据我所知,主要问题是在新的旋转坐标系中找到图片角落的坐标。这可以通过以下方式解决:
procedure DrawRotatedBitmap(const Canvas : TCanvas; const Bitmap : TBitmap;
const PointA, PointB : TPointF; const Offset : TPointF; const Scale : Single);
var
OldMatrix, TranslationAlongLineMatrix, RotationMatrix, TranslationMatrix,
ScaleMatrix, FinalMatrix: TMatrix;
W, H : Single;
SrcRect, DestRect: TRectF;
Corner: TPointF;
LineLength : Single;
LineAngleDeg : Integer;
begin
OldMatrix := Canvas.Matrix; // Original, to restore
try
{$ifdef DRAW_HELPERS}
Canvas.Fill.Color := TAlphaColorRec.Black;
Canvas.DrawLine(PointA, PointB, 0.5);
{$endif}
W := PointB.X - PointA.X;
H := PointA.Y - PointB.Y;
LineLength := abs(PointA.Distance(PointB));
// Looking for the middle of the task line
// and the coordinates of the image left upper angle
// solving the proportion width/linelength=xo/0.5requireddimensions
Corner := TPointF.Create((PointB.X + PointA.X) / 2, (PointA.Y + PointB.Y) / 2);// Middle
{$ifdef DRAW_HELPERS}
Canvas.Stroke.Color := TAlphaColorRec.Red;
Canvas.DrawEllipse(TRectF.Create(Corner,2,2),1);
{$endif}
Corner.X := Corner.X - Bitmap.Width / 2 * W / LineLength;
Corner.Y := Corner.Y + Bitmap.Width / 2 * H / LineLength;
{$ifdef DRAW_HELPERS}
Canvas.Stroke.Color := TAlphaColorRec.Green;
Canvas.DrawEllipse(TRectF.Create(Corner,2,2),1);
{$endif}
// Account for scale (if the FMX control is scaled / zoomed); translation
// (the control may not be located at (0, 0) in its parent form, so its canvas
// is offset) and rotation
ScaleMatrix := TMatrix.CreateScaling(Scale, Scale);
TranslationMatrix := TMatrix.CreateTranslation(Offset.X, Offset.Y);
RotationMatrix := TMatrix.CreateRotation(-ArcTan2(H, W));
TranslationAlongLineMatrix := TMatrix.CreateTranslation(Corner.X, Corner.Y);
FinalMatrix := ((RotationMatrix * ScaleMatrix) * TranslationMatrix) * TranslationAlongLineMatrix;
// If in the top left or top right quadrants, the image will appear
// upside down. So, rotate the image 180 degrees
// This is useful when the image contains text, ie is an annotation or similar,
// or needs to always appear "under" the line
LineAngleDeg := Round(RadToDeg(-Arctan2(H, W)));
case LineAngleDeg of
-180..-90,
90..180 : FinalMatrix := TMatrix.CreateRotation(DegToRad(180)) * TMatrix.CreateTranslation(Bitmap.Width, 0) * FinalMatrix;
end;
Canvas.SetMatrix(FinalMatrix);
// And finally draw the bitmap
DestRect := TRectF.Create(PointF(0, 0), Bitmap.Width, Bitmap.Height);
SrcRect := TRectF.Create(0, 0, Bitmap.Width, Bitmap.Height);
{$ifdef DRAW_HELPERS}
Canvas.DrawBitmap(Bitmap, SrcRect, DestRect, 0.5);
{$else}
Canvas.DrawBitmap(Bitmap, SrcRect, DestRect, 1);
{$endif}
finally
// Restore the original matrix
Canvas.SetMatrix(OldMatrix);
end;
end;
ifdef-ed 绘制的线条和点也可能对您有所帮助 - 这些绘制线条和一些有用的点(线条中心和图像左上角),对调试很有用。
DavidM 编辑:此外,还有平移和缩放矩阵。油漆盒(最终)在父窗体的画布上绘制,但可能不在 (0, 0) 处,因此需要考虑目标画布的偏移位置。此外,控件可以具有缩放功能,因此也需要将其内置到最终的旋转矩阵中。
此代码经过大量编辑,无论方向如何/quadrant the line angle is in 都可以正常工作。也就是说,当线条完全水平或垂直时,以及在右下角以外的象限中时,它应该可以工作。
一个有趣的调整是识别示例中的位图包含文本。当线位于左上角或右上角象限时——即从其原点向上然后向左或向右——位图在人眼看来是上下颠倒的。调整识别并旋转位图,使位图的“顶部”始终面向线,位图的“底部”通常指向下方,使图像看起来正确向上。如果您不需要代表可识别事物(例如符号、文本、标签等)的图像,则可以删除此调整
插图
具有不同的角度和缩放比例。