13

我正在使用 route-me 的框架来处理位置。在此代码中，两个标记（点）之间的路径将绘制为一条线。

我的问题：“如果我想在线的中间（或顶部）添加一个箭头，我应该添加什么代码，以便它指向方向”

谢谢



- (void)drawInContext:(CGContextRef)theContext
{
    renderedScale = [contents metersPerPixel];

    float scale = 1.0f / [contents metersPerPixel];

    float scaledLineWidth = lineWidth;
    if(!scaleLineWidth) {
        scaledLineWidth *= renderedScale;
    }
    //NSLog(@"line width = %f, content scale = %f", scaledLineWidth, renderedScale);

    CGContextScaleCTM(theContext, scale, scale);

    CGContextBeginPath(theContext);
    CGContextAddPath(theContext, path);

    CGContextSetLineWidth(theContext, scaledLineWidth);
    CGContextSetStrokeColorWithColor(theContext, [lineColor CGColor]);
    CGContextSetFillColorWithColor(theContext, [fillColor CGColor]);

    // according to Apple's documentation, DrawPath closes the path if it's a filled style, so a call to ClosePath isn't necessary
    CGContextDrawPath(theContext, drawingMode);
}
4

4 回答 4

22 
- (void) drawLine: (CGContextRef) context from: (CGPoint) from to: (CGPoint) to 
{
    double slopy, cosy, siny;
    // Arrow size
    double length = 10.0;  
    double width = 5.0;

    slopy = atan2((from.y - to.y), (from.x - to.x));
    cosy = cos(slopy);
    siny = sin(slopy);

    //draw a line between the 2 endpoint
    CGContextMoveToPoint(context, from.x - length * cosy, from.y - length * siny );
    CGContextAddLineToPoint(context, to.x + length * cosy, to.y + length * siny);
    //paints a line along the current path
    CGContextStrokePath(context);

    //here is the tough part - actually drawing the arrows
    //a total of 6 lines drawn to make the arrow shape
    CGContextMoveToPoint(context, from.x, from.y);
    CGContextAddLineToPoint(context,
                        from.x + ( - length * cosy - ( width / 2.0 * siny )),
                        from.y + ( - length * siny + ( width / 2.0 * cosy )));
    CGContextAddLineToPoint(context,
                        from.x + (- length * cosy + ( width / 2.0 * siny )),
                        from.y - (width / 2.0 * cosy + length * siny ) );
    CGContextClosePath(context);
    CGContextStrokePath(context);

    /*/-------------similarly the the other end-------------/*/
    CGContextMoveToPoint(context, to.x, to.y);
    CGContextAddLineToPoint(context,
                        to.x +  (length * cosy - ( width / 2.0 * siny )),
                        to.y +  (length * siny + ( width / 2.0 * cosy )) );
    CGContextAddLineToPoint(context,
                        to.x +  (length * cosy + width / 2.0 * siny),
                        to.y -  (width / 2.0 * cosy - length * siny) );
    CGContextClosePath(context);
    CGContextStrokePath(context);
}
于 2011-10-09T09:38:24.110 回答
6

一旦你的路径上有两个点，实际三角形/箭头的绘制就很容易了。

CGContextMoveToPoint( context , ax , ay );
CGContextAddLineToPoint( context , bx , by );
CGContextAddLineToPoint( context , cx , cy );
CGContextClosePath( context ); // for triangle

获得积分有点棘手。你说路径是一条线，而不是一条曲线或一系列曲线。这使它更容易。

使用 CGPathApply 选择路径上的两个点。可能这是最后两个点，其中一个可能是 kCGPathElementMoveToPoint，另一个可能是 kCGPathElementAddLineToPoint。设 mx,my 为第一个点，nx,ny 为第二个点，所以箭头将从 m 指向 n。

假设您希望箭头位于线的末端，bx,by 从上方将等于线上的 nx,ny。在 mx,my 和 nx,ny 之间选择一个点 dx,dy 来计算其他点。

现在计算 ax,ay 和 cx,cy 使得它们与 dx,dy 在一条线上并且与路径等距。以下应该很接近，尽管我可能有一些错误的迹象：

r = atan2( ny - my , nx - mx );
bx = nx;
by = ny;
dx = bx + sin( r ) * length;
dy = by + cos( r ) * length;
r += M_PI_2; // perpendicular to path
ax = dx + sin( r ) * width;
ay = dy + cos( r ) * width;
cx = dx - sin( r ) * width;
cy = dy - cos( r ) * width;

长度是箭尖到基部的距离，宽度是箭杆到倒钩的距离，或箭头宽度的一半。

如果 path 是一条曲线，则不是寻找 mx,my 作为前一个点或移动，而是最终曲线的最终控制点。每个控制点都在一条与曲线相切并通过相邻点的线上。

于 2010-04-05T06:18:50.147 回答
2

我发现了这个问题，因为我有同样的问题。我拿了drawonward的例子，它是如此接近......但是通过翻转cos和sin，我能够让它工作：

r = atan2( ny - my , nx - mx );
r += M_PI;
bx = nx;
by = ny;
dx = bx + cos( r ) * length;
dy = by + sin( r ) * length;
r += M_PI_2; // perpendicular to path
ax = dx + cos( r ) * width;
ay = dy + sin( r ) * width;
cx = dx - cos( r ) * width;
cy = dy - sin( r ) * width;

一旦我这样做了，我的箭头就完全指向错误的方向。所以我添加了第二行 ( r += M_PI;)

感谢继续绘制！

于 2010-04-16T06:32:21.797 回答
1

这是Friedhelm Brügge答案的Swift 4+版本：（我会在图片上画出来）

func drawArrow(image: UIImage, ptSrc: CGPoint, ptDest: CGPoint) {

        // create context with image size
        UIGraphicsBeginImageContext(image.size)
        let context = UIGraphicsGetCurrentContext()

        // draw current image to the context
        image.draw(in: CGRect(x: 0, y: 0, width: image.size.width, height: image.size.height))

        var slopY: CGFloat, cosY: CGFloat, sinY: CGFloat;
        // Arrow size
        let length: CGFloat = 35.0;
        let width: CGFloat = 35.0;

        slopY = atan2((ptSrc.y - ptDest.y), (ptSrc.x - ptDest.x));
        cosY = cos(slopY);
        sinY = sin(slopY);

        //here is the tough part - actually drawing the arrows
        //a total of 6 lines drawn to make the arrow shape
        context?.setFillColor(UIColor.white.cgColor)
        context?.move(to: CGPoint(x: ptSrc.x, y: ptSrc.y))
        context?.addLine(to: CGPoint(x: ptSrc.x + ( -length * cosY - ( width / 2.0 * sinY )), y: ptSrc.y + ( -length * sinY + ( width / 2.0 * cosY ))))
        context?.addLine(to: CGPoint(x: ptSrc.x + (-length * cosY + ( width / 2.0 * sinY )), y: ptSrc.y - (width / 2.0 * cosY + length * sinY )))
        context?.closePath()
        context?.fillPath()

        context?.move(to: CGPoint(x: ptSrc.x, y: ptSrc.y))
        context?.addLine(to: CGPoint(x: ptDest.x +  (length * cosY - ( width / 2.0 * sinY )), y: ptDest.y +  (length * sinY + ( width / 2.0 * cosY ))))
        context?.addLine(to: CGPoint(x: ptDest.x +  (length * cosY + width / 2.0 * sinY), y: ptDest.y -  (width / 2.0 * cosY - length * sinY)))
        context?.closePath()
        context?.fillPath()

        // draw current context to image view
        imgView.image = UIGraphicsGetImageFromCurrentImageContext()

        //close context
        UIGraphicsEndImageContext()
    }
于 2019-01-02T09:28:56.387 回答