下面是一张图片，说明了我正在尝试做的事情。
我在 c# 中旋转图像（白色矩形）。问题是我需要将绿色正方形（另一个图像）放置在相对于矩形的同一点上，而不管旋转如何。绿色方块不能随矩形旋转。
正如您在旋转白色矩形后看到的那样，我每次都会得到不同大小的画布。
在创建新的旋转图像后，我试图找到放置绿色方块的确切点。
下面是一些旋转图像并将点放在正确位置的代码。现在的问题是，如果我在 RotateImage 方法中应用转换，我会看到所有的矩形，但显然我的红点在错误的位置。我应该强调我需要知道点的位置，而不仅仅是把它放在正确的位置。
public class IconController : Controller
{
//
// GET: /Icon/
public ActionResult Index()
{
return View();
}
////Icon/Icon?connected=true&heading=320&type=logo45
public ActionResult Icon(bool connected, float heading, string type)
{
var dir = Server.MapPath("/images");
//RED SQUARE IM TRYING TO PLACE ON THE BLUE RECTANGLE.
var path = Path.Combine(dir, "mapicons/center.png");
//GREEN RECTANGLE WITH FIXED YELLOW (Actual center) AND BLUE (point im really trying to find)
var path2 = Path.Combine(dir, "mapicons/connected-marker.png");
Image innerIcon = Image.FromFile(path);
Image marker = Image.FromFile(path2);
using (marker)
{
Point orginalCenter = new Point((marker.Width / 2), (marker.Height / 2));
Bitmap markerbitmap = RotateImage(new Bitmap(marker), heading);
marker = (Image)markerbitmap;
using (var bitmap = new Bitmap(marker.Width, marker.Height))
{
using (var canvas = Graphics.FromImage(bitmap))
{
PointF newCenter = RotatePoint(orginalCenter, 80, 120, heading, marker.Width, marker.Height);
canvas.DrawRectangle(new Pen(Color.Black), 0, 0, bitmap.Width, bitmap.Height);
canvas.InterpolationMode = InterpolationMode.HighQualityBicubic;
canvas.DrawImage(marker, new Rectangle(0, 0, marker.Width, marker.Height), new Rectangle(0, 0, marker.Width, marker.Height), GraphicsUnit.Pixel);
canvas.DrawImage(innerIcon, newCenter.X - (innerIcon.Width / 2), newCenter.Y - (innerIcon.Height / 2));
canvas.Save();
}
try
{
bitmap.Save(Path.Combine(dir, "result.png"), ImageFormat.Png);
path = Path.Combine(dir, "result.png");
}
catch (Exception ex) { }
}
}
return base.File(path, "image/png");
}
public static Bitmap RotateImage(Bitmap b, float Angle)
{
// The original bitmap needs to be drawn onto a new bitmap which will probably be bigger
// because the corners of the original will move outside the original rectangle.
// An easy way (OK slightly 'brute force') is to calculate the new bounding box is to calculate the positions of the
// corners after rotation and get the difference between the maximum and minimum x and y coordinates.
float wOver2 = b.Width / 2.0F;
float hOver2 = b.Height / 2.0F;
float radians = -(float)(Angle / 180.0 * Math.PI);
// Get the coordinates of the corners, taking the origin to be the centre of the bitmap.
PointF[] corners = new PointF[]{
new PointF(-wOver2, -hOver2),
new PointF(+wOver2, -hOver2),
new PointF(+wOver2, +hOver2),
new PointF(-wOver2, +hOver2)
};
for (int i = 0; i < 4; i++)
{
PointF p = corners[i];
PointF newP = new PointF((float)(p.X * Math.Cos(radians) - p.Y * Math.Sin(radians)), (float)(p.X * Math.Sin(radians) + p.Y * Math.Cos(radians)));
corners[i] = newP;
}
// Find the min and max x and y coordinates.
float minX = corners[0].X;
float maxX = minX;
float minY = corners[0].Y;
float maxY = minY;
for (int i = 1; i < 4; i++)
{
PointF p = corners[i];
minX = Math.Min(minX, p.X);
maxX = Math.Max(maxX, p.X);
minY = Math.Min(minY, p.Y);
maxY = Math.Max(maxY, p.Y);
}
// Get the size of the new bitmap.
SizeF newSize = new SizeF(maxX - minX, maxY - minY);
// ...and create it.
Bitmap returnBitmap = new Bitmap((int)Math.Ceiling(newSize.Width), (int)Math.Ceiling(newSize.Height));
// Now draw the old bitmap on it.
using (Graphics g = Graphics.FromImage(returnBitmap))
{
g.TranslateTransform(newSize.Width / 2.0f, newSize.Height / 2.0f);
g.RotateTransform(Angle);
g.TranslateTransform(-b.Width / 2.0f, -b.Height / 2.0f);
g.DrawImage(b, 0, 0);
}
return returnBitmap;
}
public static Point RotatePoint(Point pointToRotate, Point centerPoint, double angleInDegrees)
{
double angleInRadians = angleInDegrees * (Math.PI / 180);
double cosTheta = Math.Cos(angleInRadians);
double sinTheta = Math.Sin(angleInRadians);
Point pt = new Point();
pt.X = (int)(cosTheta * (pointToRotate.X-centerPoint.X) - sinTheta * (pointToRotate.Y-centerPoint.Y) + centerPoint.X);
pt.Y = (int)(sinTheta * (pointToRotate.X - centerPoint.X) + cosTheta * (pointToRotate.Y - centerPoint.Y) + centerPoint.Y);
//p'y = sin(theta) * (px-ox) + cos(theta) * (py-oy) + oy
return pt;
}
}
所以现在我需要能够取消注释 RotateImage 方法中的 TranslateTransform 以便我看到正确的矩形并修改 RotatePoint 方法以便我得到正确的新位置