我需要使用raphael.js绘制各种大小的同心弧。我试图理解http://raphaeljs.com/polar-clock.html背后的代码，这与我想要的非常相似，但是，没有评论，很难理解。
理想情况下，我需要一个函数来创建一条距离某个中心点给定距离的路径，从某个角度开始，以某个其他角度结束。
答案 0 :(得分：53)
答案还可以，但不能动画。我为你敲了极点时钟的重要东西。这是一个激发成长的红色弧线。享受。
// Custom Arc Attribute, position x&y, value portion of total, total value, Radius
var archtype = Raphael("canvas", 200, 100);
archtype.customAttributes.arc = function (xloc, yloc, value, total, R) {
var alpha = 360 / total * value,
a = (90 - alpha) * Math.PI / 180,
x = xloc + R * Math.cos(a),
y = yloc - R * Math.sin(a),
path;
if (total == value) {
path = [
["M", xloc, yloc - R],
["A", R, R, 0, 1, 1, xloc - 0.01, yloc - R]
];
} else {
path = [
["M", xloc, yloc - R],
["A", R, R, 0, +(alpha > 180), 1, x, y]
];
}
return {
path: path
};
};
//make an arc at 50,50 with a radius of 30 that grows from 0 to 40 of 100 with a bounce
var my_arc = archtype.path().attr({
"stroke": "#f00",
"stroke-width": 14,
arc: [50, 50, 0, 100, 30]
});
my_arc.animate({
arc: [50, 50, 40, 100, 30]
}, 1500, "bounce");
答案 1 :(得分：10)
这是我如何做到的。以下代码允许您指定开始和结束角度以及内半径和外半径（用于执行那些时髦的甜甜圈样式饼图）。解决方案不依赖于使用线段逼近曲线，并且可以根据原始问题中提到的时钟示例进行动画处理。
首先创建你的Raphael绘图区;以下假设您的HTML文件中包含id为“raphael_paper”的div：
var paper = Raphael("raphael_paper", 800, 800);
对于这个Raphael对象，我们添加了一个自定义arc属性，一个取圆圈中心（x和y坐标），起始角度，结束角度，内半径和外半径的函数：
paper.customAttributes.arc = function (centerX, centerY, startAngle, endAngle, innerR, outerR) {
var radians = Math.PI / 180,
largeArc = +(endAngle - startAngle > 180);
// calculate the start and end points for both inner and outer edges of the arc segment
// the -90s are about starting the angle measurement from the top get rid of these if this doesn't suit your needs
outerX1 = centerX + outerR * Math.cos((startAngle-90) * radians),
outerY1 = centerY + outerR * Math.sin((startAngle-90) * radians),
outerX2 = centerX + outerR * Math.cos((endAngle-90) * radians),
outerY2 = centerY + outerR * Math.sin((endAngle-90) * radians),
innerX1 = centerX + innerR * Math.cos((endAngle-90) * radians),
innerY1 = centerY + innerR * Math.sin((endAngle-90) * radians),
innerX2 = centerX + innerR * Math.cos((startAngle-90) * radians),
innerY2 = centerY + innerR * Math.sin((startAngle-90) * radians);
// build the path array
var path = [
["M", outerX1, outerY1], //move to the start point
["A", outerR, outerR, 0, largeArc, 1, outerX2, outerY2], //draw the outer edge of the arc
["L", innerX1, innerY1], //draw a line inwards to the start of the inner edge of the arc
["A", innerR, innerR, 0, largeArc, 0, innerX2, innerY2], //draw the inner arc
["z"] //close the path
];
return {path: path};
};
现在我们可以使用它来绘制指定厚度的弧线，在我们想要的任何地方开始和结束，例如。
var redParams = {stroke: "#f00", "stroke-width": 1, fill:"#eee"},
greenParams = {stroke: "#0f0", "stroke-width": 1, fill:"#eee"},
blueParams = {stroke: "#00f", "stroke-width": 1, fill:"#eee"},
cx = 300, cy = 300, innerRadius = 100, outerRadius = 250,
var red = paper.path().attr(redParams).attr({arc: [cx, cy, 0, 90, innerRadius, outerRadius]});
var green = paper.path().attr(greenParams).attr({arc: [cx, cy, 270, 320, innerRadius, outerRadius]});
var blue = paper.path().attr(blueParams).attr({arc: [cx, cy, 95, 220, innerRadius, outerRadius]});
这应该会产生三个带有红色，蓝色和绿色1px边框的灰色弧段。
答案 2 :(得分：7)
实际上我自己找到了答案。我首先想到的是一些涉及贝塞尔曲线的奇特，但这只是有效。
- ＆GT;使用SVG路径语法创建路径，该路径与raphael
一样function arc(center, radius, startAngle, endAngle) {
angle = startAngle;
coords = toCoords(center, radius, angle);
path = "M " + coords[0] + " " + coords[1];
while(angle<=endAngle) {
coords = toCoords(center, radius, angle);
path += " L " + coords[0] + " " + coords[1];
angle += 1;
}
return path;
}
function toCoords(center, radius, angle) {
var radians = (angle/180) * Math.PI;
var x = center[0] + Math.cos(radians) * radius;
var y = center[1] + Math.sin(radians) * radius;
return [x, y];
}
答案 3 :(得分：7)
只是为了从user592699的答案中删除一些猜测，这是完整的代码：
<script src="raphael.js"></script>
<script>
var paper = Raphael(20, 20, 320, 320);
function arc(center, radius, startAngle, endAngle) {
angle = startAngle;
coords = toCoords(center, radius, angle);
path = "M " + coords[0] + " " + coords[1];
while(angle<=endAngle) {
coords = toCoords(center, radius, angle);
path += " L " + coords[0] + " " + coords[1];
angle += 1;
}
return path;
}
function toCoords(center, radius, angle) {
var radians = (angle/180) * Math.PI;
var x = center[0] + Math.cos(radians) * radius;
var y = center[1] + Math.sin(radians) * radius;
return [x, y];
}
paper.path(arc([100, 100], 80, 0, 270)); // draw an arc
// centered at (100, 100),
// radius 80, starting at degree 0,
// beginning at coordinate (80, 0)
// which is relative to the center
// of the circle,
// going clockwise, until 270 degree
</script>
答案 4 :(得分：2)
对于那些希望用闭路而不是中风制作弧线的人，我已经扩展了genkilabs的答案来解决问题。如果您需要为弧线提供外部笔划，这可能会有所帮助。
// Custom Arc Attribute, position x&y, value portion of total, total value, Radius, width
var archtype = Raphael("canvas", 200, 100);
archtype.customAttributes.arc = function (xloc, yloc, value, total, R, width) {
if(!width) width = R * 0.4;
var alpha = 360 / total * value,
a = (90 - alpha) * Math.PI / 180,
w = width / 2,
r1 = R + w,
r2 = R - w,
x1 = xloc + r1 * Math.cos(a),
y1 = yloc - r1 * Math.sin(a),
x2 = xloc + r2 * Math.cos(a),
y2 = yloc - r2 * Math.sin(a),
path;
if (total == value) {
path = [
["M", xloc, yloc - r1],
["A", r1, r1, 0, 1, 1, xloc - 0.01, yloc - r1],
["Z"],
["M", xloc - 0.01, yloc - r2],
["A", r2, r2, 0, 1, 0, xloc, yloc - r2],
["Z"]
];
} else {
path = [
["M", xloc, yloc - r1],
["A", r1, r1, 0, +(alpha > 180), 1, x1, y1],
["L", x2, y2],
["A", r2, r2, 0, +(alpha > 180), 0, xloc, yloc - r2],
["L", xloc, yloc - r1],
["Z"]
];
}
return {
path: path
};
};
//make an arc at 50,50 with a radius of 30 that grows from 0 to 40 of 100 with a bounce
var my_arc = archtype.path().attr({
"fill": "#00f",
"stroke": "#f00",
"stroke-width": 5,
arc: [50, 50, 0, 100, 30]
});
my_arc.animate({
arc: [50, 50, 40, 100, 30]
}, 1500, "bounce");
答案 5 :(得分：1)
您也可以在不使用循环的情况下执行此操作。以下实现了这一点并且也适用于负角度。
传递Raphael对象为 r 。角度从0度开始，这是圆圈的顶部，而不是在其他几个解决方案中列出的右侧。
function drawArc(r, centerX, centerY, radius, startAngle, endAngle) {
var startX = centerX+radius*Math.cos((90-startAngle)*Math.PI/180);
var startY = centerY-radius*Math.sin((90-startAngle)*Math.PI/180);
var endX = centerX+radius*Math.cos((90-endAngle)*Math.PI/180);
var endY = centerY-radius*Math.sin((90-endAngle)*Math.PI/180);
var flg1 = 0;
if (startAngle>endAngle)
flg1 = 1;
else if (startAngle<180 && endAngle<180)
flg1 = 0;
else if (startAngle>180 && endAngle>180)
flg1 = 0;
else if (startAngle<180 && endAngle>180)
flg1 = 0; // edited for bugfix here, previously this was 1
else if (startAngle>180 && endAngle<180)
flg1 = 1;
return r.path([['M',startX, startY],['A',radius,radius,0,flg1,1,endX,endY]]);
};
答案 6 :(得分：1)
我已经调整了genkilabs的答案，包括旋转和反转能力。此外，填充了多少环被改为单数百分比。 （反演改编自this post）。希望它有用！
paper.customAttributes.arc = function (xloc, yloc, percent, rad, rot, invert) {
var alpha = 3.6 * percent,
a = (90 - alpha) * Math.PI / 180,
x = xloc + rad * Math.cos(a),
y = yloc - rad * Math.sin(a),
path;
if (invert) {
x = xloc - rad * Math.cos(a);
}
if (percent >= 100) {
path = [
["M", xloc, yloc - rad],
["A", rad, rad, 0, 1, 1, xloc - 0.01, yloc - rad]
];
} else {
path = [
["M", xloc, yloc - rad],
["A", rad, rad, 0, +(alpha > 180), +(!invert), x, y]
];
}
return {
path: path,
transform: "r"+rot+","+xloc+","+yloc,
};
};