Android Fetch Json

Question

我很抱歉，但我刚接触android编程，所以这看起来像是一个非常愚蠢的问题。

我建立了与我的JSON数据的连接，但我只需要获取我需要的数据，而不是所有不必要的数据。我的Json看起来像这样：

  

数据 “：[{” ID “：1，” 温度 “：” 21.375" ， “湿度”： “28”}

这里我只需要id：1，温度：21.375，湿度：28

建立连接我有以下方法：

private class jsonConnection extends AsyncTask<Void, Void, Void> {

    @Override
    protected Void doInBackground(Void... voids) {

        String username = "websiteUsername";
        String password = "websitePassword";
        String credentials = username + ":" + password;

        String sensorID = "1";
        String api = "api";
        String websiteURL = "https://" + credentials + "@website.com" + "/" + sensorID + "/" + api;

        String credBase64 = Base64.encodeToString( credentials.getBytes(), Base64.DEFAULT ).replace( "\n", "" );

        try {

            URL url = new URL( websiteURL );

            System.out.println( url );
            System.out.println( credentials );
            System.out.println( websiteURL );

            connection = (HttpsURLConnection) url.openConnection();

            String basicAuth = "Basic " + new String( encodeBase64URLSafeString( credentials.getBytes() ) );

            connection.setRequestProperty( "Authorization", basicAuth );
            connection.setRequestMethod( "GET" );
            connection.setRequestProperty( "Content-Type", "application/x-www-form-urlencoded" );
            connection.setRequestProperty( "Content-Language", "en-US" );
            connection.setUseCaches( false );
            connection.setDoInput( true );
            connection.setDoOutput( true );
            connection.connect();

            InputStream stream = connection.getInputStream();

            bufferedReader = new BufferedReader( new InputStreamReader( stream ) );

            //StringBuffer buffer = new StringBuffer();

            String line = "";
            while (line != null) {
                line = bufferedReader.readLine();
                data = data + line;
            }


        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }

    @Override
    protected void onPreExecute() {
        super.onPreExecute();

        pd = new ProgressDialog( MainActivity.this );
        pd.setMessage( "Please wait" );
        pd.setCancelable( false );
        pd.show();
    }

    @Override
    protected void onPostExecute(Void result) {
        super.onPostExecute( result );
        if (pd.isShowing()) {
            pd.dismiss();
        }
            ScrollingMovementMethod());
            System.out.println( data );
            json_string = data;
            System.out.println( "json_string: " + json_string );
        }
    }

我想我必须做这样的事情，但我不确定如何继续。

        try {

        JSONObject parentObject = new JSONObject( data );
        JSONArray parentArray = parentObject.getJSONArray( "data" );

        JSONObject finalObject = parentArray.getJSONObject( 0 );

        int idName = finalObject.getInt( "id" );
        int tempName = finalObject.getInt( "temperature" );
        int humName = finalObject.getInt( "humidity" );
        String batName = finalObject.getString( "battery" );
        String modeName = finalObject.getString( "mode" );
        String dateName = finalObject.getString( "date_time" );
        String timeName = finalObject.getString( "date_time" );
        String luxName = finalObject.getString( "lux" );

        System.out.println( idName + tempName + humName + batName + modeName + dateName + timeName + luxName );

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

我怀疑它不应该是参数中的数据，但也许是别的东西，也许这会解决它？

    JSONObject parentObject = new JSONObject( data );