5

我有一个包含如下Service集合的域：tags 

@Entity
public class Service extends AbstractEntity<Long> {
            private static final long serialVersionUID = 9116959642944725990L;

        @ElementCollection(fetch = FetchType.EAGER, targetClass = java.lang.String.class)
        @CollectionTable(name = "service_tags", joinColumns = @JoinColumn(name = "s_id"))
        @Column(name = "tag")
        private Set<String> tags;
    }

我想选择Serviceswith特定KEY的Service.tags。

hql加入Service如下Service.tags： 

select s from Service s INNER JOIN s.tags t where s.status=0 and (s.serviceType=9 or t.tag in ('College'))

但是，上面的 hql 返回我有以下异常： 

Caused by: java.lang.IllegalArgumentException: org.hibernate.QueryException: cannot dereference scalar collection element: tag [select s from com.zazzercode.domain.Service s INNER JOIN s.tags t where s.status=0 and (s.serviceType=9 or t.tag in ('College')) ]

select s from Service s INNER JOIN s.tags t where s.status=0虽然有效。

查看查询非实体集合的 JPQL，我尝试如下

"select s from Service s where s.status=0 and s.priviligedUser.priviligedUserType IN (2,4) and (s.serviceType=9 or (KEY(s.tags)='tag' and (VALUE(s.tags)='College'))"

得到以下异常： 

Caused by: org.hibernate.hql.ast.QuerySyntaxException: unexpected token: null near line 1, column 188 [select s from com.esewa.server.entity.Service s where s.status=0 and (s.serviceType=9 or (KEY(s.tags)='tag' and (VALUE(s.tags)='College'))]

几个月前，我使用标准 api实现了同样的目标。

4

2 回答 2

3

感谢JPQL 查询非实体的集合

以下代码有效！

"select s from Service s INNER JOIN s.tags t where s.status=0 and and (s.serviceType=9 or  VALUE(s.tags) in ('College')) "
于 2013-09-06T12:31:34.957 回答
3 
"select s from Service s where s.status=0 and (s.serviceType=9 or 'College' in elements(s.tags))"
于 2014-04-18T10:41:39.117 回答