如何编写一个函数,接受 2D 点数组并返回贝塞尔/二次曲线,以便稍后使用 HTML5 Canvas bezierCurveTo
或 quadraticCurveTo< 重新绘制它
方法?
最佳答案
编辑:改进。
See a demo它使用下面的代码。
var makeCurveArgs = function(points) {
var copy = points.slice();
var result = [];
copy.shift(); //drop the first point, it will be handled elsewhere
var tangent;
if(copy.length >= 3) {
var cp1 = copy.shift();
var cp2 = copy.shift();
var p2 = copy.shift();
result.push([cp1[0], cp1[1], cp2[0], cp2[1], p2[0], p2[1]]);
}
while(copy.length >= 2) {
var cp1 = [2 * p2[0] - cp2[0], 2 * p2[1] - cp2[1]];
var cp2 = copy.shift();
var p2 = copy.shift();
result.push([cp1[0], cp1[1], cp2[0], cp2[1], p2[0], p2[1]]);
}
return result;
}
var notThatHard = function(points) {
var origin = points[0].slice();
var curves = makeCurveArgs(points);
var drawCurves = function(context) {
context.beginPath();
context.moveTo(origin[0], origin[1]);
for(var i = 0; i < curves.length; i++) {
var c = curves[i];
context.bezierCurveTo(c[0], c[1], c[2], c[3], c[4], c[5]);
}
};
return drawCurves;
};
一般方法是,您向我提供点和控制点的坐标,然后我返回一个函数,该函数将在 Canvas 上下文上执行该路径。
我给出的函数需要一个 2N+2 2 元素数组的数组;每个 2 元素数组都是一个 (x,y) 坐标。坐标使用如下:
points[0]: starting point of the curve
points[1]: lies on a line tangent to the beginning of the 1st bezier curve
points[2]: lies on a line tangent to the end of the 1st bezier curve
points[3]: end of 1st bezier curve, start of 2nd bezier curve
points[4]: lies on a line tangent to the end of the 2nd bezier curve
points[5]: end of 2nd bezier curve, start of 3rd curve
...
points[2*K+2]: lies on a line tangent to the end of the Kth bezier curve
points[2*K+3]: end of Kth bezier curve, start of (K+1)th
我认为 quadraticCurveTo
的类似函数并不难编写。
关于html - JavaScript 曲线生成,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7584917/