c# - 将 DrawingImage 转换为 BitmapImage

Question

我想绘制大量形状(直线、椭圆和...)，然后将它们保存为位图或 png。我画了图，问题是:如何在 C# 中将 DrawingImage 转换为 BitmapImage？代码是这样的:

DrawingGroup drawingGroup = new DrawingGroup();
using(DrawingContext context = drawingGroup.Open())
{
    //make some drawing 
}
DrawingImage drawingImage = new DrawingImage(drawingGroup)

// your suggestion? DrawingImage - > BitmapImage

Answer 1

您可以将 ImageDrawing 放入 Image 控件中并将其渲染到 RenderTargetBitmap 中，它是一个 BitmapSource，因此可以由 BitmapEncoder 序列化。 (本例中为 PngBitmapEncoder)。

public void SaveDrawingToFile(Drawing drawing, string fileName, double scale)
{
    var drawingImage = new Image { Source = new DrawingImage(drawing) };
    var width = drawing.Bounds.Width * scale;
    var height = drawing.Bounds.Height * scale;
    drawingImage.Arrange(new Rect(0, 0, width, height));

    var bitmap = new RenderTargetBitmap((int)width, (int)height, 96, 96, PixelFormats.Pbgra32);
    bitmap.Render(drawingImage);

    var encoder = new PngBitmapEncoder();
    encoder.Frames.Add(BitmapFrame.Create(bitmap));

    using (var stream = new FileStream(fileName, FileMode.Create))
    {
        encoder.Save(stream);
    }
}

请注意，您实际上不需要 BitmapImage 进行编码，因为 BitmapSource(或任何派生类，如 RenderTargetBitmap)将被接受为 BitmapFrame.Create 的参数。 .

---

稍微不同的解决方案将涉及 DrawingVisual而不是绘图图像:

public void SaveDrawingToFile(Drawing drawing, string fileName, double scale)
{
    var drawingVisual = new DrawingVisual();

    using (var drawingContext = drawingVisual.RenderOpen())
    {
        drawingContext.PushTransform(new ScaleTransform(scale, scale));
        drawingContext.PushTransform(new TranslateTransform(-drawing.Bounds.X, -drawing.Bounds.Y));
        drawingContext.DrawDrawing(drawing);
    }

    var width = drawing.Bounds.Width * scale;
    var height = drawing.Bounds.Height * scale;
    var bitmap = new RenderTargetBitmap((int)width, (int)height, 96, 96, PixelFormats.Pbgra32);
    bitmap.Render(drawingVisual);

    var encoder = new PngBitmapEncoder();
    encoder.Frames.Add(BitmapFrame.Create(bitmap));

    using (var stream = new FileStream(fileName, FileMode.Create))
    {
        encoder.Save(stream);
    }
}