代码之家 › 专栏 › 技术社区 › Sitansu

Spring Boot JPQL不适用于特定条件

jpql jpa spring-boot spring java

Sitansu · 技术社区 · 5 年前

我在spring boot应用程序中面临与JPQL相关的问题。我面临的问题是“无效参数索引!您似乎声明的查询方法参数太少!“。无法按用户名和客户端代码获取记录。请检查下面的Spring Boot应用程序代码片段。

Bean类UserClients.Java。

@Entity
@Table(name = "usersclients")
public class UserClients implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long ID;

    @JsonBackReference
    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "userName", referencedColumnName = "userName")
    private Users user;

    @ManyToOne(fetch = FetchType.EAGER)
    @JoinColumn(name = "clientCode", referencedColumnName = "code")
    private Clients client;
}

存储库类UserClientsRepository

@Repository
public interface UserClientsRepository extends CrudRepository<UserClients, Long> {

@Async
@Query(value = "from UserClients userCli join userCli.user user  join userCli.client client  where user.userName= ?0 and client.clientCode= ?1", nativeQuery = true)
UserClients fetchRecordByUserNameClient(String userName,String clientCode);

}

服务类UserClientsService

@Service
public class UserClientsService {

    @Autowired
    private UserClientsRepository userClientsRepository;


    public UserClients fetchRecordByUserNameClient(String username, String clientCode) {
        return userClientsRepository.fetchRecordByUserNameClient(username, clientCode);
    }

}

控制器类身份验证控制器

@CrossOrigin(origins = "*", maxAge = 3600)
@RestController
@RequestMapping("/token")
public class AuthenticationController {


    @Autowired
    private UserClientsService userClientsService;


    @RequestMapping(value = "/android-generate-token", method = RequestMethod.POST)
    public ApiResponse<AuthToken> loginActivity(@RequestBody LoginUserDto loginUser) {
        try {
            final UserClients userClients= userClientsService.fetchRecordByUserNameClient(loginUser.getUsername(), 
                    loginUser.getClient());
            if(userClients == null) {
                return new ApiResponse<>(401, "failed", null);
            }
            return new ApiResponse<>(200, "success", new AuthToken(token, user.getUserName()));
        } catch (AuthenticationException e) {
            return new ApiResponse<>(401, e.getMessage(), null);
        }
    }

}

1 回复 | 直到 5 年前

Simon Martinelli 5 年前

你的问题不对。

2) 您应该使用命名参数。

另外,我不确定您想用@Async实现什么。如果不返回未来对象,此查询将永远不会异步运行。

@Repository
public interface UserClientsRepository extends CrudRepository<UserClients, Long> {

    @Async
    @Query("select userCli from UserClients userCli join userCli.user user join userCli.client client "+ 
           "where user.userName= :userName and client.clientCode= :clientCode")
    Future<UserClients> fetchRecordByUserNameClient(String userName,String clientCode);

}