从资源文件/内容获取流

Question

这是从资源文件获取Stream的正确/唯一方法吗？

    Uri uri = new Uri(fullPath);

    StorageFile storageFile = 
      await Windows.Storage.StorageFile.
        GetFileFromApplicationUriAsync(uri);

    IRandomAccessStreamWithContentType randomAccessStream = 
      await storageFile.OpenReadAsync();

    IInputStream resourceStream = (IInputStream)
      randomAccessStream.GetInputStreamAt(0);

我的所有其他源(http和本地存储)都返回一个Stream对象,并且使用if-else代码使用其中一个或其他代码是很痛苦的.

我也尝试过创建一个MemoryStream,但我甚至找不到一个方法来获取字节...请帮助.

    uint size = (uint)randomAccessStream.Size;
    IBuffer buffer = new Windows.Storage.Streams.Buffer(size);
    await randomAccessStream.ReadAsync(buffer, size, 
      InputStreamOptions.None);

    Stream stream = new MemoryStream(buffer); // error takes byte[] not IBuffer

从资源读取时的IInputStream.ReadAsync():http: //msdn.microsoft.com/en-us/library/windows/apps/windows.storage.streams.iinputstream.readasync.aspx

而Stream.Read()和Stream.ReadAsync()看起来像这样:

http://msdn.microsoft.com/en-us/library/system.io.stream.read.aspx

和

http://msdn.microsoft.com/en-us/library/hh137813.aspx

谢谢

Answer 1

好的我找到了!

    StorageFile storageFile =
      await Windows.Storage.StorageFile.GetFileFromApplicationUriAsync(uri);

    var randomAccessStream = await storageFile.OpenReadAsync();
    Stream stream = randomAccessStream.AsStreamForRead();

Answer 2

你也可以少一行:

Stream stream = await storageFile.OpenStreamForReadAsync();