UWP的BitmapImage到流

Question

我有一个使用转换器在XAML加载的图像。而不是负载这一形象再次，我要采取的图像，并找到主色，以便能够使用网页上的其他图形。到目前为止，我有这个：

var himage = (BitmapImage)image_home.Source;

using (var stream = await himage.OpenReadAsync())  //**can't open himage this way**
    {
      //Create a decoder for the image
         var decoder = await BitmapDecoder.CreateAsync(stream);

      //Create a transform to get a 1x1 image
         var myTransform = new BitmapTransform { ScaledHeight = 1, ScaledWidth = 1 };

      //Get the pixel provider
         var pixels = await decoder.GetPixelDataAsync(
         BitmapPixelFormat.Rgba8,
         BitmapAlphaMode.Ignore,
         myTransform,
         ExifOrientationMode.IgnoreExifOrientation,
         ColorManagementMode.DoNotColorManage);

      //Get the bytes of the 1x1 scaled image
         var bytes = pixels.DetachPixelData();

      //read the color 
         var myDominantColor = Color.FromArgb(255, bytes[0], bytes[1], bytes[2]);
   }

很显然，我可以用OpenReadAsync无法打开的BitmapImage画佳，有什么事情，我需要在那里做才能够实现这一目标？

Answer 1

BitmapDecoder需要RandomAccessStream对象来创建新实例。除非您知道原始来源，否则BitmapImage可能无法直接提取。RandomAccessStream根据您的评论，您将图像 Uri 绑定到图像控件，因此您可以知道原始来源，并且可以按类从属性中获取图像， 不需要RandomAccessStream再次加载图像。代码如下：BitmapImageUriSourceRandomAccessStreamReference

\n\n

 var himage = (BitmapImage)image_home.Source;\n RandomAccessStreamReference random = RandomAccessStreamReference.CreateFromUri(himage.UriSour\xe2\x80\x8c\xe2\x80\x8bce);\n\n using (IRandomAccessStream stream = await random.OpenReadAsync())   \n {\n     //Create a decoder for the image\n     var decoder = await BitmapDecoder.CreateAsync(stream);\n    ...\n     //read the color \n     var myDominantColor = Color.FromArgb(255, bytes[0], bytes[1], bytes[2]);\n }\n

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