Hello Ravi,

I've solved this problem through an Input Stream I give you my code :

first how to put this stream in my imageView, mThumbnail:
mThumbnail.setImageDrawable(imageOperations(url, "name.jpg"));

now my imageOperations method :

        public static Drawable imageOperations(String url, String
saveFilename) {
                try {
                        InputStream is = (InputStream) fetch(url);
                        Drawable d = Drawable.createFromStream(is, "src");
                        return d;
                } catch (MalformedURLException e) {
                        e.printStackTrace();
                        return null;
                } catch (IOException e) {
                        e.printStackTrace();
                        return null;
                }
        }

and the fetch method :
        private static Object fetch(String address) throws
MalformedURLException,IOException {
                URL url = new URL(address);
                Object content = url.getContent();
                return content;
        }

It might not be the best way but it works. I found it on a blog...
don't remember which one.

Nice day.

Fräntz

On 23 juin, 09:32, Ravi Savaliya <[EMAIL PROTECTED]> wrote:
> I am try to display image from internet. I have url  of image , that i
> passed in constructing URI ref.
>
> screen display blank
>
> here is code,
>
> -----------------------------------------------------------------------------------------------------------------------------------
>       ImageView imageView = (ImageView) findViewById(R.id.imageView);
>
>       Uri uri=Uri.parse("http://static.px.yelp.com/upthumb/
> q0POOE3vv2LzNg1qN8MMyw/ms");
>       imageView.setImageURI(uri);
>        setContentView(imageView);
> -----------------------------------------------------------------------------------------------------------------------------------
>
> any reply is appreciable
>
> thanks

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