JPA查詢加入錯誤：org.hibernate.hql.internal.ast.QuerySyntaxException：路徑預期的加入

Question

我試圖在下面的JPA查詢加入，但出現以下錯誤：

org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [from com.crm.entity.User user join fetch Role role on role.user_id = user.id where user.deleted = false and user.enabled = true and user.username = :username]

這裏是執行：

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(); 
    } 

} 

User實體：

@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; 
... 

Role實體：

@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; 
... 

加入我的查詢時出了什麼問題？

Answer 1

這是HQL不是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); 

你必須對對象結構的工作不能在表