如何在 Spring 引导中将不同的 JSON 响应消耗到客户端的单个端点中？

Question

I am trying to consume api json response in the client side. It is for user authentication. When the correct user name and password is given the following response is achieved:

But when wrong user name or password is given, the following response can be seen.

For the valid user name and password, I have implemented the following method which works fine for that purpose. But I cannot handle the situation when the wrong user name or password is given.

public ResponseEntity<AuthenticateUserOutputModel> getAutheticateUser(
        AuthenticateUserInputModel authenticateUserInput) {

    String url = AUTHENTICATE_USER;

    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);
    
    RestTemplate restTemplate = new RestTemplate();
    
    HttpEntity<AuthenticateUserInputModel> requestEntity = new HttpEntity<>(authenticateUserInput, headers);
    
    MultiValueMap<String, String> form = new LinkedMultiValueMap<>();
    form.add("username",authenticateUserInput.getUsername() );
    form.add("password", authenticateUserInput.getPassword());
    
    
    HttpEntity<MultiValueMap<String, String>> requestEntity = new HttpEntity<MultiValueMap<String, String>>(form, headers);
    
     ResponseEntity<AuthenticateUserOutputModel> responseEntity = new ResponseEntity<>(HttpStatus.OK);
     
     try {
         
         responseEntity = restTemplate.postForEntity(url, requestEntity,AuthenticateUserOutputModel.class);
         
    } catch (Exception e) {
        
        System.out.println("Error Found");
    }finally {
        return responseEntity;
    }
    
}

How can I implement this?

Answer 1

First you have to catch this error inside catch(HttpClientErrorException e) , inside this catch statement, you can add if with condition e.getStatusCode() == HttpStatus.UNAUTHORIZED to check is this 401 Unauthorized , then you know that something with your Authorization is wrong (password or username) and inside this if statement you can do something (return particular information for user or thorw an exception)

public ResponseEntity<AuthenticateUserOutputModel> getAutheticateUser(
        AuthenticateUserInputModel authenticateUserInput) {

    String url = AUTHENTICATE_USER;

    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);

    RestTemplate restTemplate = new RestTemplate();

    HttpEntity<AuthenticateUserInputModel> requestEntity = new HttpEntity<>(authenticateUserInput, headers);

    MultiValueMap<String, String> form = new LinkedMultiValueMap<>();
    form.add("username",authenticateUserInput.getUsername() );
    form.add("password", authenticateUserInput.getPassword());


    HttpEntity<MultiValueMap<String, String>> requestEntity = new HttpEntity<MultiValueMap<String, String>>(form, headers);

    ResponseEntity<AuthenticateUserOutputModel> responseEntity = new ResponseEntity<>(HttpStatus.OK);

    try {

        responseEntity = restTemplate.postForEntity(url, requestEntity,AuthenticateUserOutputModel.class);
        
    } catch (HttpClientErrorException e) {

        if (e.getStatusCode() == HttpStatus.UNAUTHORIZED) {
            // Do something when 401
        }
        
    } catch (Exception e) {

        System.out.println("Error Found");
    }finally {
        return responseEntity;
    }

}