JPA查询连接错误：org。冬眠hql。内部的阿斯特。QuerySyntaxException:应为join的路径

Question

我试图加入以下JPA查询，但出现以下错误：

组织。冬眠hql。内部的阿斯特。QuerySyntaxException:预期加入的路径！[from com.crm.entity.User-User-join-fetch-Role-on-Role.User_-id=User.id，其中User.deleted=false，User.enabled=true，User.username=：username]

以下是实现：

import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.persistence.Query;
import javax.transaction.Transactional;

import org.springframework.stereotype.Repository;

import com.crm.entity.User;

@Transactional
@Repository
public class UserJpaDaoImpl implements UserJpaDaoCustom {

    @PersistenceContext
    private EntityManager em;

    @Override
    public User getUser(String username) {
        Query query = em.createQuery("from User user "
                                    + "join fetch Role role on role.userId = user.id "
                                    + "where user.deleted = false "
                                    + "and user.enabled = true "
                                    + "and user.username = :username", User.class);
        query.setParameter("username", username);
        return (User)query.getSingleResult();
    }

}

用户实体：

@Entity
@Table(name = "user")
public class User extends BaseEntity implements UserDetails, Visible {

    private static final long serialVersionUID = 1L;

    @Column(name = "username")
    private String username;

    @Column(name = "password")
    private String password;

    /* Spring Security fields*/
    @OneToMany
    @JoinColumn(name = "user_id")
    private List<Role> roles;
...

角色实体：

@Entity
@Table(name = "role")
public class Role implements GrantedAuthority, Identifiable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", unique = true, nullable = false)
    private Integer id;

    @Column(name = "name", nullable = false)
    private String name;

    @Column(name = "user_id")
    private Integer userId;
...

我的查询中的连接有什么问题？

Answer 1

这是HQL不SQL：

   Query query = em.createQuery("from User user "
                                + "join fetch user.role "
                                + "where user.deleted = false "
                                + "and user.enabled = true "
                                + "and user.username = :username", User.class);

您必须处理对象结构，而不是表