Question

我通过PHP获取数据并设置为listview .. 首先我打开应用程序，列表视图将显示出来，但在调用完并重新打开应用程序后，列表视图将不会出现，需要调用刷新才能重新显示..

FeedAdapter：

 public View getView(int position, View convertView, ViewGroup parent){


    View row = inflater.inflate(R.layout.row_feed, parent, false);
    TextView username = (TextView) row.findViewById(R.id.txtUsername);

    MovieFeed moviefeed = localmovies.get(position);

    username.setText(moviefeed.getUsername());


    return row;
}

带有listview的Tab1片段

 public View onCreateView(LayoutInflater inflater, @Nullable ViewGroup container, Bundle savedInstanceState) {
    View v = inflater.inflate(R.layout.tab_1, container, false);

    tweetListView = (ListView) v.findViewById(R.id.tweetList);


        urlString = "http://idonno.net/G.php"; //json url
        new ProcessJSON().execute(urlString);
        tweetItemArrayAdapter = new FeedAdapter(getActivity(), localmovies);
     tweetListView.setAdapter(tweetItemArrayAdapter);

return v;


}

Answer 1

您应该实现asyncTask并确保在实例化listview之前确实拥有数据。就像在这个带有匿名内部类的例子中一样。

 @Override
 protected void onPostExecute(String result){
    progressDialog.dismiss();
    runOnUiThread(new Runnable(){
        public void run(){
            final ListView lv1 = (ListView) findViewById(R.id.listings);
            lv1.setAdapter(new CustomListAdapter(YourActivity.this, shares));
        }
    });

}

如果您随时更新列表，请拨打适配器上的notifyDataSetChanged()。

您可以将asyncTask与此问题How to get the result of OnPostExecute() to main activity because AsyncTask is a separate class?

的已接受答案分开

ListView在调用finish（）后不会出现，并重新打开

1 个答案: