使用意圖從另一個URL打開一個網址？

Question

我想打開一個圖像的URL從圖像的另一個URL使用intent.I新來android所以這個代碼我的工作。

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); 

Answer 1

我是否認爲你有一個圖像按鈕，當你點擊它你想要去那些URL之一嗎？

嘗試添加到您的圖像按鈕。

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); 
} 

UPDATE 嘗試使用此異步任務

//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(); 
    } 
} 

，並在你的活動這樣稱呼它。

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