JPA 查询连接错误：org.hibernate.hql.internal.ast.QuerySyntaxException：连接所需的路径

Posted 2023-03-23

【中文标题】JPA 查询连接错误：org.hibernate.hql.internal.ast.QuerySyntaxException：连接所需的路径

【英文标题】：JPA query join error: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join

【发布时间】：2017-03-28 14:17:30

【问题描述】：

我正在尝试加入以下 JPA 查询，但收到以下错误：

org.hibernate.hql.internal.ast.QuerySyntaxException：路径预期为 加入！ [来自 com.crm.entity.User 用户加入 fetch Role 角色上 role.user_id = user.id 其中 user.deleted = false 和 user.enabled = true 和 user.username = :username]

这里是实现：

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();
    

User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;
...

Role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？

【问题讨论】：

JPQL 查询以“SELECT alias”开头。其他任何内容都不符合 JPA 规范

【参考方案1】：

这是 HQL 而不是 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);

你必须处理对象结构而不是表格