Android AsyncTask 无法从 php-mysql 检索 JSON 字符串

Question

我可以显示存储在“值”中的数组中的数据，但是当我尝试从数据库中获取数据时，我得到了

"E/Surface: getSlotFromBufferLocked: 未知缓冲区"

错误。我有 php 脚本和 mysql 数据库，其中有一个“问题”列。我不确定在连接建立或 json 解析或其他任何地方是否有问题。

我的 VersionsFragment.java

public class VersionsFragment extends ListFragment {
public VersionsFragment() {
}


@Override
public void onActivityCreated(Bundle savedInstanceState) {
    super.onActivityCreated(savedInstanceState);
    //String[] versionName = getResources().getStringArray(R.array.version_names); //this works fine
    new MyTask().execute();
}

class MyTask extends AsyncTask<String, Integer, String[]> {

    @Override
    protected String[] doInBackground(String... params) {
        String[] versionName = new String[0];
        ArrayList<String> scripts = new ArrayList<String>();
        try {
            URL url = new URL("http://192.168.1.105/android/phpcall.php");
            HttpURLConnection conn = (HttpURLConnection) url.openConnection();
            String strJson = "";
            String line;
            BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            while ((line = br.readLine()) != null) {
                strJson += line;
            }
            //Log.d(VersionsFragment.class.getSimpleName(), line);
            JSONArray jArray = new JSONArray(strJson);

            for (int i = 0; i < jArray.length(); i++) {

                JSONObject jObject = jArray.getJSONObject(i);

                String name = jObject.getString("question");

                scripts.add(name);
            }
            versionName = scripts.toArray(new String[scripts.size()]);

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

    protected void onPostExecute(String[] result){
        //for(int i=0;i<result.length;i++)
            //Log.d(VersionsFragment.class.getSimpleName(), result[i]);
        ArrayAdapter<String> adapter = new ArrayAdapter<String>(getActivity(), android.R.layout.simple_list_item_1, result);
        setListAdapter(adapter);
    }
}

@Override
public void onListItemClick(ListView l, View v, int position, long id) {
    OnVersionNameSelectionChangeListener listener = (OnVersionNameSelectionChangeListener) getActivity();
     listener.OnSelectionChanged(position);
   }
}

Answer 1

将上面的代码编辑如下：

JSONObject jsonObj = new JSONObject(strJson);

// Getting JSON Array node
JSONArray jArray = jsonObj.getJSONArray(YOUR_JSON_ARRAY_NAME);