use*_*083 5 java hibernate spring-data-jpa
我正在使用spring-data-jpa + hibernate。1.我遇到以下异常......
Caused by: java.lang.IllegalArgumentException: Parameter with that position [1] did not exist
at org.hibernate.jpa.spi.BaseQueryImpl.findParameterRegistration(BaseQueryImpl.java:518) ~[BaseQueryImpl.class:4.3.7.Final]
at org.hibernate.jpa.spi.BaseQueryImpl.setParameter(BaseQueryImpl.java:674) ~[BaseQueryImpl.class:4.3.7.Final]
at org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:198) ~[AbstractQueryImpl.class:4.3.7.Final]
at org.hibernate.jpa.spi.AbstractQueryImpl.setParameter(AbstractQueryImpl.java:49) ~[AbstractQueryImpl.class:4.3.7.Final]
at org.springframework.data.jpa.repository.query.ParameterBinder.bind(ParameterBinder.java:165) ~[ParameterBinder.class:?]
at org.springframework.data.jpa.repository.query.StringQueryParameterBinder.bind(StringQueryParameterBinder.java:66) ~[StringQueryParameterBinder.class:?]
Run Code Online (Sandbox Code Playgroud)
......
我相信例外来自
public interface FamousExperienceDao extends PagingAndSortingRepository<FamousExperience,
Long>,JpaSpecificationExecutor<FamousExperience>
{
@Query( value =
"select new com.tujia.community.entity.BriefInfomation(f.id,f.title,f.summary,f.thumbnail,f.author, f.issueDate, f.counter) from FamousExperience f"
,countQuery ="select count(f.id) from FamousExperience f")
public Page<BriefInfomation> findExps(Specification<FamousExperience> spec, Pageable pgbl);
}
Run Code Online (Sandbox Code Playgroud)因为在我从函数 findExps 的参数中删除规范规范后,它工作得很好,我只想为查询添加规范。
BYW,FamousExperience 类扩展了 BriefInfomation。我使用了 JPA 查询的“构造函数表达式”功能,当我尝试查询时,我只是不需要属性“内容”。
@Entity
@Table(name = "famous_experience")
public class FamousExperience extends BriefInfomation
{
private String content;
/**
* @return the content
*/
public String getContent()
{
return content;
}
/**
* @param content the content to set
*/
public void setContent(String content)
{
this.content = content;
}
}
Run Code Online (Sandbox Code Playgroud)
请帮助我!
如果您在那里有规范规范,则需要在查询中的某个地方引用它...重要的是,当您在那里有可分页时,计数器似乎不会从 ?0 开始,而是从 ?1 开始
所以这样的事情应该有效
@Query( value =
"from FamousExperience f where f.spec = ?1"
,countQuery ="select count(f.id) from FamousExperience f")
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
8118 次 |
| 最近记录: |