我想使用画布标签 javascript 绘制一个箭头。我已经使用二次函数完成了它，但是我在计算箭头的旋转角度时遇到了问题......
有人对此有任何线索吗？
谢谢
José Leal Domingues Neto
问问题
84593 次
14 回答
108
尽可能简单。您必须自己添加 context.beginPath() 并添加 context.stroke() ：
ctx = document.getElementById("c").getContext("2d");
ctx.beginPath();
canvas_arrow(ctx, 10, 30, 200, 150);
canvas_arrow(ctx, 100, 200, 400, 50);
canvas_arrow(ctx, 200, 30, 10, 150);
canvas_arrow(ctx, 400, 200, 100, 50);
ctx.stroke();
function canvas_arrow(context, fromx, fromy, tox, toy) {
var headlen = 10; // length of head in pixels
var dx = tox - fromx;
var dy = toy - fromy;
var angle = Math.atan2(dy, dx);
context.moveTo(fromx, fromy);
context.lineTo(tox, toy);
context.lineTo(tox - headlen * Math.cos(angle - Math.PI / 6), toy - headlen * Math.sin(angle - Math.PI / 6));
context.moveTo(tox, toy);
context.lineTo(tox - headlen * Math.cos(angle + Math.PI / 6), toy - headlen * Math.sin(angle + Math.PI / 6));
}<html>
<body>
<canvas id="c" width="500" height="500"></canvas>
</body>于 2011-06-13T17:13:28.853 回答
32
好的，所以当我试图自己解决这个问题时，这个页面上的第一个答案对我有很大帮助，尽管正如其他人已经说过的，如果你的线宽大于 1px，你会得到有趣的形状。其他人建议的修复方法几乎奏效了，但是在尝试使用更粗的箭头时我仍然遇到了一些问题。在玩了几个小时之后，我能够将上述解决方案与我自己的一些修补程序结合起来，提出以下代码，可以在不扭曲箭头形状的情况下以你想要的任何厚度绘制箭头。
function drawArrow(fromx, fromy, tox, toy){
//variables to be used when creating the arrow
var c = document.getElementById("myCanvas");
var ctx = c.getContext("2d");
const width = 22;
var headlen = 10;
// This makes it so the end of the arrow head is located at tox, toy, don't ask where 1.15 comes from
tox -= Math.cos(angle) * ((width*1.15));
toy -= Math.sin(angle) * ((width*1.15));
var angle = Math.atan2(toy-fromy,tox-fromx);
//starting path of the arrow from the start square to the end square and drawing the stroke
ctx.beginPath();
ctx.moveTo(fromx, fromy);
ctx.lineTo(tox, toy);
ctx.strokeStyle = "#cc0000";
ctx.lineWidth = width;
ctx.stroke();
//starting a new path from the head of the arrow to one of the sides of the point
ctx.beginPath();
ctx.moveTo(tox, toy);
ctx.lineTo(tox-headlen*Math.cos(angle-Math.PI/7),toy-headlen*Math.sin(angle-Math.PI/7));
//path from the side point of the arrow, to the other side point
ctx.lineTo(tox-headlen*Math.cos(angle+Math.PI/7),toy-headlen*Math.sin(angle+Math.PI/7));
//path from the side point back to the tip of the arrow, and then again to the opposite side point
ctx.lineTo(tox, toy);
ctx.lineTo(tox-headlen*Math.cos(angle-Math.PI/7),toy-headlen*Math.sin(angle-Math.PI/7));
//draws the paths created above
ctx.strokeStyle = "#cc0000";
ctx.lineWidth = width;
ctx.stroke();
ctx.fillStyle = "#cc0000";
ctx.fill();
}
现在这是我在程序中使用的代码。我发现消除失真问题的关键是继续从箭头的尖端到一侧点，到另一侧点，回到尖端，然后回到第一个侧点，然后做一个充满。这修正了箭头的形状。
希望这可以帮助！
于 2014-09-28T00:27:06.113 回答
16
这是另一种绘制箭头的方法。它使用此处的三角形方法：https ://stackoverflow.com/a/8937325/1828637
一个小助手功能。
function canvas_arrow(context, fromx, fromy, tox, toy, r){
var x_center = tox;
var y_center = toy;
var angle;
var x;
var y;
context.beginPath();
angle = Math.atan2(toy-fromy,tox-fromx)
x = r*Math.cos(angle) + x_center;
y = r*Math.sin(angle) + y_center;
context.moveTo(x, y);
angle += (1/3)*(2*Math.PI)
x = r*Math.cos(angle) + x_center;
y = r*Math.sin(angle) + y_center;
context.lineTo(x, y);
angle += (1/3)*(2*Math.PI)
x = r*Math.cos(angle) + x_center;
y = r*Math.sin(angle) + y_center;
context.lineTo(x, y);
context.closePath();
context.fill();
}
这是一个在行首和行尾绘制箭头的演示。
var can = document.getElementById('c');
var ctx = can.getContext('2d');
ctx.lineWidth = 10;
ctx.strokeStyle = 'steelblue';
ctx.fillStyle = 'steelbllue'; // for the triangle fill
ctx.lineJoin = 'butt';
ctx.beginPath();
ctx.moveTo(50, 50);
ctx.lineTo(150, 150);
ctx.stroke();
canvas_arrow(ctx, 50, 50, 150, 150, 10);
canvas_arrow(ctx, 150, 150, 50, 50, 10);
function canvas_arrow(context, fromx, fromy, tox, toy, r){
var x_center = tox;
var y_center = toy;
var angle;
var x;
var y;
context.beginPath();
angle = Math.atan2(toy-fromy,tox-fromx)
x = r*Math.cos(angle) + x_center;
y = r*Math.sin(angle) + y_center;
context.moveTo(x, y);
angle += (1/3)*(2*Math.PI)
x = r*Math.cos(angle) + x_center;
y = r*Math.sin(angle) + y_center;
context.lineTo(x, y);
angle += (1/3)*(2*Math.PI)
x = r*Math.cos(angle) + x_center;
y = r*Math.sin(angle) + y_center;
context.lineTo(x, y);
context.closePath();
context.fill();
}<canvas id="c" width=300 height=300></canvas>于 2016-04-23T01:22:34.210 回答
8
你可以做：
ctx.save();
ctx.translate(xOrigin, yOrigin);
ctx.rotate(angle);
// draw your arrow, with its origin at [0, 0]
ctx.restore();
于 2009-04-30T23:35:47.860 回答
5
Typescript 版本，当行宽 >> 1 时有固定的箭头提示
function canvas_arrow( context, fromx, fromy, tox, toy ) {
const dx = tox - fromx;
const dy = toy - fromy;
const headlen = Math.sqrt( dx * dx + dy * dy ) * 0.3; // length of head in pixels
const angle = Math.atan2( dy, dx );
context.beginPath();
context.moveTo( fromx, fromy );
context.lineTo( tox, toy );
context.stroke();
context.beginPath();
context.moveTo( tox - headlen * Math.cos( angle - Math.PI / 6 ), toy - headlen * Math.sin( angle - Math.PI / 6 ) );
context.lineTo( tox, toy );
context.lineTo( tox - headlen * Math.cos( angle + Math.PI / 6 ), toy - headlen * Math.sin( angle + Math.PI / 6 ) );
context.stroke();
}
于 2020-11-09T17:21:34.267 回答
5
var canvas = document.getElementById('canvas');
var ctx = canvas.getContext('2d');
ctx.clearRect(0, 0, canvas.width, canvas.height);
arrow({x: 10, y: 10}, {x: 100, y: 170}, 10);
arrow({x: 40, y: 250}, {x: 10, y: 70}, 5);
function arrow (p1, p2, size) {
var angle = Math.atan2((p2.y - p1.y) , (p2.x - p1.x));
var hyp = Math.sqrt((p2.x - p1.x) * (p2.x - p1.x) + (p2.y - p1.y) * (p2.y - p1.y));
ctx.save();
ctx.translate(p1.x, p1.y);
ctx.rotate(angle);
// line
ctx.beginPath();
ctx.moveTo(0, 0);
ctx.lineTo(hyp - size, 0);
ctx.stroke();
// triangle
ctx.fillStyle = 'blue';
ctx.beginPath();
ctx.lineTo(hyp - size, size);
ctx.lineTo(hyp, 0);
ctx.lineTo(hyp - size, -size);
ctx.fill();
ctx.restore();
}<canvas id = "canvas" width = "300" height = "400"></canvas>于 2017-10-03T01:35:29.600 回答
4
给定大小和起始位置，以下代码将为您绘制箭头。
function draw_arrow(context, startX, startY, size) {
var arrowX = startX + 0.75 * size;
var arrowTopY = startY - 0.707 * (0.25 * size);
var arrowBottomY = startY + 0.707 * (0.25 * size);
context.moveTo(startX, startY);
context.lineTo(startX + size, startX);
context.lineTo(arrowX, arrowTopY);
context.moveTo(startX + size, startX);
context.lineTo(arrowX, arrowBottomY);
context.stroke();
}
window.onload = function() {
var canvas = document.getElementById("myCanvas");
var context = canvas.getContext("2d");
var startX = 50;
var startY = 50;
var size = 100;
context.lineWidth = 2;
draw_arrow(context, startX, startY, size);
};body {
margin: 0px;
padding: 0px;
}
#myCanvas {
border: 1px solid #9C9898;
}<!DOCTYPE HTML>
<html>
<body onmousedown="return false;">
<canvas id="myCanvas" width="578" height="200"></canvas>
</body>
</html>于 2011-12-26T09:44:06.907 回答
2
这段代码类似于 Titus Cieslewski 的解决方案，也许箭头更好一些：
function canvasDrawArrow(context, fromx, fromy, tox, toy) {
var headlen = 10.0;
var back = 4.0;
var angle1 = Math.PI / 13.0;
var angle2 = Math.atan2(toy - fromy, tox - fromx);
var diff1 = angle2 - angle1;
var diff2 = angle2 + angle1;
var xx = getBack(back, fromx, fromy, tox, toy);
var yy = getBack(back, fromy, fromx, toy, tox);
context.moveTo(fromx, fromy);
context.lineTo(tox, toy);
context.moveTo(xx, yy);
context.lineTo(xx - headlen * Math.cos(diff1), yy - headlen * Math.sin(diff1));
context.moveTo(xx, yy);
context.lineTo(xx - headlen * Math.cos(diff2), yy - headlen * Math.sin(diff2));
}
function getBack(len, x1, y1, x2, y2) {
return x2 - (len * (x2 - x1) / (Math.sqrt(Math.pow(y2 - y1, 2) + Math.pow(x2 - x1, 2))));
}
这适用于lineWidth > 1. x画轴的时候可以派上y用场
于 2016-04-23T14:20:37.890 回答
1
function RTEShape()
{
this.x = 50;
this.y = 50;
this.w = 100; // default width and height?
this.h = 100;
this.fill = '#444444';
this.text = "Test String";
this.type;
this.color;
this.size = 6;
// The selection color and width. Right now we have a red selection with a small width
this.mySelColor = '#CC0000';
this.mySelWidth = 2;
this.mySelBoxColor = 'darkred';// New for selection boxes
this.mySelBoxSize = 6;
}
RTEShape.prototype.buildArrow = function(canvas)
{
this.type = "arrow";
// Make sure we don't execute when canvas isn't supported
if (canvas.getContext){
// use getContext to use the canvas for drawing
var ctx = canvas.getContext('2d');
var oneThirdX = this.x + (this.w/3);
var twoThirdX = this.x + ((this.w*2)/3);
var oneFifthY = this.y - (this.y/5);
var twoFifthY = this.y - ((this.y*3)/5);
/**/
//ctx.beginPath();
ctx.moveTo(oneThirdX,this.y); // 125,125
ctx.lineTo(oneThirdX,oneFifthY); // 125,105
ctx.lineTo(this.x*2,oneFifthY); // 225,105
ctx.lineTo(this.x*2,twoFifthY); // 225,65
ctx.lineTo(oneThirdX,twoFifthY); // 125,65
ctx.lineTo(oneThirdX,(this.y/5)); // 125,45
ctx.lineTo(this.x,(this.y+(this.y/5))/2); // 45,85
ctx.fillStyle = "green";
ctx.fill();
ctx.fillStyle = "yellow";
ctx.fillRect(this.x,this.y,this.w,this.h);
} else {
alert('Error on buildArrow!\n'+err.description);
}
}
于 2010-12-23T08:59:17.663 回答
1
您好，非常感谢您的建议。
我可以建议你放弃笨重的 atan 吗？您也可以使用线性代数来增加或减少角度：
var cospix=0.866025404; //cosinus of pi/6
function canvas_arrow(context, fromx, fromy, tox, toy) {
ctx.strokeStyle = '#AA0000';
var headlen = 10; // length of head in pixels
var dx = tox - fromx;
var dy = toy - fromy;
var length = Math.sqrt(dy*dy + dx*dx); //length of arrow
var sina = dy/length, cosa = dx/length; //computing sin and cos of arrow angle
var cosp=cosa*cospix-0.5*sina, cosm=cosa*cospix+0.5*sina,
sinp=cosa*0.5+cospix*sina, sinm=cospix*sina-cosa*0.5;
//computing cos and sin of arrow angle plus pi/6, respectively minus pi/6
//(p for plus, m for minus at the end of variable's names)
context.moveTo(fromx, fromy);
context.lineTo(tox, toy);
context.lineTo(tox - headlen * cosm, toy - headlen * sinm); //computing coordinates using the cos and sin computed above
context.moveTo(tox, toy);
context.lineTo(tox - headlen * cosp, toy - headlen * sinp); //computing coordinates using the cos and sin computed above
}
于 2021-01-31T07:54:47.100 回答
0
你可以推动你的矩阵，旋转它，画你的箭头，然后弹出矩阵。
于 2009-04-30T21:29:29.803 回答
0
我已经为此苦苦挣扎了一段时间。我需要在 javascript 和 c# 中都这样做。对于 javascript，我发现了一个不错的库jCanvas。
我的主要问题是绘制漂亮的箭头，jCanvas 做得很好。对于我的 c# 项目，我对 jCanvas 代码进行了逆向工程。
希望这可以帮助某人
于 2018-10-26T11:19:14.900 回答
0
虽然这个问题大部分都得到了回答，但我发现答案很缺乏。最佳答案会产生难看的箭头，当使用 1 以外的宽度时，许多箭头会超出点，而其他的则有不必要的步骤。
这是最简单的答案，它绘制了一个漂亮的箭头（用颜色填充的适当三角形），并缩回箭头的点以考虑线条的宽度。
ctx = document.getElementById('canvas').getContext('2d');
/* Draw barrier */
ctx.beginPath();
ctx.moveTo(50, 30);
ctx.lineTo(450, 30);
ctx.stroke();
draw_arrow(50, 180, 150, 30);
draw_arrow(250, 180, 250, 30);
draw_arrow(450, 180, 350, 30);
function draw_arrow(x0, y0, x1, y1) {
const width = 8;
const head_len = 16;
const head_angle = Math.PI / 6;
const angle = Math.atan2(y1 - y0, x1 - x0);
ctx.lineWidth = width;
/* Adjust the point */
x1 -= width * Math.cos(angle);
y1 -= width * Math.sin(angle);
ctx.beginPath();
ctx.moveTo(x0, y0);
ctx.lineTo(x1, y1);
ctx.stroke();
ctx.beginPath();
ctx.lineTo(x1, y1);
ctx.lineTo(x1 - head_len * Math.cos(angle - head_angle), y1 - head_len * Math.sin(angle - head_angle));
ctx.lineTo(x1 - head_len * Math.cos(angle + head_angle), y1 - head_len * Math.sin(angle + head_angle));
ctx.closePath();
ctx.stroke();
ctx.fill();
}<canvas id="canvas" width="500" height="180"></canvas>于 2022-01-27T05:16:44.870 回答
0
这是工作解决方案
function draw_arrow(ctx,fx,fy,tx,ty){ //ctx is the context
var angle=Math.atan2(ty-fy,tx-fx);
ctx.moveTo(fx,fy); ctx.lineTo(tx,ty);
var w=3.5; //width of arrow to one side. 7 pixels wide arrow is pretty
ctx.strokeStyle="#4d4d4d"; ctx.fillStyle="#4d4d4d";
angle=angle+Math.PI/2; tx=tx+w*Math.cos(angle); ty=ty+w*Math.sin(angle);
ctx.lineTo(tx,ty);
//Drawing an isosceles triangle of sides proportional to 2:7:2
angle=angle-1.849096; tx=tx+w*3.5*Math.cos(angle); ty=ty+w*3.5*Math.sin(angle);
ctx.lineTo(tx,ty);
angle=angle-2.584993; tx=tx+w*3.5*Math.cos(angle); ty=ty+w*3.5*Math.sin(angle);
ctx.lineTo(tx,ty);
angle=angle-1.849096; tx=tx+w*Math.cos(angle); ty=ty+w*Math.sin(angle);
ctx.lineTo(tx,ty);
ctx.stroke(); ctx.fill();
}
于 2019-01-15T10:22:03.197 回答