Spring JPA的分页排序问题

我和SpringJPA讨论了一个关于排序和分页的问题。目前我收到以下警告：

WARN [NamedQuery] (NamedQuery.java:65) - Finder method public abstract     org.springframework.data.domain.Page loc.starterkit.business.entities.application.repository.ApplicationRepository.findByNameLike(java.lang.String,org.springframework.data.domain.Pageable) 
is backed by a NamedQuery but contains a Pageable parameter! Sorting delivered via this Pageable will not be applied!

我的JPA接口代码如下所示：

public interface ApplicationRepository extends PagingAndSortingRepository<Application,Long>{

Page<Application> findByNameLike(String name, Pageable pageable);

}

将PagingAndSortingRepository更改为JPARepository似乎会引发相同的警告。但是，经过一些测试，我发现这个警告毫无意义。事情似乎进展顺利：

Hibernate: 
/* select
    count(*) 
from
    Application as generatedAlias0 
where
    generatedAlias0.name like :param0 */ select
        count(*) as col_0_0_ 
    from
        APPLICATION applicatio0_ 
    where
        applicatio0_.NAME like ?
Hibernate: 
/* select
    generatedAlias0 
from
    Application as generatedAlias0 
where
    generatedAlias0.name like :param0 
order by
    generatedAlias0.name asc */ select
        applicatio0_.ID as ID0_,
        applicatio0_.DESCRIPTION as DESCRIPT2_0_,
        applicatio0_.NAME as NAME0_ 
    from
        APPLICATION applicatio0_ 
    where
        applicatio0_.NAME like ? 
    order by
        applicatio0_.NAME asc limit ?

---

所以我认为这个警告没有任何意义，但我想知道为什么会发出它

你能发布你的

应用程序

实体吗？另外，为什么您要扩展

PagingAndSortingRepository

而不是

JpaRepository

（这是

PagingAndSortingRepository

的扩展）？遇到了相同的问题。对此有任何解决方法吗？您如何知道可分页返回内容/为空？

@Entity
@Table(name="APPLICATION")
public class Application {

@Id
@Column(name="ID")
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;

@Column(name="NAME")
private String name;

@Column(name="DESCRIPTION")
private String description;

public Application() {
    super();
}

.... (getters and setters here)

Hibernate: 
/* select
    count(*) 
from
    Application as generatedAlias0 
where
    generatedAlias0.name like :param0 */ select
        count(*) as col_0_0_ 
    from
        APPLICATION applicatio0_ 
    where
        applicatio0_.NAME like ?
Hibernate: 
/* select
    generatedAlias0 
from
    Application as generatedAlias0 
where
    generatedAlias0.name like :param0 
order by
    generatedAlias0.name asc */ select
        applicatio0_.ID as ID0_,
        applicatio0_.DESCRIPTION as DESCRIPT2_0_,
        applicatio0_.NAME as NAME0_ 
    from
        APPLICATION applicatio0_ 
    where
        applicatio0_.NAME like ? 
    order by
        applicatio0_.NAME asc limit ?