如何将JSON解析为脱机使用

Question

我已经成功地解析了JSON，但是现在我想缓存它以供脱机使用，甚至internet都不可用，如果有任何新条目出现，我也想缓存它。

缓存数据的最佳选择是什么？SharedPreferences或SQLite database

下面是我的代码，用于分析JSON：

public class MainActivity extends Activity {

    ArrayList<Actors> actorsList;   
    ActorAdapter adapter;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        actorsList = new ArrayList<Actors>();
        new JSONAsyncTask().execute("http://microblogging.wingnity.com/JSONParsingTutorial/jsonActors");

        ListView listview = (ListView)findViewById(R.id.list);
        adapter = new ActorAdapter(getApplicationContext(), R.layout.row, actorsList);

        listview.setAdapter(adapter);

        listview.setOnItemClickListener(new OnItemClickListener() {

            @Override
            public void onItemClick(AdapterView<?> arg0, View arg1, int position,
                    long id) {
                // TODO Auto-generated method stub
                Toast.makeText(getApplicationContext(), actorsList.get(position).getName(), Toast.LENGTH_LONG).show();              
            }
        });
    }


    class JSONAsyncTask extends AsyncTask<String, Void, Boolean> {

        ProgressDialog dialog;

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            dialog = new ProgressDialog(MainActivity.this);
            dialog.setMessage("Loading, please wait");
            dialog.setTitle("Connecting server");
            dialog.show();
            dialog.setCancelable(false);
        }

        @Override
        protected Boolean doInBackground(String... urls) {
            try {

                //------------------>>
                HttpGet httppost = new HttpGet(urls[0]);
                HttpClient httpclient = new DefaultHttpClient();
                HttpResponse response = httpclient.execute(httppost);

                // StatusLine stat = response.getStatusLine();
                int status = response.getStatusLine().getStatusCode();

                if (status == 200) {
                    HttpEntity entity = response.getEntity();
                    String data = EntityUtils.toString(entity);


                    JSONObject jsono = new JSONObject(data);
                    JSONArray jarray = jsono.getJSONArray("actors");

                    for (int i = 0; i < jarray.length(); i++) {
                        JSONObject object = jarray.getJSONObject(i);

                        Actors actor = new Actors();

                        actor.setName(object.getString("name"));
                        actor.setDescription(object.getString("description"));

                        actorsList.add(actor);
                    }
                    return true;
                }

                //------------------>>

            } catch (ParseException e1) {
                e1.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {
                e.printStackTrace();
            }
            return false;
        }

        protected void onPostExecute(Boolean result) {
            dialog.cancel();
            adapter.notifyDataSetChanged();
            if(result == false)
                Toast.makeText(getApplicationContext(), "Unable to fetch data from server", Toast.LENGTH_LONG).show();

        }
    }

}

Answer 1

为什么不使用这样的方法将其保存到应用程序的缓存文件夹中：

String path = Environment.getExternalStorageDirectory() + File.separator + "cache" + File.separator;
File dir = new File(path);
if (!dir.exists()) {
    dir.mkdirs();
}
path += "data";
File data = new File(path);
if (!data.createNewFile()) {
    data.delete();
    data.createNewFile();
}
ObjectOutputStream objectOutputStream = new ObjectOutputStream(new FileOutputStream(data));
objectOutputStream.writeObject(actorsList);
objectOutputStream.close();

在此之后，您可以在任何时候使用：

List<?> list = null;
File data = new File(path);
try {
    if(data.exists()) {
        ObjectInputStream objectInputStream = new ObjectInputStream(new FileInputStream(data));
        list = (List<Object>) objectInputStream.readObject();
        objectInputStream.close();
    }
} catch (IOException e) {
    e.printStackTrace();
} catch (ClassNotFoundException e) {
    e.printStackTrace();
}

更新：好，使类名为ObjectToFileUtil，将此代码粘贴到创建的类中

package <yourpackagehere>;

import android.os.Environment;

import java.io.*;

public class ObjectToFileUtil {

    public static String objectToFile(Object object) throws IOException {
        String path = Environment.getExternalStorageDirectory() + File.separator + "cache" + File.separator;
        File dir = new File(path);
        if (!dir.exists()) {
            dir.mkdirs();
        }
        path += "data";
        File data = new File(path);
        if (!data.createNewFile()) {
            data.delete();
            data.createNewFile();
        }
        ObjectOutputStream objectOutputStream = new ObjectOutputStream(new FileOutputStream(data));
        objectOutputStream.writeObject(object);
        objectOutputStream.close();
        return path;
    }

    public static Object objectFromFile(String path) throws IOException, ClassNotFoundException {
        Object object = null;
        File data = new File(path);
        if(data.exists()) {
            ObjectInputStream objectInputStream = new ObjectInputStream(new FileInputStream(data));
            object = objectInputStream.readObject();
            objectInputStream.close();
        }
        return object;
    }
}

将< yourpackagehere >更改为您的包名，并且不要忘记向WRITE_EXTERNAL_STORAGE添加AndroidManifest.xml.权限在您的MainActivity添加字段中

private String dataPath;

并将onPostExecute类的JSONAsyncTask方法替换为

protected void onPostExecute(Boolean result) {
    dialog.cancel();
    adapter.notifyDataSetChanged();
    if(result) {
        try {
            dataPath = objectToFile(arrayList);
        } catch (IOException e) {
            e.printStackTrace();
        }
    } else {
        Toast.makeText(getApplicationContext(), "Unable to fetch data from server", Toast.LENGTH_LONG).show();
    }
}

现在，您可以在任何时候从文件中访问actorsList，方法是

try {
    actorsList = (ArrayList<Actors>)objectFromFile(dataPath);
} catch (IOException e) {
    e.printStackTrace();
} catch (ClassNotFoundException e) {
    e.printStackTrace();
}

如果要在关闭应用程序后保存文件路径，则必须保存dataPath字符串(并在应用程序启动时加载)，例如，使用SharedPreferences。

Answer 2

And what would be the best option to cache data ? SharedPreferences or SQLite database

这完全是基于你收到的数据。

1. 如果数据是小的，非结构化数据，那么使用共享Pref。
2. 如果数据很大，则使用SQLite。

但是为了更好地存储完整的数据，您可以使用档案概念。将字符串数据存储在代码String data = EntityUtils.toString(entity);中，您必须将来自服务器的数据中的任何更改保存到file.If中，如果没有internet，则将其添加到file.And中检索数据。从上面的链接获取文件操作的示例代码。