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);

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);

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");

Android Query可以为你处理这个问题，还有更多(比如缓存和加载进度)。

看看这里。

我认为这是最好的方法。

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;
}
}

嗨，我有最简单的代码试试这个

    public class ImageFromUrlExample extends Activity {
    @Override
    public void onCreate(Bundle savedInstanceState) {
            super.onCreate(savedInstanceState);
            setContentView(R.layout.main);  
            ImageView imgView =(ImageView)findViewById(R.id.ImageView01);
            Drawable drawable = LoadImageFromWebOperations("http://www.androidpeople.com/wp-content/uploads/2010/03/android.png");
            imgView.setImageDrawable(drawable);

    }

    private Drawable LoadImageFromWebOperations(String url)
    {
          try{
        InputStream is = (InputStream) new URL(url).getContent();
        Drawable d = Drawable.createFromStream(is, "src name");
        return d;
      }catch (Exception e) {
        System.out.println("Exc="+e);
        return null;
      }
    }
   }

main。xml

  <LinearLayout 
    android:id="@+id/LinearLayout01"
    android:layout_width="fill_parent"
    android:layout_height="fill_parent"
    xmlns:android="http://schemas.android.com/apk/res/android">
   <ImageView 
       android:id="@+id/ImageView01"
       android:layout_height="wrap_content" 
       android:layout_width="wrap_content"/>

试试这个