如何解析URL请求的JSON结果?

我是android和java的新手。我想获得一个url请求(结果是JSON)并对其进行解析(例如从yahoo api获得JSON天气)。

public static String getStringFromURL(String urlString) throws IOException {
    HttpURLConnection urlConnection;
    URL url = new URL(urlString);
    urlConnection = (HttpURLConnection) url.openConnection();
    urlConnection.setRequestMethod("GET");
    urlConnection.setReadTimeout(10000 /* milliseconds */);
    urlConnection.setConnectTimeout(15000 /* milliseconds */);
    urlConnection.setDoOutput(true);
    urlConnection.connect();
    BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(url.openStream()));
    char[] buffer = new char[1024];
    String outputString;
    StringBuilder builder = new StringBuilder();
    String line;
    while ((line = bufferedReader.readLine()) != null) {
        builder.append(line).append("\n");
    }
    bufferedReader.close();
    outputString = builder.toString();
    return outputString;
}

public void setWeather (View view) throws IOException, JSONException {
    String json = getStringFromURL("https://query.yahooapis.com/v1/public/yql?q=select * from weather.forecast where woeid in (select woeid from geo.places(1) where text='Esfahan')&format=json");
    JSONObject jso = new JSONObject(json);
    JSONObject query = jso.getJSONObject("query");
    JSONObject result = query.getJSONObject("results");
    JSONObject channel = result.getJSONObject("channel");
    JSONObject windI = channel.getJSONObject("wind");
    JSONObject location = channel.getJSONObject("location");
    String last = "";
    last = location.getString("city");
    TextView tv = (TextView) findViewById(R.id.textView);
    tv.setText(last);
}

当我在设备上运行此应用程序时,应用程序崩溃。 在Android Monitor上写入错误: