Jak przekonwertować prosty strumień (http web odpowiedzi) na bitmapimage w oknie C# 8?

Question

staram 1000 razy, w celu przekształcenia prostego strumienia (http WebResponse) do BitmapImage, ale nikt poradnik pracuje w C# Windows 8.

Przykład:

BitmapImage image = new BitmapImage(); 
image.SetSource(stream); 
image1.Source = image; 

Dziękuję za wszystkie odpowiedzi.

Rozwiązanie

InMemoryRandomAccessStream randomAccessStream = new InMemoryRandomAccessStream(); 
DataWriter writer = new DataWriter(randomAccessStream.GetOutputStreamAt(0)); 
writer.WriteBytes((byte[])command); 
await writer.StoreAsync(); 
BitmapImage image = new BitmapImage(); 
image.SetSource(randomAccessStream); 

Answer 1

Próbował pan jest?

InMemoryRandomAccessStream randomAccessStream = new InMemoryRandomAccessStream(); 
DataWriter writer = new DataWriter(randomAccessStream.GetOutputStreamAt(0)); 
writer.WriteBytes(response.Content.ReadAsByteArray()); 
BitmapImage image = new BitmapImage(); 
image.SetSource(randomAccessStream); 

Answer 2

Spróbuj kod:

private async Task GetLocalImageAsync(string internetUri, string folderName, 
             string uniqueName) 
{ 
    using (var response = await HttpWebRequest.CreateHttp(internetUri) 
           .GetResponseAsync()) 
    { 
     using (var stream = response.GetResponseStream()) 
     { 
      var folder = await ApplicationData.Current.LocalFolder 
           .CreateFolderAsync(folderName, 
             CreationCollisionOption.OpenIfExists); 
      var file = await folder.CreateFileAsync(
            string.Format("{0}", uniqueName), 
            CreationCollisionOption.ReplaceExisting); 
      using (var filestream = await file.OpenStreamForWriteAsync()) 
      { 
       await stream.CopyToAsync(filestream); 
       await filestream.FlushAsync(); 
      } 
     } 
    } 
}