3

我刚刚经历了一堆东西，试图弄清楚如何让图像甚至旋转。这有效，但现在它正在剪辑，我不确定如何让它停止......我正在使用这个 rotateImage 方法：

public static Image RotateImage(Image img, float rotationAngle)
    {
        //create an empty Bitmap image
        Bitmap bmp = new Bitmap(img.Width, img.Height);

        //turn the Bitmap into a Graphics object
        Graphics gfx = Graphics.FromImage(bmp);

        //now we set the rotation point to the center of our image
        gfx.TranslateTransform((float)bmp.Width / 2, (float)bmp.Height / 2);

        //now rotate the image
        gfx.RotateTransform(rotationAngle);

        gfx.TranslateTransform(-(float)bmp.Width / 2, -(float)bmp.Height / 2);

        //set the InterpolationMode to HighQualityBicubic so to ensure a high
        //quality image once it is transformed to the specified size
        gfx.InterpolationMode = InterpolationMode.HighQualityBicubic;

        //now draw our new image onto the graphics object
        gfx.DrawImage(img, new System.Drawing.Point(0, 0));

        //dispose of our Graphics object
        gfx.Dispose();

        //return the image
        return bmp;
    }

我尝试使空位图更大，但这仅适用于一侧，因为图像固定在位图的左上角。任何想法，将不胜感激！

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4 回答 4

22

我从另一个网站找到了一些帮助。这是我最终为那些想知道的人所做的：

// Rotates the input image by theta degrees around center.
    public static Bitmap RotateImage(Bitmap bmpSrc, float theta)
    {
        Matrix mRotate = new Matrix();
        mRotate.Translate(bmpSrc.Width / -2, bmpSrc.Height / -2, MatrixOrder.Append);
        mRotate.RotateAt(theta, new System.Drawing.Point(0, 0), MatrixOrder.Append);
        using (GraphicsPath gp = new GraphicsPath())
        {  // transform image points by rotation matrix
            gp.AddPolygon(new System.Drawing.Point[] { new System.Drawing.Point(0, 0), new System.Drawing.Point(bmpSrc.Width, 0), new System.Drawing.Point(0, bmpSrc.Height) });
            gp.Transform(mRotate);
            System.Drawing.PointF[] pts = gp.PathPoints;

            // create destination bitmap sized to contain rotated source image
            Rectangle bbox = boundingBox(bmpSrc, mRotate);
            Bitmap bmpDest = new Bitmap(bbox.Width, bbox.Height);

            using (Graphics gDest = Graphics.FromImage(bmpDest))
            {  // draw source into dest
                Matrix mDest = new Matrix();
                mDest.Translate(bmpDest.Width / 2, bmpDest.Height / 2, MatrixOrder.Append);
                gDest.Transform = mDest;
                gDest.DrawImage(bmpSrc, pts);
                return bmpDest;
            }
        }
    }

    private static Rectangle boundingBox(Image img, Matrix matrix)
    {
        GraphicsUnit gu = new GraphicsUnit();
        Rectangle rImg = Rectangle.Round(img.GetBounds(ref gu));

        // Transform the four points of the image, to get the resized bounding box.
        System.Drawing.Point topLeft = new System.Drawing.Point(rImg.Left, rImg.Top);
        System.Drawing.Point topRight = new System.Drawing.Point(rImg.Right, rImg.Top);
        System.Drawing.Point bottomRight = new System.Drawing.Point(rImg.Right, rImg.Bottom);
        System.Drawing.Point bottomLeft = new System.Drawing.Point(rImg.Left, rImg.Bottom);
        System.Drawing.Point[] points = new System.Drawing.Point[] { topLeft, topRight, bottomRight, bottomLeft };
        GraphicsPath gp = new GraphicsPath(points,
                                                            new byte[] { (byte)PathPointType.Start, (byte)PathPointType.Line, (byte)PathPointType.Line, (byte)PathPointType.Line });
        gp.Transform(matrix);
        return Rectangle.Round(gp.GetBounds());
    }
于 2010-03-01T02:18:05.870 回答
4

跟着这些步骤：

  1. 创建空的目标图像，其宽度和高度 = 旋转图像的边界框
  2. 将源图像绘制到目标图像上。源的 (0,0) 将映射到目标的 (ox,oy)
  3. 现在执行旋转目标图像的步骤。

完成上述步骤的详细信息：

W,H = 目的地的宽度和高度

w,h = 源的宽度和高度

c=|cos(θ)|

s=|sin(θ)|

w * c + h * s = W

w * s + h * c = H

牛 = ( W - w ) / 2

oy = ( H - h ) / 2

于 2010-02-28T22:03:59.813 回答
4 
public Bitmap rotateImage(Bitmap b, float angle)
    {
        if (angle > 0)
        {
            int l = b.Width;
            int h = b.Height;
            double an = angle * Math.PI / 180;
            double cos = Math.Abs(Math.Cos(an));
            double sin = Math.Abs(Math.Sin(an));
            int nl = (int)(l * cos + h * sin);
            int nh = (int)(l * sin + h * cos);
            Bitmap returnBitmap = new Bitmap(nl, nh);
            Graphics g = Graphics.FromImage(returnBitmap);
            g.TranslateTransform((float)(nl-l) / 2, (float)(nh-h) / 2);
            g.TranslateTransform((float)b.Width / 2, (float)b.Height / 2);
            g.RotateTransform(angle);
            g.TranslateTransform(-(float)b.Width / 2, -(float)b.Height / 2);
            g.DrawImage(b, new Point(0, 0));
            return returnBitmap;
        }
        else return b;
    }
于 2012-02-05T01:15:44.843 回答
1

旋转的图像可能需要包含更大的位图而不进行裁剪。计算边界的一种相当简单的方法是将变换矩阵应用于原始边界矩形的角。由于新的位图更大，因此必须将原始图像绘制成居中，而不是在 (0,0) 处。

这在您的代码的以下修订中得到了证明：

public static Image RotateImage(Image img, float rotationAngle)
{
    int minx = int.MaxValue, maxx = int.MinValue, miny = int.MaxValue, maxy = int.MinValue;

    using (Bitmap bmp = new Bitmap(1, 1)) // Dummy bitmap, so we can use TransformPoints to figure out the correct size.
    {
        using (Graphics g = Graphics.FromImage(bmp))
        {
            g.TranslateTransform((float)img.Width / 2, (float)img.Height / 2);
            g.RotateTransform(rotationAngle);
            g.TranslateTransform(-(float)img.Width / 2, -(float)img.Height / 2);

            Point[] pts = new Point[4];
            pts[0] = new Point(0, 0);
            pts[1] = new Point(img.Width, 0);
            pts[2] = new Point(img.Width, img.Height);
            pts[3] = new Point(0, img.Height);
            g.TransformPoints(CoordinateSpace.Device, CoordinateSpace.World, pts);

            foreach (Point pt in pts)
            {
                minx = Math.Min(minx, pt.X);
                maxx = Math.Max(maxx, pt.X);
                miny = Math.Min(miny, pt.Y);
                maxy = Math.Max(maxy, pt.Y);
            }
        }
    }

    Bitmap bmp2 = new Bitmap(maxx - minx, maxy - miny);
    using (Graphics g = Graphics.FromImage(bmp2))
    {
        g.TranslateTransform((float)bmp2.Width / 2, (float)bmp2.Height / 2);
        g.RotateTransform(rotationAngle);
        g.TranslateTransform(-(float)bmp2.Width / 2, -(float)bmp2.Height / 2);
        g.InterpolationMode = InterpolationMode.HighQualityBicubic;
        g.DrawImage(img, bmp2.Width / 2 - img.Width / 2, bmp2.Height / 2 - img.Height / 2);
    }

    return bmp2;
}
于 2010-02-28T22:09:05.667 回答