2

我有一个显示逐项叠加的地图视图,每个 onTap 事件都显示一个对话框,其中包含有关旅行社和三个按钮的信息,其中一个应该在浏览器中打开旅行社网站,但是当我单击按钮时,我得到了: 没有找到处理 Intent 的活动。

这是我的代码:

protected boolean onTap(int i){
float[] results = new float[1];
        Location.distanceBetween(decdegrees(appState.getMyLocation().getLatitudeE6()), decdegrees(appState.getMyLocation().getLongitudeE6()), decdegrees(points.get(i).getPoint().getLatitudeE6()), decdegrees(points.get(i).getPoint().getLongitudeE6()), results);
        float distance = results[0]/1000;
        DecimalFormat maxDigitsFormatter = new DecimalFormat("#.#");
        String infos=points.get(i).getSnippet() + "#" + String.valueOf(maxDigitsFormatter.format(distance));
        final String [] AInfos = infos.split("#");

        final Dialog ADialog = new Dialog(c);
        ADialog.setContentView(R.layout.travelagency);

        TextView txtAgence = (TextView)ADialog.findViewById(R.id.txtAgence);
        TextView txtAddress = (TextView)ADialog.findViewById(R.id.txtAddress);
        TextView txtDistance = (TextView)ADialog.findViewById(R.id.txtDistance);
        TextView txtFax = (TextView)ADialog.findViewById(R.id.txtFax);
        Button btnCall = (Button)ADialog.findViewById(R.id.btnCall);
        Button btnWebSite = (Button)ADialog.findViewById(R.id.btnWebSite);
        Button btnCancel = (Button)ADialog.findViewById(R.id.btnCancel);

        ADialog.setTitle(AInfos[0]);
        btnCall.setText("Appeler : " + AInfos[2]);
        txtAgence.setText(AInfos[1]);
        txtDistance.setText("Approximativement à : " +AInfos[6] + " Km");
        txtAddress.setText("Adresse : " + AInfos[3]);
        txtFax.setText("Fax : " + AInfos[4]);
        ADialog.show();

        btnCancel.setOnClickListener(new OnClickListener() {
            public void onClick(View v) {
                ADialog.dismiss();
            }
        });

        btnWebSite.setOnClickListener(new OnClickListener() {
            public void onClick(View v) {
                Intent myIntent = new Intent(Intent.ACTION_VIEW,Uri.parse(AInfos[5]));
                v.getContext().startActivity(myIntent);
            }
        });

        return (true);
    }

我在这里这里找到了例子,但突然对我不起作用..

谢谢

4

2 回答 2

3

这将创建在默认浏览器中打开网页的正确意图:

Uri url = Uri.parse("http://www.someUrl.com");
        Intent intent = new Intent(Intent.ACTION_VIEW, url);
        startActivity(intent);
于 2011-04-22T10:47:56.590 回答
0

很可能您AInfos[5]的网址不正确。硬编码一个 url http://www.google.com,看看它是否首先工作。还打印AInfos[5]包含的内容。

于 2011-04-22T10:53:15.597 回答