AsyncTask doInBackground返回多个字符串

Question

我正试图在android工作室中建立一个非常基本的天气应用程序。我正在使用AsyncClass返回多个字符串。

正如您在代码中看到的，我使用了一个名为“wrapper”的类，用于存储字符串，这样我就可以返回一个类对象，并在AsyncTask的onPostExecute方法中使用它。我面临的问题是，当我测试应用程序时，所有返回的字符串都是未定义的（包装类的默认值）。这意味着没有在doInBackground方法中更新字符串，我似乎想不出原因！

我得活动

    @Override
    public void onConnectionFailed(ConnectionResult connectionResult) {
        Log.i(MainActivity.class.getSimpleName(), "Can't connect to Google Play Services!");
    }
    private class Wrapper
    {
          String Temperature = "UNDEFINED";
         String city = "UNDEFINED";
         String country = "UNDEFINED";
    }

    private class GetWeatherTask extends AsyncTask<String, Void, Wrapper> {
        private TextView textView;

        public GetWeatherTask(TextView textView) {
            this.textView = textView;
        }

        @Override
        protected Wrapper doInBackground(String... strings) {
            Wrapper w = new Wrapper();
            String Temperature = "x";
            String city = "y";
            String country = "z";

            try {
                URL url = new URL(strings[0]);
                HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();

                InputStream stream = new BufferedInputStream(urlConnection.getInputStream());
                BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(stream));
                StringBuilder builder = new StringBuilder();

                String inputString;
                while ((inputString = bufferedReader.readLine()) != null) {
                    builder.append(inputString);
                }

                JSONObject topLevel = new JSONObject(builder.toString());
                JSONObject main = topLevel.getJSONObject("main");
                JSONObject cityobj = topLevel.getJSONObject("city");
                Temperature = String.valueOf(main.getDouble("temp"));
                city = cityobj.getString("name");
               country = cityobj.getString("country");

                w.Temperature= Temperature;
                w.city= city;
                w.country=country;

                urlConnection.disconnect();
            } catch (IOException | JSONException e) {
                e.printStackTrace();
            }
            return w;
        }

        @Override
        protected void onPostExecute(Wrapper w) {
            textView.setText("Current Temperature: " + w.Temperature + " C" + (char) 0x00B0
                       +"\n" + "Current Location: "+ w.country +"\n" + "City: "+  w.city );
        }
    }
}

更新：

结果我在代码中使用了错误的url，我使用的是：http://api.openweathermap.org/data/2.5/weather？lat=%f&lon=%f&units=%s&appid=%s

相反，我应该用：

-AKA我应该用预测而不是天气

Answer 1

您的错误从这里开始

JSONObject main = topLevel.getJSONObject("main");

可能是因为TopLevel对象没有“main”键。

{  
   "city":{  },
   "cod":"200",
   "message":0.1859,
   "cnt":40,
   "list":[  ]
}

将您的JSON扔到这里。https://jsonformatter.curiousconcept.com/

String city = "undefined";
String country = "undefined";
List<Double> temperatures = new ArrayList<Double>();

try {
    JSONObject object = new JSONObject(builder.toString());
    JSONObject jCity = object.getJSONObject("city");
    city = jCity.getString("name");
    country = jCity.getString("country");

    JSONArray weatherList = object.getJSONArray("list");
    for (int i = 0; i < weatherList.length(); i++) {
        JSONObject listObject = weatherList.getJSONObject(i);
        double temp = listObject.getJSONObject("main").getDouble("temp");
        temperatures.add(temp);
    }

} catch (JSONException e) {
    e.printStackTrace();
}

return new Wrapper(city, country, temperatures);