Question

嘿，我是初学者，使用json和volley解析mysql数据库中的数据，问题是数据只想显示在片段上，而在列表消失后仅显示几秒钟，就像这张图片Trouble Problem并且列表数据不会在片段中显示，我希望它在片段上显示

这是我用于片段显示的代码

fragment.java

sequelize.query()

那么如何使数据可以在片段上显示呢？因此从数据库数据存储到数据库中的数据将显示在碎片上

fragment.xml

public class KomentarFragment extends Fragment {
TextView nama,komen;
View v;
ArrayList<Komen> kmn = new ArrayList<>();
public static String url ="http://surveyclickon.000webhostapp.com/android_register_login/komen.php";

@Nullable
@Override
public View onCreateView(@NonNull LayoutInflater inflater, @Nullable ViewGroup container, @Nullable Bundle savedInstanceState) {
    View kom =  inflater.inflate(R.layout.fragment_komentar,container,false);
    nama = kom.findViewById(R.id.nm2);
    komen = kom.findViewById(R.id.kmn2);

    new AsyncFetch().execute();

    return kom;
}

@Override
public void onCreate(@Nullable Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
}

    private class AsyncFetch extends AsyncTask<String, String, String> {

    ProgressDialog pDialog = new ProgressDialog(getActivity());
    HttpURLConnection conn;
    URL url = null;

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

        pDialog.setMessage("Loading list ...");
        pDialog.setCancelable(false);
        pDialog.show();
    }

    @Override
    protected String doInBackground(String... strings) {
        try {

            // Enter URL address where your json file resides
            // Even you can make call to php file which returns json data
            url = new URL("http://surveyclickon.000webhostapp.com/android_register_login/back.php");

        } catch (MalformedURLException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
            return e.toString();
        }
        try {

            // Setup HttpURLConnection class to send and receive data from php and mysql
            conn = (HttpURLConnection) url.openConnection();
            conn.setRequestMethod("POST");

            // setDoOutput to true as we recieve data from json file
            conn.setDoOutput(true);

        } catch (IOException e1) {
            // TODO Auto-generated catch block
            e1.printStackTrace();
            return e1.toString();
        }
        try {

            int response_code = conn.getResponseCode();

            // Check if successful connection made
            if (response_code == HttpURLConnection.HTTP_OK) {

                // Read data sent from server
                InputStream input = conn.getInputStream();
                BufferedReader reader = new BufferedReader(new InputStreamReader(input));
                StringBuilder result = new StringBuilder();
                String line;

                while ((line = reader.readLine()) != null) {
                    result.append(line);
                }

                // Pass data to onPostExecute method
                return (result.toString());

            } else {

                return ("unsuccessful");
            }

        } catch (IOException e) {
            e.printStackTrace();
            return e.toString();
        } finally {
            conn.disconnect();
        }
    }

    @Override
    protected void onPostExecute(String result) {
        pDialog.dismiss();

        try{
            JSONObject object = new JSONObject(result);
            int code = object.getInt("refund");
            if(code == 0) {
                JSONArray jsonArray = object.getJSONArray("refund");
                ArrayList<Refund> ref = new ArrayList<>();
                for (int i = 0; i < jsonArray.length(); i++) {
                    JSONObject jsonObject = jsonArray.getJSONObject(i);
                    Refund refund = new Refund();
                    refund.setNm(jsonObject.getString("nama"));
                    refund.setNo(jsonObject.getString("no"));
                    refund.setAlam(jsonObject.getString("alamat"));
                    refund.setKode(jsonObject.getString("kode"));
                    refund.setAlas(jsonObject.getString("alasan"));
                    ref.add(refund);
                }
            }
        }catch (JSONException e) {

            e.printStackTrace();
            Toast.makeText(getActivity(), "Json error: " + e.getMessage(), Toast.LENGTH_LONG).show();
        }

    }
}

Answer 1

您基本上没有更新UI，所以一无所获

public class KomentarFragment extends Fragment {
    List<Refund> refundList = new ArrayList<>();
    ....
}

然后在您的postExecute方法中

for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject jsonObject = jsonArray.getJSONObject(i);
            Refund refund = new Refund();
            refund.setNm(jsonObject.getString("nama"));
            refund.setNo(jsonObject.getString("no"));
            refund.setAlam(jsonObject.getString("alamat"));
            refund.setKode(jsonObject.getString("kode"));
            refund.setAlas(jsonObject.getString("alasan"));
            refundList.add(refund);
        }
//Update the UI
//Store to the database

然后，您可以根据需要使用recyclerview或数据库更新UI，也可以在postExecute方法中进行

数据不会显示在片段上

1 个答案: