Android:将JSON数据保存到SharedPreferences
问题描述:
我有这个ListView从Web上的JSON获取其数据(图像+文本)。 现在我有一个任务,使ListView无需互联网连接即可访问。我的想法是,当应用程序第一次运行互联网时,通过从网络上保存JSON数据,当它无法找到互联网连接时,它将从持久存储中获取数据。Android:将JSON数据保存到SharedPreferences
有人可以帮我吗?我还是一个初学者无法找到与JSON SharedPreferences的例子。 非常感谢
public class ProjectsList extends Activity {
/** Called when the activity is first created. */
//ListView that will hold our items references back to main.xml
ListView lstTest;
//Array Adapter that will hold our ArrayList and display the items on the ListView
ProjectAdapter arrayAdapter;
//List that will host our items and allow us to modify that array adapter
ArrayList<Project> prjcts=null;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.projects_list);
//Initialize ListView
lstTest= (ListView)findViewById(R.id.lstText);
//Initialize our ArrayList
prjcts = new ArrayList<Project>();
//Initialize our array adapter notice how it references the listitems.xml layout
arrayAdapter = new ProjectAdapter(ProjectsList.this, R.layout.listitems,prjcts,ProjectsList.this);
//Set the above adapter as the adapter of choice for our list
//lstTest.setAdapter(arrayAdapter);
lstTest.setAdapter(arrayAdapter);
if (isOnline())
{
//Instantiate the Web Service Class with he URL of the web service not that you must pass
WebService webService = new WebService("http://liebenwald.spendino.net/admanager/dev/android/projects.json");
//Pass the parameters if needed , if not then pass dummy one as follows
Map<String, String> params = new HashMap<String, String>();
params.put("var", "");
//Get JSON response from server the "" are where the method name would normally go if needed example
// webService.webGet("getMoreAllerts", params);
String response = webService.webGet("", params);
try
{
//Parse Response into our object
Type collectionType = new TypeToken<ArrayList<Project>>(){}.getType();
//JSON expects an list so can't use our ArrayList from the lstart
List<Project> lst= new Gson().fromJson(response, collectionType);
//Now that we have that list lets add it to the ArrayList which will hold our items.
for(Project l : lst)
{
prjcts.add(l);
ConstantData.projectsList.add(l);
}
//Since we've modified the arrayList we now need to notify the adapter that
//its data has changed so that it updates the UI
arrayAdapter.notifyDataSetChanged();
}
catch(Exception e)
{
Log.d("Error: ", e.getMessage());
}
}
lstTest.setOnItemClickListener(new OnItemClickListener() {
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
Intent care = new Intent(ProjectsList.this, ProjectDetail.class);
care.putExtra("spendino.de.ProjectDetail.position",position);
startActivity(care);
}
});
}
@Override
public void onDestroy()
{
yAdapter.imageLoader.stopThread();
lstTest.setAdapter(null);
super.onDestroy();
}
protected boolean isOnline() {
ConnectivityManager cm = (ConnectivityManager) getSystemService(Context.CONNECTIVITY_SERVICE);
NetworkInfo netInfo = cm.getActiveNetworkInfo();
if (netInfo != null && netInfo.isConnected()) {
return true;
} else {
AlertDialog.Builder alertbox = new AlertDialog.Builder(this);
alertbox.setTitle("spendino Helfomat");
alertbox.setMessage ("Please check your internet connection");
alertbox.setPositiveButton("OK", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
//Main.this.finish();
}
});
alertbox.show();
return false;
}
}
}
答
SharedPreferences有保存JSON对象是,你必须设法将其转换为字符串没有方法。然后,当你得到它时,你必须把这个字符串解析回JSON。祝你好运!
JSON为String:
JSONObject o = new JSONObject(data.trim());
String name = o.getString(Constants.NAME);
long date = o.getLong(Constants.DATE);
String mes = o.getString(Constants.MESSAGE);
StringBuilder buf = new StringBuilder(text.getText());
buf.append(name).append(" (").append(dfTime.format(new Date(date))).append(")\n").append(mes).append("\n");
text.setText(buf.toString());
从字符串制作一个JSON是不是一个艰巨的任务,使用的StringTokenizer。祝你好运!
+1打我吧:) – MByD 2011-05-12 12:44:03
你能告诉我一个我上面的代码的例子吗? thx – hectichavana 2011-05-12 14:57:20
我会将它添加到我上面的答案中。 – Egor 2011-05-12 15:29:41