这是一个迟到的回复，正如上面所建议的AsyncTask将会和谷歌一点后，我发现了一个解决这个问题的方法。

Drawable Drawable = Drawable. createfromstream ((InputStream) new URL(" URL ").getContent()， "src");

imageView.setImageDrawable(可拉的);

这是完整的功能:

public void loadMapPreview () {
    //start a background thread for networking
    new Thread(new Runnable() {
        public void run(){
            try {
                //download the drawable
                final Drawable drawable = Drawable.createFromStream((InputStream) new URL("url").getContent(), "src");
                //edit the view in the UI thread
                imageView.post(new Runnable() {
                    public void run() {
                        imageView.setImageDrawable(drawable);
                    }
                });
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
    }).start();
}

不要忘记在你的AndroidManifest.xml中添加以下权限来访问互联网。

< uses-permission android: name = " android.permission。互联网" / >

我自己也试过，还没有遇到任何问题。

public class LoadWebImg extends Activity {

String image_URL=
 "http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png";

   /** Called when the activity is first created. */
   @Override
   public void onCreate(Bundle savedInstanceState) {
       super.onCreate(savedInstanceState);
       setContentView(R.layout.main);

       ImageView bmImage = (ImageView)findViewById(R.id.image);
    BitmapFactory.Options bmOptions;
    bmOptions = new BitmapFactory.Options();
    bmOptions.inSampleSize = 1;
    Bitmap bm = LoadImage(image_URL, bmOptions);
    bmImage.setImageBitmap(bm);
   }

   private Bitmap LoadImage(String URL, BitmapFactory.Options options)
   {       
    Bitmap bitmap = null;
    InputStream in = null;       
       try {
           in = OpenHttpConnection(URL);
           bitmap = BitmapFactory.decodeStream(in, null, options);
           in.close();
       } catch (IOException e1) {
       }
       return bitmap;               
   }

private InputStream OpenHttpConnection(String strURL) throws IOException{
 InputStream inputStream = null;
 URL url = new URL(strURL);
 URLConnection conn = url.openConnection();

 try{
  HttpURLConnection httpConn = (HttpURLConnection)conn;
  httpConn.setRequestMethod("GET");
  httpConn.connect();

  if (httpConn.getResponseCode() == HttpURLConnection.HTTP_OK) {
   inputStream = httpConn.getInputStream();
  }
 }
 catch (Exception ex)
 {
 }
 return inputStream;
}
}

要做到这一点，一个简单而干净的方法是使用开源库Prime。

不管怎样，人们问我的评论，把它作为答案。我正在发帖。

URL newurl = new URL(photo_url_str); 
mIcon_val = BitmapFactory.decodeStream(newurl.openConnection().getInputStream());
profile_photo.setImageBitmap(mIcon_val);

    String img_url= //url of the image
    URL url=new URL(img_url);
    Bitmap bmp; 
    bmp=BitmapFactory.decodeStream(url.openConnection().getInputStream());
    ImageView iv=(ImageView)findviewById(R.id.imageview);
    iv.setImageBitmap(bmp);

我最近在这里找到了一个线程，因为我必须为带有图像的列表视图做类似的事情，但原理很简单，正如您可以在那里显示的第一个示例类中读到的那样(由jleedev)。 你得到图像的输入流(从网络)

private InputStream fetch(String urlString) throws MalformedURLException, IOException {
    DefaultHttpClient httpClient = new DefaultHttpClient();
    HttpGet request = new HttpGet(urlString);
    HttpResponse response = httpClient.execute(request);
    return response.getEntity().getContent();
}

然后你将图像存储为可绘制的，你可以将它传递给ImageView(通过setImageDrawable)。同样从上面的代码片段来看一下整个线程。

InputStream is = fetch(urlString);
Drawable drawable = Drawable.createFromStream(is, "src");