Spring Boot JPQL 不适用于特定条件答案

【问题标题】：Spring Boot JPQL not working for Specific ConditionSpring Boot JPQL 不适用于特定条件
【发布时间】：2019-10-19 23:38:13
【问题描述】：

我在 Spring Boot 应用程序中遇到了与 JPQL 相关的问题。我面临问题“无效的参数索引！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。请检查我下面的 Spring Boot 应用程序代码 sn-p。

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;
}

Repository 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 AuthenticationController

@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);
        }
    }

}

【问题讨论】：

标签： java spring spring-boot jpa jpql

【解决方案1】：

您的查询错误。

1) 你设置了 native = true ，这意味着你想使用 SQL。但查询看起来像 HQL

2) 你应该使用命名参数。

另外，我不确定你想用@Async 实现什么。如果不返回 Future 对象，此查询将永远不会异步运行。

所以你的查询应该是这样的：

@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);

}

【讨论】：