我已经使用 json 创建了一个 android 应用程序，用于从 ror 网站获取并显示在列表视图中，现在我想从我们的应用程序中添加数据，它也必须显示在我们应用程序的列表视图中，然后它也必须显示在网站中。如何在我们的应用程序中使用 post 方法和显示。
得到我这样使用的方法
public class MainActivity extends ListActivity implements FetchDataListener
{
private ProgressDialog dialog;
@Override
protected void onCreate(Bundle savedInstanceState)
{
super.onCreate(savedInstanceState);
//setContentView(R.layout.activity_list_item);
initView();
}
private void initView()
{
// show progress dialog
dialog = ProgressDialog.show(this, "", "Loading...");
String url = "http://floating-wildwood-1154.herokuapp.com/posts.json";
FetchDataTask task = new FetchDataTask(this);
task.execute(url);
}
@Override
public void onFetchComplete(List<Application> data)
{
// dismiss the progress dialog
if ( dialog != null )
dialog.dismiss();
// create new adapter
ApplicationAdapter adapter = new ApplicationAdapter(this, data);
// set the adapter to list
setListAdapter(adapter);
}
@Override
public void onFetchFailure(String msg)
{
// dismiss the progress dialog
if ( dialog != null )
dialog.dismiss();
// show failure message
Toast.makeText(this, msg, Toast.LENGTH_LONG).show();
}
}
fetchdatatask.java
public class FetchDataTask extends AsyncTask<String, Void, String>
{
private final FetchDataListener listener;
private String msg;
public FetchDataTask(FetchDataListener listener)
{
this.listener = listener;
}
@Override
protected String doInBackground(String... params)
{
if ( params == null )
return null;
// get url from params
String url = params[0];
try
{
// create http connection
HttpClient client = new DefaultHttpClient();
HttpGet httpget = new HttpGet(url);
// connect
HttpResponse response = client.execute(httpget);
// get response
HttpEntity entity = response.getEntity();
if ( entity == null )
{
msg = "No response from server";
return null;
}
// get response content and convert it to json string
InputStream is = entity.getContent();
return streamToString(is);
}
catch ( IOException e )
{
msg = "No Network Connection";
}
return null;
}
@Override
protected void onPostExecute(String sJson)
{
if ( sJson == null )
{
if ( listener != null )
listener.onFetchFailure(msg);
return;
}
try
{
// convert json string to json object
JSONObject jsonObject = new JSONObject(sJson);
JSONArray aJson = jsonObject.getJSONArray("post");
// create apps list
List<Application> apps = new ArrayList<Application>();
for ( int i = 0; i < aJson.length(); i++ )
{
JSONObject json = aJson.getJSONObject(i);
Application app = new Application();
app.setContent(json.getString("content"));
// add the app to apps list
apps.add(app);
}
//notify the activity that fetch data has been complete
if ( listener != null )
listener.onFetchComplete(apps);
}
catch ( JSONException e )
{
e.printStackTrace();
msg = "Invalid response";
if ( listener != null )
listener.onFetchFailure(msg);
return;
}
}
/**
* This function will convert response stream into json string
*
* @param is
* respons string
* @return json string
* @throws IOException
*/
public String streamToString(final InputStream is) throws IOException
{
BufferedReader reader = new BufferedReader(new InputStreamReader(is));
StringBuilder sb = new StringBuilder();
String line = null;
try
{
while ( (line = reader.readLine()) != null )
{
sb.append(line + "\n");
}
}
catch ( IOException e )
{
throw e;
}
finally
{
try
{
is.close();
}
catch ( IOException e )
{
throw e;
}
}
return sb.toString();
}
}
像这样我正在使用 get 方法并显示，出于同样的目的，我想添加 post 方法以在 android listview 中显示并在网站中显示。
如果我单击菜单按钮（如添加）将创建一个按钮，在该按钮中它将显示一页，在该页面中我必须添加数据并单击保存，它必须显示在列表视图中并在网站上发布
我怎么能做到这一点。