使用Intent从其他网址打开网址?

时间:2013-08-29 16:05:24

标签: android url android-intent

我想使用intent从图像的另一个url打开图像的url。我是android的新手,所以我的这段代码不起作用。

 url_1 = new URL("http://garooh.905pm.com"+com.org.constant.Helper.Gadd_list.get(0).thisEvent.getImage_four_thumb());

 url_2 = new URL("http://garooh.905pm.com"+ com.org.constant.Helper.Gadd_list.get(0).thisEvent.getImage_four_thumb());
    Intent intent = new Intent();
    intent.setAction(Intent.ACTION_VIEW);
    intent.setDataAndType(url_1, url_2);
    startActivity(intent);

1 个答案:

答案 0 :(得分:0)

我认为您有一个图像按钮是正确的,当您点击它时,您想要转到其中一个网址吗?

尝试将此添加到您的imageButton。

imageButton.setOnClickListener(new locatorButtonClickListener());

private class imageButtonListener implements OnClickListener 
{
    @Override
    public void onClick(View button) {
        new DisplayImageFromUrl((ImageButton) findViewById(R.id.imageButtonEnd), this).execute(//Enter your link here);
}

<强>更新 尝试使用此异步任务

    //Display image to bitmap using URL
public class DisplayImageFromUrl extends AsyncTask<String, Void, Bitmap> {
    ImageView bmImage;
    Context context1;
    ProgressDialog pd;
    @Override
    protected void onPreExecute() {
        super.onPreExecute();
        pd = new ProgressDialog(context1);
        pd.setMessage("Loading Images...");
        pd.show();
    }
    public DisplayImageFromUrl(ImageView bmImage, Context context) {
        this.bmImage = bmImage;
        this.context1 = context;
    }
    protected Bitmap doInBackground(String... urls) {
        String urldisplay = urls[0];
        Bitmap mIcon11 = null;
        try {
            InputStream in = new java.net.URL(urldisplay).openStream();
            mIcon11 = BitmapFactory.decodeStream(in);
        } catch (Exception e) {
            Log.e("Error", e.getMessage());
            e.printStackTrace();
        }
        return mIcon11;
    }
    protected void onPostExecute(Bitmap result) {
        bmImage.setImageBitmap(result);
        pd.dismiss();
    }
}

你在活动中就这样称呼它。

        new DisplayImageFromUrl((ImageButton) findViewById(R.id.imageButtonEnd), this).execute(//Enter your link here);