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我有一个 URL,我想转换为位图,但我没有得到 o/p 。我应用了一些方法来对这件事进行排序,但仍然没有得到任何东西。

我在 Stack Overflow 上的上一个答案中搜索了解决方案,但我也被它卡住了。我在做什么 :

URL url = new URl("http://www.XYZ.com/image.jpg"); 

并将这个东西放入

BitmapFactory.decodeStream(_ _ _); 

我应该怎么做才能解决它?

4

3 回答 3

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private Drawable loadImageFromWeb(String url) {
        try {   
            InputStream is = (InputStream) new URL(url).getContent();
            Drawable d = Drawable.createFromStream(is, "src name");
            return d;
        } catch (Exception e) {
            return null;
        }
    }   
于 2013-06-08T06:47:38.560 回答
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一个漫长但实用的解决方案

    try 
    {    
        InputStream inn = OpenConnection(URL);
        Bitmap bmp = BitmapFactory.decodeStream(inn);
        inn.close();    
    } 
    catch (IOException e1) {    
        e1.printStackTrace();
    }

    public InputStream OpenHttpConnection(String url) {
       InputStream in = null;
       int response = -1;

       URL url = new URL(url);
       URLConnection conn = url.openConnection();

       try {
           HttpURLConnection httpConn = (HttpURLConnection) conn;
           httpConn.setAllowUserInteraction(false);
           httpConn.setInstanceFollowRedirects(true);
           httpConn.setRequestMethod("GET");
           httpConn.connect();
           response = httpConn.getResponseCode();

           if (response == HttpURLConnection.HTTP_OK) {
            in = httpConn.getInputStream();
           }
       } 
       catch (Exception ex) 
       {
        ex.printStackTrace();
       }
    return in;
  }
于 2013-07-26T12:25:07.497 回答
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这个怎么样?当我尝试时它有效。

URL url = new URL("http://www.XYZ.com/image.jpg");
Object content = url.getContent();
InputStream is = (InputStream) content;
//drawable
Drawable drawable = Drawable.createFromStream(is, "src");
//bitmap
Bitmap bitmap = BitmapFactory.decodeStream(is);
于 2013-06-09T07:44:39.240 回答