获取错误非法尝试使用元素属性引用[lmsRoleLmsFeature]取消引用集合[vu360user0_.id.lmsRoles]

时间:2015-09-09 12:25:14

标签: hql jpql jpa-2.1 hibernate-5.x

我搜索了这个但无法更正我的查询。需要帮助。这是我的VU360User

use Zend\Session\Container;

public function indexAction()
    {
        $session = new Container('sessionsample');
        $session->offsetSet('id',1);
    }

这是我的LmsRole类

@Entity
public class VU360User extends AuditedEntity implements UserDetails, CredentialsContainer, Cloneable {
    ...
    private Learner learner;
    private Set<LmsRole> lmsRoles;
    ...

    @OneToOne(mappedBy = "vu360User", fetch=FetchType.LAZY)
    public Learner getLearner() {
        return learner;
    }
    //Setter

    @ManyToMany(fetch=FetchType.LAZY)
    @JoinTable(
        name = "VU360USER_ROLE", 
        joinColumns = { 
            @JoinColumn(name = "USER_ID", referencedColumnName = "ID") 
        }, 
        inverseJoinColumns = { 
            @JoinColumn(name = "ROLE_ID", referencedColumnName = "ID") 
        }
    )
    public Set<LmsRole> getLmsRoles() {
        return lmsRoles;
    }
    //Setter

}

这是我的LmsRoleLmsFeature类

@Entity
public class LmsRole extends BaseEntity implements GrantedAuthority  {
    ....
    private List<LmsRoleLmsFeature> lmsRoleLmsFeature;

    @OneToMany(mappedBy = "lmsRole", fetch=FetchType.LAZY)
    public List<LmsRoleLmsFeature> getLmsRoleLmsFeature() {
        return lmsRoleLmsFeature;
    }

    public void setLmsRoleLmsFeature(List<LmsRoleLmsFeature> lmsRoleLmsFeature) {
        this.lmsRoleLmsFeature = lmsRoleLmsFeature;
    }
    ...
}

现在这里是我想要进行的查询但是获得异常

@Entity
public class LmsRoleLmsFeature extends BaseEntity implements Serializable{
    ...
    private LmsRole lmsRole;
    private LmsFeature lmsFeature;

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="LMSROLE_ID")
    public LmsRole getLmsRole() {
        return lmsRole;
    }

    @ManyToOne(fetch=FetchType.LAZY)
    @JoinColumn(name="LMSFEATURE_ID")
    public LmsFeature getLmsFeature() {
        return lmsFeature;
    }
    ...
}

我遇到了异常

@Query("select distinct u, l, c, d, ta, la, p, ra, i from #{#entityName} u, IN (u.lmsRoles) lr "
        + "join fetch u.learner l "
        + "join fetch l.customer c "
        + "join fetch c.distributor d "
        + "join fetch u.trainingAdministrator ta "
        + "join fetch u.lmsAdministrator la "
        + "join fetch u.proctor p "
        + "join fetch u.regulatoryAnalyst ra "
        + "join fetch u.instructor i "
        + "join u.lmsRoles.lmsRoleLmsFeature lrlf "
        //+ "join lr.lmsRoleLmsFeature lrlf "
        + "where u.username = :userName and c.activeTf = true and d.status = true")
VU360User findByUserNameWithAllEntitiesAssociations(@Param("userName")String userName);

我试过加入,但我无法做到。我尝试了不同的设置但无法解决它。请帮忙。

由于

0 个答案:

没有答案