Java Spring Boot JPQL在特定条件下不工作
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“参数索引无效!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应用程序代码段 Bean类UserClients.JavaJava Spring Boot JPQL在特定条件下不工作,java,spring,spring-boot,jpa,jpql,Java,Spring,Spring Boot,Jpa,Jpql,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“参数索引无效!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应用程序代码段 Bean类UserClients.Java @Entity @Table(name = "usersclients") public class UserClients implements Serializable { @Id @GeneratedValue(stra
@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);
}
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);
}
}
}
@CrossOrigin(origins=“*”,maxAge=3600)
@RestController
@请求映射(“/token”)
公共类身份验证控制器{
@自动连线
私有用户客户端服务用户客户端服务;
@RequestMapping(value=“/android generate token”,method=RequestMethod.POST)
公共ApiResponse loginActivity(@RequestBody LoginUserTo loginUser){
试一试{
final UserClients UserClients=userClientsService.fetchRecordByUserNameClient(loginUser.getUsername(),
loginUser.getClient());
if(userClients==null){
返回新的ApiResponse(401,“失败”,null);
}
返回新的ApiResponse(200,“success”,新的AuthToken(token,user.getUserName());
}捕获(身份验证异常e){
返回新的ApiResponse(401,e.getMessage(),null);
}
}
}
您的查询是错误的
1) 您已设置native=true,这意味着您希望使用SQL。但是查询看起来像HQL
2) 您应该使用命名参数
此外,我不确定您希望通过@Async实现什么。如果不返回将来的对象,此查询将永远不会异步运行
因此,您的查询应该如下所示:
@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);
}
@存储库
公共接口UserClientsRepository扩展了Crudepository{
@异步的
@查询(“从UserClients选择userCli userCli join userCli.user user join userCli.client client”+
“其中user.userName=:userName和client.clientCode=:clientCode”)
未来fetchRecordByUserNameClient(字符串用户名、字符串客户端代码);
}
您的查询是错误的
1) 您已设置native=true,这意味着您希望使用SQL。但是查询看起来像HQL
2) 您应该使用命名参数
此外,我不确定您希望通过@Async实现什么。如果不返回将来的对象,此查询将永远不会异步运行
因此,您的查询应该如下所示:
@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);
}
@存储库
公共接口UserClientsRepository扩展了Crudepository{
@异步的
@查询(“从UserClients选择userCli userCli join userCli.user user join userCli.client client”+
“其中user.userName=:userName和client.clientCode=:clientCode”)
未来fetchRecordByUserNameClient(字符串用户名、字符串客户端代码);
}