JSON 结果返回 null

Question

以下是通过输入城市名称或地名来查找天气的代码。结果是来自doInBackground方法的字符串...但不幸的是它返回null...

@Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);

        if(result == null){
            Toast.makeText(MainActivity.this,"Place not found",Toast.LENGTH_LONG).show();
        }
        else{
            try {

                String message = "";

                JSONObject jsonObject = new JSONObject(result);

                String weatherInfo = jsonObject.getString("weather");

                Log.i("Weather content", weatherInfo);

                JSONArray arr = new JSONArray(weatherInfo);

                for (int i = 0; i < arr.length(); i++) {

                    JSONObject jsonPart = arr.getJSONObject(i);

                    String main = "";
                    String description = "";

                    main = jsonPart.getString("main");
                    description = jsonPart.getString("description");

                    if (main != "" && description != "") {

                        message += main + ": " + description + "\r\n";

                    }

                }

                if (message != "") {

                    weatherReport.setText(message);

                } else {

                    Toast.makeText(getApplicationContext(), "Could not find weather", Toast.LENGTH_LONG).show();

                }


            } catch (JSONException e) {

                Toast.makeText(getApplicationContext(), "Could not find weather", Toast.LENGTH_LONG).show();

            }

doInBackground() 方法..

protected String doInBackground(String... urls) {

        String result = "";
        URL url;
        HttpURLConnection httpURLConnection;

        try {
            url = new URL(urls[0]);
            httpURLConnection = (HttpURLConnection) url.openConnection();
            InputStream in = httpURLConnection.getInputStream();
            InputStreamReader reader = new InputStreamReader(in);
            int data = reader.read();
            while(data != -1){
                char current = (char) data;
                result += current;

                data = reader.read();
            }
            return result;

        }
        //combined the exceptions MalformedURL and IOException to a common to display a toast msg
        catch (Exception e) {
            e.printStackTrace();
        }

        return null;

    }

下载 url ..DownloadTask 是从 AsyncTask 扩展的类的名称，并具有方法 doInBackground 和 onPostExecute().. 请帮助我了解为什么结果字符串返回 null..

weather.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {

            try {

                //to hide the keyboard after pressing the button
                InputMethodManager manager =
                        (InputMethodManager) getSystemService(Context.INPUT_METHOD_SERVICE);
                manager.hideSoftInputFromInputMethod(weatherInput.getWindowToken(),0);

                DownloadTask downloadTask = new DownloadTask();

                //used to encode the entered input for url.. for example San Fransisco appears in url
                //as San%20Fransisco ... and to enable that we use the encoder...
                String encodedCity = URLEncoder.encode(city,"UTF-8");

                downloadTask.execute("http://api.openweathermap.org/data/2.5/weather?q=" + encodedCity +
                        "&appid=cd66504ca815ddc1971662a9f2147f84\n");

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

        }
    });

Answer 1

我在您的代码中发现了一些问题。试试这个。

String message = ""; 
JSONObject jsonObject = new JSONObject(result); 
JSONArray arr = jsonObject.optJSONArray("weather"); 
for (int i = 0; i < arr.length(); i++) { 
    JSONObject jsonPart = arr.optJSONObject(i); 
    String main = jsonPart.optString("main");
    String description = jsonPart.optString("description"); 
    if (!TextUtils.isEmpty(main) && !TextUtils.isEmpty(description)) { 
        message += main + ": " + description + "\r\n"; 
     } 
 } 
 if (!TextUtils.isEmpty(message)) { 
       weatherReport.setText(message); 
  } else { 
      Toast.makeText(getApplicationContext(), "Could not find weather", Toast.LENGTH_LONG).show(); 
     }

Answer 2

请在编写任何 Java 代码之前测试您的 API。

---

您的网址中有一个尾随 \n。

只需在您的代码中打开实际的 URL，或者将其 cURL...

curl 'http://api.openweathermap.org/data/2.5/weather?q=San%20Antonio&appid=cd66504ca815ddc1971662a9f2147f84\n' | jq 

{
  "cod": 401,
  "message": "Invalid API key. Please see http://openweathermap.org/faq#error401 for more info."
}

去掉末尾的\n

{
  "coord": {
    "lon": -98.49,
    "lat": 29.42
  },
  "weather": [
    {
      "id": 800,
      "main": "Clear",
      "description": "clear sky",
      "icon": "01n"
    }
  ],
  "base": "stations",
  "main": {
    "temp": 288.15,
    "pressure": 1023,
    "humidity": 27,
    "temp_min": 285.15,
    "temp_max": 291.15
  },
  "visibility": 16093,
  "wind": {
    "speed": 3.6,
    "deg": 30
  },
  "clouds": {
    "all": 1
  },
  "dt": 1488423180,
  "sys": {
    "type": 1,
    "id": 2713,
    "message": 0.0075,
    "country": "US",
    "sunrise": 1488459476,
    "sunset": 1488501264
  },
  "id": 4726206,
  "name": "San Antonio",
  "cod": 200
}