我已成功解析JSON但现在我想将其缓存以供离线使用,即使互联网不可用,如果有任何新条目,我也想要缓存它。
什么是缓存数据的最佳选择? SharedPreferences
或SQLite database
这是我的代码,我用它来解析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();
}
}
}
答案 0 :(得分:8)
为什么不使用以下内容将其保存到应用的缓存文件夹中:
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();
之后,您可以随时使用以下内容阅读:
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();
}
UPDATE:好的,将类命名为ObjectToFileUtil,将此代码粘贴到创建的类</ p>
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;
}
}
更改&lt; yourpackagehere&gt;到您的软件包名称,不要忘记将 WRITE_EXTERNAL_STORAGE 权限添加到 AndroidManifest.xml 。在MainActivity中添加字段
private String dataPath;
并将JSONAsyncTask类的onPostExecute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,您可以随时使用
从文件中访问获取actorsListtry {
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。
答案 1 :(得分:0)
And what would be the best option to cache data ? SharedPreferences or SQLite database
这纯粹基于您收到的数据。
但是为了更好地存储完整数据,您可以使用file concept。将字符串数据存储在您必须保存到文件的代码String data = EntityUtils.toString(entity);
data
中。如果服务器中的数据发生任何变化,请将其添加到文件中。如果互联网不存在,则检索数据。从上面的链接获取文件操作的示例代码。