我如何在Spring数据存储库中编写SELECT TOP 25 sql查询

Question

一个简单的问题,因为我确信这是愚蠢的.我有以下查询,我可以在NetBeans sql命令窗口中执行:

SELECT TOP 25 * FROM ARCUST_BIG  WHERE arcustno<='300000' ORDER BY arcustno DESC

我的目标是将它放在我的ArcustRepository类中:

公共接口ArcustRepository扩展了JpaRepository {

Arcust findByPrimaryKey(String id);

@Query("SELECT COUNT(a) FROM Arcust a")
Long countAll();

@Query("SELECT TOP 25 a FROM Arcust a WHERE a.arcustno<='?1' ORDER BY a.arcustno DESC")
List<Arcust> findByTop(String arcustno);
}

但是,findBytop查询似乎不起作用,当我用tomcat7启动我的服务时返回:

2013-08-15 08:15:20 ERROR ContextLoader:319 - Context initialization failed
org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'arcustService': Injection of autowired dependencies failed; nested exception is org.springframework.beans.factory.BeanCreationException: Could not autowire field: private com.waudware.pics.repository.ArcustRepository com.waudware.pics.service.ArcustService.arcustRepository; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'arcustRepository': FactoryBean threw exception on object creation; nested exception is java.lang.IllegalArgumentException: Validation failed for query for method public abstract java.util.List com.waudware.pics.repository.ArcustRepository.findByTop(java.lang.String)!
Caused by: java.lang.IllegalArgumentException: Validation failed for query for method public abstract java.util.List com.waudware.pics.repository.ArcustRepository.findByTop(java.lang.String)!
Caused by: java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: 25 near line 1, column 12 [SELECT TOP 25 a FROM com.waudware.pics.domain.Arcust a WHERE a.arcustno<='?1' ORDER BY a.arcustno DESC]

Answer 1

我会说你需要

List<Arcust> findTop25ByArcustnoLessThanOrderByArcustnoDesc(String arcustno);

这将使用JPA并可能适用于所有数据库,将启动SPRING JPA 1.7.0(Evans发布列车)

我实现CrudRepository而不是JpaRepository

Answer 2

纯SQL

使用"限制"

SELECT * FROM ARCUST_BIG 
WHERE arcustno<='300000' ORDER BY arcustno DESC Limit 0, 25

JPA

List<Arcust> findTop25ByArcustnoLessThanOrderByArcustnoDesc(String arcustno);

Answer 3

我不确定拉克什的答案是否正确.他似乎在编写SQL,而不是JPA查询语法.
我尝试在JPA @Query中使用LIMIT并得到一个例外,说"限制"无法识别.

@Query("select d from Device d where d.deviceName like CONCAT('%', :deviceName, '%') and d.deviceId not in :notList ORDER BY deviceName DESC Limit 1001")

Caused by: java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: unexpected token: Limit near line 1, column 162

另外,请参阅Hari Shankar的回答,其中说JPA不支持"限制": JPA不支持"限制"

Answer 4

您可以在 Repository 中使用您的查询，无需 TOP 25：

@Query("SELECT a FROM Arcust a WHERE a.arcustno<='?1' ORDER BY a.arcustno DESC")
    List<Arcust> findByTop(String arcustno, Pageable pageable);
}

在 Service 中，使用 PageRequest，返回一个 Page 对象：

Page<Arcust> arcusts = arcustRepository.findByTop(arcustno, PageRequest.of(0, 25));
List<Arcust> arcust = arcusts.getContent();