java - Android:POST请求，显示特定文本

Question

好吧，我对 POST 和 GET 请求等概念很陌生。所以，如果这是一个愚蠢的问题，我深表歉意。

我在这个 TvrdjavaFragment.java 中有一个异步任务，它获取并显示 JSON 对象中的所有内容作为注释(toast)。例如，我只需要 temperatura 并将其显示为注释，而不是所有内容。我的问题是如何从 JSON 对象获取特定内容？

这是我的 TvrdjavaFragment.java 文件:

public class TvrdjavaFragment extends Fragment {

    Button btnIdinaperiod;
    TextView pokaziServer;
    String rezultat = "";
    String strURL = "http://MYLINK";

    public TvrdjavaFragment() {
        // Required empty public constructor
    }


    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container,
                             Bundle savedInstanceState) {

        View view = inflater.inflate(R.layout.fragment_tvrdjava, container, false);
        // Inflate the layout for this fragment

        pokaziServer = (TextView) view.findViewById(R.id.testServer);
        //int i = Integer.parseInt(pokaziServer.getText().toString());

        //Log.d("TAG", "TestLogIvan");
        new NabaviServer().execute();

        btnIdinaperiod = (Button) view.findViewById(R.id.buttonIdinaperiod);

        btnIdinaperiod.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                PeriodFragment periodFragment = new PeriodFragment();
                FragmentTransaction periodFragmentTransaction = getActivity().getSupportFragmentManager().beginTransaction();
                periodFragmentTransaction.replace(R.id.frame, periodFragment);
                periodFragmentTransaction.addToBackStack(null); //Kada pretisne BACK, vrati se nazad
                periodFragmentTransaction.commit();
            }
        });

        return view;
    }

    public class NabaviServer extends AsyncTask<String, String, String>
    {
        @Override
        protected void onPreExecute() {
            super.onPreExecute();
        }

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

            Toast.makeText(getActivity(), "Izlazak je: " + rezultat,Toast.LENGTH_LONG).show();
            Log.i("IVANTAG", rezultat);
        }

        @Override
        protected String doInBackground(String... params) {

            try{
                URL url = new URL(strURL);
                HttpURLConnection con = (HttpURLConnection)url.openConnection();
                con.setRequestMethod("POST");
                con.connect();

                BufferedReader bf = new BufferedReader(new InputStreamReader(con.getInputStream())); //??s

                String value = bf.readLine();
                System.out.println("test " + value);
                rezultat = value;


            }
            catch(Exception e)
            {
                System.out.println(e);
            }

            return null;
        }
    }

}

这就是我的 JSON 对象的样子:

[
  {
    "id": 1,
    "vPritisak": "1",
    "vVazduha": "0",
    "nVisina": "0",
    "temperatura": "0",
    "metan": "0",
    "uDioksid": "1",
    "lokacija": "Kicevo",
    "created_at": null,
    "updated_at": null
  },
  {
    "id": 2,
    "vPritisak": "0",
    "vVazduha": "2",
    "nVisina": "0",
    "temperatura": "0",
    "metan": "0",
    "uDioksid": "0",
    "lokacija": "Cair",
    "created_at": null,
    "updated_at": null
  },
etc...

编辑:当我尝试添加 JSONObject jObj = new JSONObject(s); 时，我只收到错误“Unhandled exception: org.json.JSONException ”。当我尝试添加 Try - Catch (就像它所建议的那样)时，应用程序崩溃了:

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

        try {
            JSONObject jObj = new JSONObject(s);
            String temperatura = jObj.getString("temperatura");//get ur temperatura here

            //try to toast it out,to see the value
            Toast.makeText(getActivity(), "Izlazak je: " + temperatura,Toast.LENGTH_LONG).show();

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


        Toast.makeText(getActivity(), "Izlazak je: " + rezultat,Toast.LENGTH_LONG).show();
        Log.i("IVANTAG", rezultat);
    }

Answer 1

您的响应是 JSONArray

JSONArray array_data = new JSONArray(value);
JSONObject object_data = array_data.getJSONObject(0); // 0, 1, 2 index of array

您可以从对象中获取任何数据

object_data.getString("temperatura");