomerEntity> getCustomers(LocaleDate validDate, Pageable pageable);

的orderBy方法：

query

要按几个参数排序，您可以将它们传递给逗号或

Specification<CustomerEntity> joins = (customer, query, cb) ->  {

    Join<CustomerEntity, CustomerItemEntity> items = customer.join("items");
    Join<CustomerItemEntity, DocumentEntity> documents = items.join("documents");

    // Ascending order by 'Document.itemNumber'
    query.orderBy(cb.asc(documents.get("itemNumber")));

    return cb.and( 
            customer.equal(customer.get("customerNumber", customerNumber)),
            items.equal(items.get("itemNumber", itemNumber)), 
            documents.equal(documents.get("validDate", validDate))
    );
};

Page<CustomerEntity> customerPage = CustomerRepo.findAll(joins, new PageRequest(0, 2));

分隔的方法：

List