Question

我试图做一个简单的天气应用程序，允许用户输入他们的城市，它将返回天气数据。我使用了一个带有EditText的片段，并且我使用接口将字符串传递给MainActivity，这需要在URL字符串中连接，这就是问题所在。获取异常错误，我相信没有找到协议。我确实检查了我输入http：或https的URL。任何建议将URL传递给我的AsyncTask

//接口方法

  @Override
    public void displayCityChosen(String cityName) {

    Log.i(TAG, "city name= " +cityName);
    String weatherURL = "http://www.api.openweathermap.org/data/2.5/weather?q=" + cityName + "&APPID=32c831d8a6937f237acff8eef3d4a58c";
    try{

        String base = URLEncoder.encode(weatherURL,"UTF-8");
        URL finalWeatherURL = new URL(base);
       new GetWeather().execute(finalWeatherURL);
    }catch (Exception e) {
        Log.i(TAG, "ERROR");
    }
}

private class GetWeather extends AsyncTask<URL, Void, JSONObject> {

    @Override
    protected JSONObject doInBackground(URL... urls) {

        int statusCode = -1;
        String jsonString = "";

        for (URL queryURL : urls) {
            try {
                URLConnection conn = queryURL.openConnection();
                jsonString = IOUtils.toString(conn.getInputStream());
                jsonResponse = new JSONObject(jsonString);

                jsonArray = jsonResponse.getJSONArray("weather");
                Log.i(TAG, "Array= " + jsonArray);
                break;
            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }


        return jsonResponse;
    }


}

Answer 1

我认为您还需要一个步骤来将URL更改为HttpURLConnection，我认为这个url不需要对您的使用进行编码。所有编码都搞乱了。

        URL url = new URL("http://blah");

        HttpURLConnection httpURLConnection = (HttpURLConnection) url
                .openConnection();
        httpURLConnection.setRequestMethod("GET");
        httpURLConnection.setDoInput(true);

        if (httpURLConnection.getResponseCode() != 200) {
            return null;
        }

        InputStream result = httpURLConnection.getInputStream();
        InputStreamReader r = new InputStreamReader(result);

FYI将字符串传递给asynctask并在asynctask中进行所有转换，它将为响应更快的应用程序做出。

将URL字符串传递给AsyncTask

1 个答案: