我必须在路径上移动小矩形。在画布内单击后,矩形会移动。
我无法为它设置动画,因为对象只是跳到所需的点。
请在Fiddle上找到代码。
HTML
<canvas id="myCanvas" width=578 height=200></canvas>
CSS
#myCanvas {
width:578px;
height:200px;
border:2px thin;
}
的JavaScript
var myRectangle = {
x: 100,
y: 20,
width: 25,
height: 10,
borderWidth: 1
};
$(document).ready(function () {
$('#myCanvas').css("border", "2px solid black");
var canvas = document.getElementById('myCanvas');
var context = canvas.getContext('2d');
var cntxt = canvas.getContext('2d');
drawPath(context);
drawRect(myRectangle, cntxt);
$('#myCanvas').click(function () {
function animate(myRectangle, canvas, cntxt, startTime) {
var time = (new Date()).getTime() - startTime;
var linearSpeed = 10;
var newX = Math.round(Math.sqrt((100 * 100) + (160 * 160)));
if (newX < canvas.width - myRectangle.width - myRectangle.borderWidth / 2) {
myRectangle.x = newX;
}
context.clearRect(0, 0, canvas.width, canvas.height);
drawPath(context);
drawRect(myRectangle, cntxt);
// request new frame
requestAnimFrame(function () {
animate(myRectangle, canvas, cntxt, startTime);
});
}
drawRect(myRectangle, cntxt);
myRectangle.x = 100;
myRectangle.y = 121;
setTimeout(function () {
var startTime = (new Date()).getTime();
animate(myRectangle, canvas, cntxt, startTime);
}, 1000);
});
});
$(document).keypress(function (e) {
if (e.which == 13) {
$('#myCanvas').click();
}
});
function drawRect(myRectangle, cntxt) {
cntxt.beginPath();
cntxt.rect(myRectangle.x, myRectangle.y, myRectangle.width, myRectangle.height);
cntxt.fillStyle = 'cyan';
cntxt.fill();
cntxt.strokeStyle = 'black';
cntxt.stroke();
};
function drawPath(context) {
context.beginPath();
context.moveTo(100, 20);
// line 1
context.lineTo(200, 160);
// quadratic curve
context.quadraticCurveTo(230, 200, 250, 120);
// bezier curve
context.bezierCurveTo(290, -40, 300, 200, 400, 150);
// line 2
context.lineTo(500, 90);
context.lineWidth = 5;
context.strokeStyle = 'blue';
context.stroke();
};
答案 0 :(得分:57)
以下是如何沿特定路径移动对象
动画涉及随时间的移动。因此,对于动画的每个“帧”,您需要知道XY坐标在哪里绘制移动对象(矩形)。
此代码采用百分比完成(0.00到1.00)并返回XY坐标,即沿路径段的百分比。例如:
以下是沿着一条线以指定的百分比获取XY的代码:
// line: percent is 0-1
function getLineXYatPercent(startPt,endPt,percent) {
var dx = endPt.x-startPt.x;
var dy = endPt.y-startPt.y;
var X = startPt.x + dx*percent;
var Y = startPt.y + dy*percent;
return( {x:X,y:Y} );
}
以下是沿着二次贝塞尔曲线获得指定百分比的XY的代码:
// quadratic bezier: percent is 0-1
function getQuadraticBezierXYatPercent(startPt,controlPt,endPt,percent) {
var x = Math.pow(1-percent,2) * startPt.x + 2 * (1-percent) * percent * controlPt.x + Math.pow(percent,2) * endPt.x;
var y = Math.pow(1-percent,2) * startPt.y + 2 * (1-percent) * percent * controlPt.y + Math.pow(percent,2) * endPt.y;
return( {x:x,y:y} );
}
以下是沿着三次贝塞尔曲线获得指定百分比的XY的代码:
// cubic bezier percent is 0-1
function getCubicBezierXYatPercent(startPt,controlPt1,controlPt2,endPt,percent){
var x=CubicN(percent,startPt.x,controlPt1.x,controlPt2.x,endPt.x);
var y=CubicN(percent,startPt.y,controlPt1.y,controlPt2.y,endPt.y);
return({x:x,y:y});
}
// cubic helper formula at percent distance
function CubicN(pct, a,b,c,d) {
var t2 = pct * pct;
var t3 = t2 * pct;
return a + (-a * 3 + pct * (3 * a - a * pct)) * pct
+ (3 * b + pct * (-6 * b + b * 3 * pct)) * pct
+ (c * 3 - c * 3 * pct) * t2
+ d * t3;
}
以下是您如何将所有内容组合在一起,为您路径的各个部分制作动画
// calculate the XY where the tracking will be drawn
if(pathPercent<25){
var line1percent=pathPercent/24;
xy=getLineXYatPercent({x:100,y:20},{x:200,y:160},line1percent);
}
else if(pathPercent<50){
var quadPercent=(pathPercent-25)/24
xy=getQuadraticBezierXYatPercent({x:200,y:160},{x:230,y:200},{x:250,y:120},quadPercent);
}
else if(pathPercent<75){
var cubicPercent=(pathPercent-50)/24
xy=getCubicBezierXYatPercent({x:250,y:120},{x:290,y:-40},{x:300,y:200},{x:400,y:150},cubicPercent);
}
else {
var line2percent=(pathPercent-75)/25
xy=getLineXYatPercent({x:400,y:150},{x:500,y:90},line2percent);
}
// draw the tracking rectangle
drawRect(xy);
这是工作代码和小提琴:http://jsfiddle.net/m1erickson/LumMX/
<!doctype html>
<html lang="en">
<head>
<style>
body{ background-color: ivory; }
canvas{border:1px solid red;}
</style>
<link rel="stylesheet" href="http://code.jquery.com/ui/1.10.3/themes/smoothness/jquery-ui.css" />
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
<script>
$(function() {
var canvas=document.getElementById("canvas");
var ctx=canvas.getContext("2d");
// set starting values
var fps = 60;
var percent=0
var direction=1;
// start the animation
animate();
function animate() {
// set the animation position (0-100)
percent+=direction;
if(percent<0){ percent=0; direction=1; };
if(percent>100){ percent=100; direction=-1; };
draw(percent);
// request another frame
setTimeout(function() {
requestAnimationFrame(animate);
}, 1000 / fps);
}
// draw the current frame based on sliderValue
function draw(sliderValue){
// redraw path
ctx.clearRect(0,0,canvas.width,canvas.height);
ctx.lineWidth = 5;
ctx.beginPath();
ctx.moveTo(100, 20);
ctx.lineTo(200, 160);
ctx.strokeStyle = 'red';
ctx.stroke();
ctx.beginPath();
ctx.moveTo(200, 160);
ctx.quadraticCurveTo(230, 200, 250, 120);
ctx.strokeStyle = 'green';
ctx.stroke();
ctx.beginPath();
ctx.moveTo(250,120);
ctx.bezierCurveTo(290, -40, 300, 200, 400, 150);
ctx.strokeStyle = 'blue';
ctx.stroke();
ctx.beginPath();
ctx.moveTo(400, 150);
ctx.lineTo(500, 90);
ctx.strokeStyle = 'gold';
ctx.stroke();
// draw the tracking rectangle
var xy;
if(sliderValue<25){
var percent=sliderValue/24;
xy=getLineXYatPercent({x:100,y:20},{x:200,y:160},percent);
}
else if(sliderValue<50){
var percent=(sliderValue-25)/24
xy=getQuadraticBezierXYatPercent({x:200,y:160},{x:230,y:200},{x:250,y:120},percent);
}
else if(sliderValue<75){
var percent=(sliderValue-50)/24
xy=getCubicBezierXYatPercent({x:250,y:120},{x:290,y:-40},{x:300,y:200},{x:400,y:150},percent);
}
else {
var percent=(sliderValue-75)/25
xy=getLineXYatPercent({x:400,y:150},{x:500,y:90},percent);
}
drawRect(xy,"red");
}
// draw tracking rect at xy
function drawRect(point,color){
ctx.fillStyle="cyan";
ctx.strokeStyle="gray";
ctx.lineWidth=3;
ctx.beginPath();
ctx.rect(point.x-13,point.y-8,25,15);
ctx.fill();
ctx.stroke();
}
// draw tracking dot at xy
function drawDot(point,color){
ctx.fillStyle=color;
ctx.strokeStyle="black";
ctx.lineWidth=3;
ctx.beginPath();
ctx.arc(point.x,point.y,8,0,Math.PI*2,false);
ctx.closePath();
ctx.fill();
ctx.stroke();
}
// line: percent is 0-1
function getLineXYatPercent(startPt,endPt,percent) {
var dx = endPt.x-startPt.x;
var dy = endPt.y-startPt.y;
var X = startPt.x + dx*percent;
var Y = startPt.y + dy*percent;
return( {x:X,y:Y} );
}
// quadratic bezier: percent is 0-1
function getQuadraticBezierXYatPercent(startPt,controlPt,endPt,percent) {
var x = Math.pow(1-percent,2) * startPt.x + 2 * (1-percent) * percent * controlPt.x + Math.pow(percent,2) * endPt.x;
var y = Math.pow(1-percent,2) * startPt.y + 2 * (1-percent) * percent * controlPt.y + Math.pow(percent,2) * endPt.y;
return( {x:x,y:y} );
}
// cubic bezier percent is 0-1
function getCubicBezierXYatPercent(startPt,controlPt1,controlPt2,endPt,percent){
var x=CubicN(percent,startPt.x,controlPt1.x,controlPt2.x,endPt.x);
var y=CubicN(percent,startPt.y,controlPt1.y,controlPt2.y,endPt.y);
return({x:x,y:y});
}
// cubic helper formula at percent distance
function CubicN(pct, a,b,c,d) {
var t2 = pct * pct;
var t3 = t2 * pct;
return a + (-a * 3 + pct * (3 * a - a * pct)) * pct
+ (3 * b + pct * (-6 * b + b * 3 * pct)) * pct
+ (c * 3 - c * 3 * pct) * t2
+ d * t3;
}
}); // end $(function(){});
</script>
</head>
<body>
<canvas id="canvas" width=600 height=300></canvas>
</body>
</html>
答案 1 :(得分:19)
如果您要使用画布的内置Bezier曲线,您仍需要自己进行数学运算。
您可以使用此cardinal spline的实现,并为您预先返回所有返回的点数。
使用的一个例子是这个小香肠 - 移动沿着斜坡移动(由上面的基数样条生成):
Full demo here (随意剪切和复制)。
你需要的主要内容是你有点数组是要找到你想要用于对象的两个点。这将为我们提供对象的角度:
cPoints = quantX(pointsFromCardinalSpline); //see below
//get points from array (dx = current array position)
x1 = cPoints[dx];
y1 = cPoints[dx + 1];
//get end-points from array (dlt=length, must be an even number)
x2 = cPoints[dx + dlt];
y2 = cPoints[dx + dlt + 1];
为避免在陡峭的斜坡上拉伸,我们会根据角度重新计算长度。为了获得近似角度,我们使用原始终点来获得一个角度,然后我们根据所需长度和这个角度计算一条新的线条长度:
var dg = getLineAngle(x1, y1, x2, y2);
var l = ((((lineToAngle(x1, y2, dlt, dg).x - x1) / 2) |0) * 2);
x2 = cPoints[dx + l];
y2 = cPoints[dx + l + 1];
现在我们可以通过从y位置减去它的垂直高度来绘制斜坡上的“汽车”。
你会注意到这样做的是“汽车”以不同的速度移动。这是由于基数样条的插值。
我们可以将其平滑,因此通过量化x轴,速度看起来更均匀。它仍然不是完美的,因为在陡峭的斜坡上,点之间的y距离将大于平面上的距离 - 我们确实需要二次量化,但为此我们只做x轴。
这为我们提供了一个新的数组,每个x位置都有新的点:
function quantX(pts) {
var min = 99999999,
max = -99999999,
x, y, i, p = pts.length,
res = [];
//find min and max of x axis
for (i = 0; i < pts.length - 1; i += 2) {
if (pts[i] > max) max = pts[i];
if (pts[i] < min) min = pts[i];
}
max = max - min;
//this will quantize non-existng points
function _getY(x) {
var t = p,
ptX1, ptX2, ptY1, ptY2, f, y;
for (; t >= 0; t -= 2) {
ptX1 = pts[t];
ptY1 = pts[t + 1];
if (x >= ptX1) {
//p = t + 2;
ptX2 = pts[t + 2];
ptY2 = pts[t + 3];
f = (ptY2 - ptY1) / (ptX2 - ptX1);
y = (ptX1 - x) * f;
return ptY1 - y;
}
}
}
//generate new array per-pixel on the x-axis
//note: will not work if curve suddenly goes backwards
for (i = 0; i < max; i++) {
res.push(i);
res.push(_getY(i));
}
return res;
}
我们需要的另外两个函数是计算一条线的角度,另一个是根据角度和长度计算终点:
function getLineAngle(x1, y1, x2, y2) {
var dx = x2 - x1,
dy = y2 - y1,
th = Math.atan2(dy, dx);
return th * 180 / Math.PI;
}
function lineToAngle(x1, y1, length, angle) {
angle *= Math.PI / 180;
var x2 = x1 + length * Math.cos(angle),
y2 = y1 + length * Math.sin(angle);
return {x: x2, y: y2};
}