Question

我正在使用 spring-data-jpa + hibernate 。我碰到下面的例外 ......

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:?]

...

我认为异常来自

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);
}

因为在我从函数findExps的params中删除了规范规范后，它运行良好，我只是想为查询添加规范。

BYW，FamousExperience类扩展了BriefInfomation。我使用了JPA Query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“content”。

@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;
    }
}

请帮助我！

Answer 1

如果你有规范规范，你需要在你的查询中的某个地方引用它...重要的是，当你有一个可分页的时候，那么计数器不会从？0开始但是在？1 < / p>

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")

＆＃34; java.lang.IllegalArgumentException：具有该位置[1]的参数不存在＆＃34;当我使用spring-data-jpa时

1 个答案: