Spring Data JPA:使用连接表进行排序和分页
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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?
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
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