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我想在 mousemove 上画一个箭头，下面的代码画了一个水平箭头。但我希望它从开始绘图点旋转到结束绘图点。我用谷歌搜索了很多，但没有找到解决方案，可能是因为我不熟悉在画布上绘画。

drawArrowMove: function(e, _self)
    {
      var width    = e.w ;
      var height   = e.h ;
      var arrowW   = 0.35 * width;
      var arrowH   = 0.75 * height;
      var p1       = {x: 0,              y: (height-arrowH)/2};
      var p2       = {x: (width-arrowW), y: (height-arrowH)/2};
      var p3       = {x: (width-arrowW), y: 0};
      var p4       = {x: width,          y: height/2};
      var p5       = {x: (width-arrowW), y: height};
      var p6       = {x: (width-arrowW), y: height-((height-arrowH)/2)};
      var p7       = {x: 0,              y: height-((height-arrowH)/2)};

      _self.ctxTemp.beginPath();
      _self.ctxTemp.moveTo(p1.x, p1.y);
      _self.ctxTemp.lineTo(p2.x, p2.y); // end of main block
      _self.ctxTemp.lineTo(p3.x, p3.y); // topmost point
      _self.ctxTemp.lineTo(p4.x, p4.y); // endpoint
      _self.ctxTemp.lineTo(p5.x, p5.y); // bottommost point
      _self.ctxTemp.lineTo(p6.x, p6.y); // end at bottom point
      _self.ctxTemp.lineTo(p7.x, p7.y);
      _self.ctxTemp.fillStyle = 'black';
      _self.ctxTemp.fill();
      _self.ctxTemp.stroke();
    }
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1 回答 1

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这是在斜线上绘制箭头的方法

在此处输入图像描述

以任何角度绘制箭头的最简单方法是使用画布的旋转功能。

您可以使用该Math.atan函数根据直线的斜率计算适当的角度。

// calculate the radian angle of the line from [x1,y1] to [x2,y2]
var radianAngle=Math.atan((y2-y1)/(x2-x1));

// adjust the angle based on line slope 
radianAngle+=((x2>x1)?90:-90)*Math.PI/180;

然后你可以画线：

// draw the line
ctx.beginPath();
ctx.moveTo(x1,y1);
ctx.lineTo(x2,y2);
ctx.stroke();

最后通过旋转上下文来附加箭头。

// rotate the canvas context to the appropriate angle
ctx.rotate(radianAngle);

请注意，由于您已经旋转了画布本身，因此您只需将箭头绘制成水平的。简单的！

// save the un-transformed state of the context
ctx.save();

ctx.beginPath();

// translate to the end of the line
ctx.translate(x2,y2);

// rotate to the appropriate angle
ctx.rotate(radianAngle);

// draw the arrowhead
ctx.moveTo(0,0);
ctx.lineTo(8,20);
ctx.lineTo(-8,20);
ctx.closePath();
ctx.fill();

// when done drawing on the rotated context, set it back to its untransformed state
ctx.restore();

这是代码和小提琴：http: //jsfiddle.net/m1erickson/CQDww/

<!doctype html>
<html>
<head>
<link rel="stylesheet" type="text/css" media="all" href="css/reset.css" /> <!-- reset css -->
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>

<style>
    body{ background-color: ivory; }
    canvas{border:1px solid red;}
</style>

<script>
$(function(){

    var canvas=document.getElementById("canvas");
    var context=canvas.getContext("2d");

    function Line(x1,y1,x2,y2){
        this.x1=x1;
        this.y1=y1;
        this.x2=x2;
        this.y2=y2;
    }
    Line.prototype.drawWithArrowheads=function(ctx){

        // arbitrary styling
        ctx.strokeStyle="blue";
        ctx.fillStyle="blue";
        ctx.lineWidth=3;

        // draw the line
        ctx.beginPath();
        ctx.moveTo(this.x1,this.y1);
        ctx.lineTo(this.x2,this.y2);
        ctx.stroke();

        // draw the starting arrowhead
        var startRadians=Math.atan((this.y2-this.y1)/(this.x2-this.x1));
        startRadians+=((this.x2>this.x1)?-90:90)*Math.PI/180;
        this.drawArrowhead(ctx,this.x1,this.y1,startRadians);
        // draw the ending arrowhead
        var endRadians=Math.atan((this.y2-this.y1)/(this.x2-this.x1));
        endRadians+=((this.x2>this.x1)?90:-90)*Math.PI/180;
        this.drawArrowhead(ctx,this.x2,this.y2,endRadians);

    }
    Line.prototype.drawArrowhead=function(ctx,x,y,radians){
        ctx.save();
        ctx.beginPath();
        ctx.translate(x,y);
        ctx.rotate(radians);
        ctx.moveTo(0,0);
        ctx.lineTo(8,20);
        ctx.lineTo(-8,20);
        ctx.closePath();
        ctx.fill();
        ctx.restore();
    }

    // create a new line object
    var line=new Line(50,50,150,150);
    // draw the line
    line.drawWithArrowheads(context);

}); // end $(function(){});
</script>

</head>

<body>
    <canvas id="canvas" width=300 height=300></canvas>
</body>
</html>
于 2013-06-28T05:25:24.400 回答