Question

http://192.168.0.101:8080/messenger/webapi/login/?name=zeeshan&password=123

以上是我的网址，我得到回复“登录成功”，我试图在我的Android应用程序上获得响应，但应用程序崩溃

Log, i have also tried @Query instead of @Path but its giving me same error

我正在使用此注释

String BASE_URL = "http://192.168.0.101:8080/messenger/webapi/";

 @GET("login/?name={name}&password={password}")
    Call<String> login(@Path("name") String name, @Path("password") String password);

LoginActivity

private void loginUser() {
    Retrofit retrofit = new Retrofit.Builder()
            .baseUrl(Api.BASE_URL)
            .addConverterFactory(ScalarsConverterFactory.create())
            .build();

    Api api = retrofit.create(Api.class);
    Log.e("TAG", name + " " + password);
    try {
        Call<String> call = api.login(name,password);
        call.enqueue(new Callback<String>() {
            @Override
            public void onResponse(Call<String> call, Response<String> response) {
                String res = response.body();
                Log.e("TAG", res);
            }

            @Override
            public void onFailure(Call<String> call, Throwable t) {

                Log.e("TAG", t.getMessage());

            }
        });
    }catch(Exception e){
        Log.e("TAG",e.toString());
    }


}

Answer 1

df1 = pd.DataFrame(np.random.rand(10,4),columns=list('ABCD'), index = list      ('IJKLMNOPQR')) 



df2 = pd.DataFrame({'1': [1,2,3,4,5,6,7], '2':[3,4,5,6,7,8,9]}, index =list('IQLMRFW') )

使用改造的用户名和密码的android注释

1 个答案: