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...但不幸的是它返回了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＆lt;＆gt;扩展的类的名称。并有方法doInBackground和onPostExecute（）..请帮助我为什么结果String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

请在编写任何Java代码之前测试您的API 。

您的网址中有一个结尾\n。

只需打开代码中的实际网址，或者cURL it ...

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." }

删除最后的\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 }

Answer 2

我发现您的代码几乎没有问题。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(); 
     }

JSON结果返回null

2 个答案: