Question

遇到错误的请求错误，不知道为什么能帮助我解决它我现在他们的错误很小，但无法理解问题所在

Api Interface
@POST("token")
    Call<LoginResponse> loginresponse(@Body RequestBody login);


Api CLient 
public class ApiClient {

    private static final String BASE_URL = "http://falkentyre.in/";
    private static Retrofit retrofit;
    private static Retrofit retrofit1;

    private ApiClient(Context context) {

        //Add gson to retrofit
        Gson gson = new GsonBuilder()
                .setLenient()
                .create();

        // Add Ok http client to retrofit
        OkHttpClient.Builder builder = new OkHttpClient.Builder();
        builder.addInterceptor(chain -> {
            Request original = chain.request();
            Request.Builder requestBuilder = original.newBuilder();
            return chain.proceed(requestBuilder.build());
        });

        OkHttpClient mOkHttpClient = builder.build();

        retrofit = new Retrofit.Builder()
                .baseUrl(BASE_URL)
                .addConverterFactory(GsonConverterFactory.create(gson))
                .client(mOkHttpClient)
                .build();
    }


    public static Retrofit getClient() {

        if (retrofit1 == null) {
            OkHttpClient client = new OkHttpClient().newBuilder()
                    .readTimeout(1000, TimeUnit.SECONDS)
                    .connectTimeout(1000, TimeUnit.SECONDS)
                    .build();

            Gson gson = new GsonBuilder()
                    .setLenient()
                    .create();

            retrofit1 = new Retrofit.Builder()
                    .baseUrl(BASE_URL)
                    .addCallAdapterFactory(RxJavaCallAdapterFactory.create())
                    .addConverterFactory(GsonConverterFactory.create(gson))
                    .client(client)
                    .build();
        }

        return retrofit1;
    }

    public <S> S createService(Class<S> serviceClass) {
        return retrofit.create(serviceClass);
    }
}
///////////////////////////////////////////////////// 

public void take_Login(String no,String pass) {
        showProgressBar();


       String login = String.format("grant_type=password&username=%s&password=%s&device=android&notificationtoken=hgh",
                no, pass);
        RequestBody body = RequestBody.create(MediaType.parse("text/plain"), login);
        ApiInterface apiInterface = ApiClient.getClient().create(ApiInterface.class);
        Call<LoginResponse> userCall = apiInterface.loginresponse(body);

        userCall.enqueue(new Callback<LoginResponse>() {
            @Override
            public void onResponse(Call<LoginResponse> call, Response<LoginResponse> response) {
                if (response.isSuccessful()) {
                    hideProgressBar();
                  Toast.makeText(MainActivity.this,response.body().getAccess_token(),Toast.LENGTH_SHORT).show();

                } else {
                    Toast.makeText(MainActivity.this,response.message(),Toast.LENGTH_SHORT).show();
                }
            }

            @Override
            public void onFailure(Call<LoginResponse> call, Throwable t) {
                Toast.makeText(MainActivity.this,t.getMessage(),Toast.LENGTH_SHORT).show();
            }
        });
    }

Answer 1

尝试一下

在ApiInterface中

@FormUrlEncoded
@POST("token")
Call<LoginResponse> loginresponse(@FieldMap HashMap<String, String> login);

对于这样的通话用途，

HashMap<String,String> map = new HashMap<>();
    map.put("grant_type","password");
    map.put("username","username want to pass");
    map.put("password","password want to pass");
    map.put("device","android");
    map.put("notificationtoken","hgh");

    ApiInterface apiInterface = ApiClient.getClient().create(ApiInterface.class);
    Call<LoginResponse> userCall = apiInterface.loginresponse(map);

对于GET方法，您可以尝试

@GET("token")
Call<LoginResponce> loginresponse(@Query("grant_type") String grant_type, @Query("username") String username, @Query("password") String password,@Query("device") String device, @Query("notificationtoken") String notificationtoken);

您可以这样称呼它

ApiInterface apiInterface = ApiClient.getClient().create(ApiInterface.class);
    Call<LoginResponse> userCall = apiInterface.loginresponse("grandType you want","username you want","password you want","android","hgh");

得到了错误的请求错误

1 个答案: