如何从StorageFile获取IRandomAccessStream

Question

首先我宣布

private MediaCapture _mediaCapture;
StorageFile capturedPhoto;
IRandomAccessStream imageStream;

第二，我捕捉到了

var lowLagCapture = await _mediaCapture.PrepareLowLagPhotoCaptureAsync(ImageEncodingProperties.CreateUncompressed(MediaPixelFormat.Bgra8));

var capturedPhoto = await lowLagCapture.CaptureAsync();


await lowLagCapture.FinishAsync();

第三，我正在设置图像源：

var softwareBitmap = capturedPhoto.Frame.SoftwareBitmap;
SoftwareBitmap softwareBitmapBGRB = SoftwareBitmap.Convert(softwareBitmap, BitmapPixelFormat.Bgra8, BitmapAlphaMode.Premultiplied);
SoftwareBitmapSource bitmapSource = new SoftwareBitmapSource();
await bitmapSource.SetBitmapAsync(softwareBitmapBGRB);

image.Source = bitmapSource;

如何获取imageStream？我在xaml中使用了CaptureElement工具。

Answer 1

这很简单，问题是你想用这个IRandomAccessStreem做什么。下面是我认为你需要的一些代码：

public void HandleImageFileOperations(StorageFile file)
{
    if (file != null)
    {
         //converts the StorageFile to IRandomAccessStream
         var stream = await file.OpenAsync(FileAccessMode.Read);
         //creates the stream to an Image just in-case you want to show it
         var image = new Windows.UI.Xaml.Media.Imaging.BitmapImage();
         image.SetSource(stream);

         //creates the image into byte array just in-case you need it to store the image
         byte[] bitmapImageBytes = null;
         var reader = new Windows.Storage.Streams.DataReader(stream.GetInputStreamAt(0));
         bitmapImageBytes = new byte[stream.Size];
         await reader.LoadAsync((uint)stream.Size);
         reader.ReadBytes(bitmapImageBytes);
    }
}