Share an image file from url

Question

I want to share an image from a URL. The code is 

           imageUrl = "http://yukoart.com/wp-content/uploads/2012/05/unwritten_19.jpg";
            var deferral = request.getDeferral();
            var imageSource = Windows.Storage.Streams.RandomAccessStreamReference.createFromUri(new Windows.Foundation.Uri(imageUrl));
            imageSource.openReadAsync().done(function (randomAccessStream) {
                request.data.properties.thumbnail = randomAccessStream;
                //request.data.setStorageItems(randomAccessStream);
                deferral.complete();
            });

However, the returned attibute "randomAccessStream" is not a type of "RandomAccessStrearm". It's type of FileRandomAccessStrearm,

So I get a error like " No such interface supported". 

Answer 1

Hi Nick,

The object returned by "RandomAccessStreamReference.CreateFromUri" is already a RandomAccessStreamReference which you can use as the thumbnail or bitmap source. You don't need to create a file reference from it further. For example:

function shareDataHandler(e) {
        var request = e.request;
        var deferral = request.getDeferral();
        request.data.properties.title = "Share Text Example";
        request.data.properties.description = "Demonstrates how to share text.";
        request.data.setText("Hello World!");
        // In this example, we use the imageFile for the thumbnail as well.
        var rawUrl = "http://i.s-microsoft.com/global/ImageStore/PublishingImages/logos/hp/logo-lg-1x.png";
        var urlToImage = new Windows.Foundation.Uri(rawUrl);
        request.data.properties.thumbnail = Windows.Storage.Streams.RandomAccessStreamReference.createFromUri(urlToImage);
        request.data.setBitmap(Windows.Storage.Streams.RandomAccessStreamReference.createFromUri(urlToImage));
        deferral.complete();
    }

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Answer 2

Dear steven,

Thanks a lot.