Question

我有一个用于rss阅读器的Android应用程序。
它读取URL并在listview中显示信息。现在我尝试点击项目时显示webview中的链接但是没有用。

这是代码示例。

感谢您的帮助。

public class RssReaderActivity extends Activity implements AsyncTaskCompletionListener{
        private ListView listview;


        @Override
        protected void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.rss_reader);
            listview=(ListView)findViewById(R.id.rssfeed_listview);
            if(Utility.determineConnectivity(this))
            new RssFeedAsyncTask(this).execute(Utility.url);
            else
                Toast.makeText(this, "No Internet Connection", Toast.LENGTH_SHORT).show();
            listview.setVerticalFadingEdgeEnabled(true);
            listview.setOnItemClickListener(onItemClickListener);
    }

    private OnItemClickListener onItemClickListener = new OnItemClickListener() {

            @Override
            public void onItemClick(AdapterView<?> arg0, View arg1, int arg2,
    long arg3) {
    RssFeedStructure data = (RssFeedStructure) listview.getItemAtPosition(arg2);
    postInfo.putString("content", data.getUrl().toString());    
                Toast.makeText(getApplicationContext(), data.getUrl().toString(), Toast.LENGTH_SHORT).show();
                Intent postviewIntent = new Intent(getApplicationContext(), DisPlayWebPageActivity.class);
                postviewIntent.putExtras(postInfo);
                startActivity(postviewIntent);
            }
        };
    @Override
        public void onTaskComplete(List<RssFeedStructure> result) {
            RssReaderListAdapter _adapter= new RssReaderListAdapter(RssReaderActivity.this,
                    result);
            listview.setAdapter(_adapter);
        }

Answer 1

这可能有用

在项目点击监听器上，您可以直接传递您的网址。

Intent i = new Intent(Intent.ACTION_VIEW, 
       Uri.parse(data.getUrl().toString()));
startActivity(i);

Android程序rss Reader

1 个答案: