我有一个与Role实体具有多对多关系的User实体 .
@Entity
@Table(name = "auth_user")
public class OAuthUser {
// @Autowired
// @Transient
// private PasswordEncoder passwordEncoder;
//
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
@Column(name = "username")
private String userName;
@Column(name = "password")
@JsonIgnore
private String password;
@Column(name = "first_name")
private String firstName;
@Column(name = "last_name")
private String lastName;
@Column(name = "email")
private String email;
@Column(name = "is_enabled")
private boolean isEnabled;
/**
* Reference:
* https://github.com/nydiarra/springboot-jwt/blob/master/src/main/java/com/nouhoun/springboot/jwt/integration/domain/User.java
* Roles are being eagerly loaded here because they are a fairly small
* collection of items for this example.
*/
@ManyToMany(fetch = FetchType.EAGER)
@Fetch(value = FetchMode.SUBSELECT)
@JoinTable(name = "user_role", joinColumns = @JoinColumn(name = "user_id", referencedColumnName = "id"), inverseJoinColumns = @JoinColumn(name = "role_id", referencedColumnName = "id"))
private List<Role> roles;
@ManyToMany(fetch = FetchType.EAGER)
@Fetch(value = FetchMode.SUBSELECT)
@JoinTable(name = "user_properties", joinColumns = @JoinColumn(name = "AuthID", referencedColumnName = "id"), inverseJoinColumns = @JoinColumn(name = "PropertyID", referencedColumnName = "id"))
private List<Property> properties;
我正在使用Spring Data JPA存储库,并且能够创建一个自定义 @Query
,它根据特定的角色ID返回用户列表 .
@Query("SELECT u FROM auth_user as u WHERE u.isEnabled AND u.id IN"
+ " (SELECT r.user_id FROM user_role as r WHERE r.role_id = ?1)")
public List<OAuthUser> findByRole(int roleID);
上面的代码导致错误 auth_user is not mapped
. 我明白为什么我得到错误;框架使用实体名称(OAuthUser)而不是表(auth_user)来执行查询 . 这通常不会成为问题,除非 user_role
没有实体;它只是一个包含两列的连接表:'user_id'和'role_id' .
实现这个目标的适当方法是什么?
谢谢 .
2 回答
错误说:
它指的是查询中使用的
auth_user
,如SELECT u FROM auth_user
. 它必须是OAuthUser
,而不是在查询中 .您正在使用jpql @Query中的表名(auth_user) . 您应该使用类名 OAuthUser 代替:
如果要使用SQL而不是jpql,则需要使用如下所示:
这样你就可以提到表名和列 .