旋转时图像调整大小

时间:2013-03-14 17:00:38

标签: c# winforms visual-studio

我正在尝试旋转图像..我有一个pictureBox 369x276。但是当我旋转时,这个尺寸减小了。

pictureBox sizeMode是 PictureBoxSizeMode.StretchImage

这是我的代码:

        Bitmap oldBitmap = (Bitmap)pictureBox1.Image;
        float angle = 90;
        var newBitmap = new Bitmap(oldBitmap.Width, oldBitmap.Height);

        var graphics = Graphics.FromImage(newBitmap);
        graphics.TranslateTransform((float)oldBitmap.Width  / 2, (float)oldBitmap.Height / 2);
        graphics.RotateTransform(angle);
        graphics.TranslateTransform(-(float)oldBitmap.Width / 2, -(float)oldBitmap.Height / 2);
        graphics.DrawImage(oldBitmap, new Point(0, 0));
        pictureBox1.Image = newBitmap;

2 个答案:

答案 0 :(得分:4)

只需使用RotateFlip:

Bitmap oldBitmap = (Bitmap)pictureBox1.Image;
oldBitmap.RotateFlip(RotateFlipType.Rotate90FlipNone);
pictureBox1.Image = oldBitmap;

正如@ Dan-o指出的那样,这允许在System.Drawing.RotateFlipType枚举中旋转任何度数。

要以任意角度旋转位图而不会丢失大小,您可以执行以下操作,但这有点令人费解!

一个 - 将WriteableBitmapEx库添加到您的项目

两个 - 将XAML,WindowsBase和PresentationCore库添加到项目中

- 使用以下内容旋转您的位图任意数量的度数:

class Program
{
    static void Main(string[] args)
    {
        Bitmap oldBitmap = (Bitmap)pictureBox1.Image;;

        var bitmapAsWriteableBitmap = new WriteableBitmap(BitmapToBitmapImage(oldBitmap));
        bitmapAsWriteableBitmap.RotateFree(23);

        var rotatedImageAsMemoryStream = WriteableBitmapToMemoryStream(bitmapAsWriteableBitmap);
        oldBitmap = new Bitmap(rotatedImageAsMemoryStream);
    }

    public static BitmapImage BitmapToBitmapImage(Bitmap bitmap)
    {
        var memStream = BitmapToMemoryStream(bitmap);
        return MemoryStreamToBitmapImage(memStream);
    }

    public static MemoryStream BitmapToMemoryStream(Bitmap image)
    {
        var memoryStream = new MemoryStream();
        image.Save(memoryStream, ImageFormat.Bmp);

        return memoryStream;
    }

    public static BitmapImage MemoryStreamToBitmapImage(MemoryStream ms)
    {
        ms.Position = 0;
        var bitmap = new BitmapImage();

        bitmap.BeginInit();

        bitmap.StreamSource = ms;
        bitmap.CacheOption = BitmapCacheOption.OnLoad;

        bitmap.EndInit();
        bitmap.Freeze();

        return bitmap;
    }

    private static MemoryStream WriteableBitmapToMemoryStream(WriteableBitmap writeableBitmap)
    {
        var ms = new MemoryStream();

        var encoder = new JpegBitmapEncoder();
        encoder.Frames.Add(BitmapFrame.Create(writeableBitmap));

        encoder.Save(ms);

        return ms;
    }
}

痛苦的屁股,但有效!

答案 1 :(得分:2)

预期的图像尺寸较小。我从来没有想过为什么,但是Graphics.DrawImage真的只有在你不仅提供起始位置而且还提供大小时才有效。其中一个重载允许您包含大小。