我需要使用 raphael.js 绘制各种大小的同心圆弧。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，以某个角度开始并以某个其他角度结束。
问问题
20114 次
7 回答
53
这个答案可以，但不能动画。我为你从极地时钟中撕下了重要的东西。这是一个红色的弧线，动画生长。请享用。
// 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");
于 2011-03-29T22:04:31.510 回答
10
这就是我的做法。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（对于制作那些流行的圆环式饼图很有用）。该解决方案不依赖于用线段近似曲线，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 文件中有一个 id 为“raphael_paper”的 div：
var paper = Raphael("raphael_paper", 800, 800);
在这个 Raphael 对象上，我们添加了一个自定义arc属性，一个接受圆心（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]});
这应该会产生三个灰色的弧段，带有红色、蓝色和绿色的 1px 边框。
于 2012-02-17T15:15:48.570 回答
7
只是为了从 user592699 的答案中消除一些猜测，这是有效的完整代码：
<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>
于 2011-04-13T08:29:15.650 回答
7
实际上自己找到了答案。我首先想到了一些涉及贝塞尔曲线的花哨的东西，但这很有效。
-> 使用 SVG 路径语法创建路径，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];
}
于 2011-02-23T13:08:15.320 回答
2
对于那些希望用闭合路径而不是笔划来制作弧线的人，我已经扩展了 genkilabs 的答案以做出解决方案。如果您需要为弧线提供外部笔触，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");
于 2013-12-18T09:55:02.017 回答
1
您也可以这样做，而不必使用循环。以下实现了这一点，并且也适用于负角。
将 Raphael 对象作为r 传入。角度从 0 度开始，这是圆的顶部，而不是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]]);
};
于 2012-12-24T15:29:48.160 回答
1
我已经调整了 genkilabs 的答案以包括旋转和反转能力。此外，环的填充量已更改为单个数字百分比。（倒置改编自这篇文章）。希望它有帮助！
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,
};
};
于 2014-12-10T04:26:05.980 回答