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我正在尝试实现一些逻辑来分享我的 webView 的图像和一些额外的信息。如果我在没有捕获屏幕的情况下这样做，它会完美运行：

   private async void OnDataRequested(DataTransferManager sender, DataRequestedEventArgs e)
    {
        if (await GetShareContent(e.Request))
        {
            if (String.IsNullOrEmpty(e.Request.Data.Properties.Title))
            {
                e.Request.FailWithDisplayText("Nenhum título adicionado");
            }
        }
    }

private async Task<bool> GetShareContent(DataRequest request, StorageFile file)
        {
            bool succeeded = false;
            string text = "Dados do Arquivo:" + Environment.NewLine + webViewModel.Name;
            string dataPackageText = text;

            if (!String.IsNullOrEmpty(dataPackageText))
            {
                DataPackage requestData = request.Data;
                requestData.Properties.Title = "Target";
                requestData.Properties.Description = webViewModel.Name;
                requestData.SetText(dataPackageText);
                succeeded = true;
            }
            else
            {
                request.FailWithDisplayText("Não há nada para compartilhar");
            }
            return succeeded;
        }

但是，如果我尝试相同的方法只是添加捕获的图像，它不起作用，没有显示任何异常，只是消息：“Não há nada para compartilhar agora”（现在没有什么可分享的）

我不知道发生了什么事。不起作用的代码：

private async void OnDataRequested(DataTransferManager sender, DataRequestedEventArgs e)
{
    StorageFile file = await captureScreen();

    if (await GetShareContent(e.Request, file))
    {
        if (String.IsNullOrEmpty(e.Request.Data.Properties.Title))
        {
            e.Request.FailWithDisplayText("Nenhum título adicionado");
        }
    }
}

 private async Task<StorageFile> captureScreen()
        {
            RenderTargetBitmap renderTargetBitmap = new RenderTargetBitmap();
            await renderTargetBitmap.RenderAsync(webView, (int)webView.Width, (int)webView.Height);

            Image myImage = new Image();
            myImage.Source = renderTargetBitmap;

            var file = await App.rootDir.CreateFileAsync("screenCapture.jpg", CreationCollisionOption.ReplaceExisting);
            var pixels = await renderTargetBitmap.GetPixelsAsync();

            using (IRandomAccessStream stream = await file.OpenAsync(FileAccessMode.ReadWrite))
            {
                var encoder = await BitmapEncoder.CreateAsync(BitmapEncoder.JpegEncoderId, stream);
                byte[] bytes = pixels.ToArray();
                encoder.SetPixelData(BitmapPixelFormat.Bgra8,
                                     BitmapAlphaMode.Ignore,
                                     (uint)webView.Width, (uint)webView.Height,
                                     96, 96, bytes);

                await encoder.FlushAsync();
            }

            return file;
        }

我虽然可能会发生这种情况，因为在调用共享时未读取图像，但我正在使用等待。我的 jpeg 是完美创建的。

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1 回答 1

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调用异步 API 时，OnDataRequested回调需要延迟，使用。DataRequest.GetDeferral

private async void OnDataRequested(DataTransferManager sender, DataRequestedEventArgs e)
{
    DataRequestDeferral deferral = e.Request.GetDeferral();

    // Code to do screen capture...

    deferral.Complete();
}

但是，根据 MSDN，“[共享操作] 函数必须在 200 毫秒内返回一个 DataPackage 对象，以防止操作超时”。截屏时间绝对有可能超过 200 毫秒。用于DataPackage.SetDataProvider可能需要较长时间的操作，例如屏幕捕获。

private void OnDataRequested(DataTransferManager sender, DataRequestedEventArgs e)
{
    DataPackage requestData = e.request.Data;
    requestData.Properties.Title = "Target";
    requestData.Properties.Description = webViewModel.Name;
    // Set up the data provider for a long running share operation...
    requestData.SetDataProvider(StandardDataFormats.Bitmap, OnDeferredRequestedHandler);
}

private async void OnDeferredRequestedHandler(DataProviderRequest providerRequest)
{
    // Again, get a deferral as an asynchronous method is called
    DataProviderDeferral deferral = providerRequest.GetDeferral();

    // Code to do screen capture...

    deferral.Complete();
}

MSDN 上的Share content source sample详细说明了如何执行此操作。

于 2013-10-03T23:43:10.767 回答