来自Android开发者:

// show The Image in a ImageView
new DownloadImageTask((ImageView) findViewById(R.id.imageView1))
            .execute("http://java.sogeti.nl/JavaBlog/wp-content/uploads/2009/04/android_icon_256.png");

public void onClick(View v) {
    startActivity(new Intent(this, IndexActivity.class));
    finish();

}

private class DownloadImageTask extends AsyncTask<String, Void, Bitmap> {
    ImageView bmImage;

    public DownloadImageTask(ImageView bmImage) {
        this.bmImage = bmImage;
    }

    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);
    }
}

确保你在AndroidManifest.xml中设置了以下权限来访问互联网。

<uses-permission android:name="android.permission.INTERNET" />

    private Bitmap getImageBitmap(String url) {
        Bitmap bm = null;
        try {
            URL aURL = new URL(url);
            URLConnection conn = aURL.openConnection();
            conn.connect();
            InputStream is = conn.getInputStream();
            BufferedInputStream bis = new BufferedInputStream(is);
            bm = BitmapFactory.decodeStream(bis);
            bis.close();
            is.close();
       } catch (IOException e) {
           Log.e(TAG, "Error getting bitmap", e);
       }
       return bm;
    } 

如果你是在点击按钮的基础上加载图像，上面接受的答案是很棒的，但是如果你是在一个新的活动中做这件事，它会冻结UI一到两秒钟。环顾四周，我发现一个简单的asynctask消除了这个问题。

要使用asynctask，在activity的末尾添加这个类:

private class DownloadImageTask extends AsyncTask<String, Void, Bitmap> {
    ImageView bmImage;

    public DownloadImageTask(ImageView bmImage) {
        this.bmImage = bmImage;
    }

    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);
    }    
}

从你的onCreate()方法调用使用:

new DownloadImageTask((ImageView) findViewById(R.id.imageView1))
        .execute(MY_URL_STRING);

结果是一个快速加载的活动和一个稍后根据用户的网络速度显示的imageview。

在我看来，最适合完成这项任务的现代图书馆是毕加索广场图书馆。它允许通过URL加载图像到ImageView，只需一行代码:

Picasso.with(context).load("http://i.imgur.com/DvpvklR.png").into(imageView);

要做到这一点，一个简单而干净的方法是使用开源库Prime。

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;
}
}