[英]The given id must not be null!; nested exception is java.lang.IllegalArgumentException
i'm creating a simple crud with spring boot.我正在用 spring 引导创建一个简单的 crud。 All function well but I have a little problem with the method findOne(Long id)所有 function 都很好,但我对方法 findOne(Long id) 有一点问题
in postman when i put this url: http://localhost:8080/api/user/id/?id=13 , i get this exception:在 postman 当我把这个 url: http://localhost:8080/api/user/id/?id=13 ,我得到这个异常:
" error ": "Internal Server Error", " error ": "内部服务器错误",
" exception ": "org.springframework.dao.InvalidDataAccessApiUsageException", "异常": "org.springframework.dao.InvalidDataAccessApiUsageException",
" message ": "The given id must not be null;; nested exception is java.lang.IllegalArgumentException: The given id must not be null,", " message ": "给定的 id 不能是 null;;嵌套异常是java.lang.IllegalArgumentException: 给定的 id 不能是 Z37A6259CC0C1DAE299A7866489DFF0,
here is my code:这是我的代码:
Repository存储库
@SuppressWarnings("unused")
@Repository
public interface UserRepository extends JpaRepository<Utilisateur, Long> {
}
Service服务
public interface UserService {
Utilisateur save(Utilisateur utilisateur);
List<Utilisateur> findAll();
Utilisateur findOne(Long id);
}
ServiceImpl服务实现
@Service
public class UserServiceImpl implements UserService{
@Autowired
UserRepository userRepository;
public UserServiceImpl() {
}
public UserServiceImpl(UserRepository userRepository) {
this.userRepository = userRepository;
}
@Override
public Utilisateur save(Utilisateur utilisateur) {
return userRepository.save(utilisateur);
}
@Override
public List<Utilisateur> findAll() {
return userRepository.findAll();
}
public Utilisateur findOne(Long id) {
return userRepository.findOne(id);
}
}
Controller Controller
@RestController
@RequestMapping("/api")
@CrossOrigin(origins="http://localhost:4200",allowedHeaders="*")
public class UserController {
@Autowired
private UserService userService;
public UserController(UserService userService) {
this.userService = userService;
}
public UserController() {
super();
// TODO Auto-generated constructor stub
}
@GetMapping("/users")
public List<Utilisateur> getUsers() {
return userService.findAll();
}
@GetMapping("/user/id/{id}")
public ResponseEntity<Utilisateur> getUser(@PathVariable Long id) {
Utilisateur utilisateur = userService.findOne(id);
return ResponseEntity.ok().body(utilisateur);
}
/*
@DeleteMapping("/user/{id}")
public Boolean deleteUser(@PathVariable Long id) {
userRepository.delete(id);
return true;
} */
@PostMapping("/user")
public ResponseEntity<Utilisateur> saveUser(@RequestBody Utilisateur utilisateur) throws URISyntaxException {
Utilisateur result = userService.save(utilisateur);
return ResponseEntity.created(new URI("/api/user/" + result.getId()))
.body(result);
}
@PutMapping("/user")
public ResponseEntity<Utilisateur> updateUser(@RequestBody Utilisateur utilisateur) throws URISyntaxException {
Utilisateur result = userService.save(utilisateur);
return ResponseEntity.ok().body(result);
}
}
You have your mapping set as a URL path variable您将映射设置为 URL 路径变量
@GetMapping("/user/id/{id}")
but the URL you tried has a query parameter: ?id=13但是您尝试的 URL 有一个查询参数: ?id=13
Try using: http://localhost:8080/api/user/id/13尝试使用: http://localhost:8080/api/user/id/13
Here is a good comparison of the two on stackoverflow这是stackoverflow上两者的一个很好的比较
The URL is incorrect. URL 不正确。 You have set it up as a path variable in your code.您已在代码中将其设置为路径变量。 In which case, instead of hitting localhost:8080/api/user/id/?id=13 in postman, you should hit localhost:8080/api/user/id/3 instead,在这种情况下,您应该点击 localhost:8080/api/user/id/3 而不是在 postman 中点击 localhost:8080/api/user/id/?id=13,
But if you were following REST standards, a better URL would look like this (no need to have the "id" in the URL).但是,如果您遵循 REST 标准,则更好的 URL 看起来像这样(不需要在 URL 中包含“id”)。 localhost:8080/api/user/3本地主机:8080/api/user/3
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