HTML5画布 - 旋转对象而不移动坐标
use*_*365 22 javascript html5-canvas
我有的是这个:
我想要的是旋转红色矩形,例如20度,但这是我最终得到的:
如您所见,矩形完美旋转,但它已移动并且不再与黑色物体匹配.
我的代码是:
context.save();
context.rotate(angle * Math.PI / 180); // in the screenshot I used angle = 20
context.translate(angle * 4, 2);
context.fillStyle = "red";
context.fillRect(x + 14, y - 16, 5, 16);
context.restore();
我想让矩形转动而不移动它的位置.
谢谢.
mar*_*rkE 53
听起来你想围绕它的中心点旋转矩形.
红色矩形是原始的,黄色矩形围绕中心点旋转.
要做到这一点,你需要context.translate先旋转到矩形的中心点然后再旋转.
// move the rotation point to the center of the rect
ctx.translate( x+width/2, y+height/2 );
// rotate the rect
ctx.rotate(degrees*Math.PI/180);
请注意,上下文现在处于旋转状态.
这意味着绘制位置[0,0]在视觉上为[x +宽度/ 2,y +高度/ 2].
因此,您必须在[-width/2,-height/2]处绘制旋转的矩形(而不是原始未旋转的x/y).
// draw the rect on the transformed context
// Note: after transforming [0,0] is visually [-width/2, -height/2]
// so the rect needs to be offset accordingly when drawn
ctx.rect( -width/2, -height/2, width,height);
这是代码和小提琴:http: //jsfiddle.net/m1erickson/z4p3n/
<!doctype html>
<html>
<head>
<link rel="stylesheet" type="text/css" media="all" href="css/reset.css" /> <!-- reset css -->
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<style>
body{ background-color: ivory; }
canvas{border:1px solid red;}
</style>
<script>
$(function(){
var canvas=document.getElementById("canvas");
var ctx=canvas.getContext("2d");
var startX=50;
var startY=80;
// draw an unrotated reference rect
ctx.beginPath();
ctx.rect(startX,startY,100,20);
ctx.fillStyle="blue";
ctx.fill();
// draw a rotated rect
drawRotatedRect(startX,startY,100,20,45);
function drawRotatedRect(x,y,width,height,degrees){
// first save the untranslated/unrotated context
ctx.save();
ctx.beginPath();
// move the rotation point to the center of the rect
ctx.translate( x+width/2, y+height/2 );
// rotate the rect
ctx.rotate(degrees*Math.PI/180);
// draw the rect on the transformed context
// Note: after transforming [0,0] is visually [x,y]
// so the rect needs to be offset accordingly when drawn
ctx.rect( -width/2, -height/2, width,height);
ctx.fillStyle="gold";
ctx.fill();
// restore the context to its untranslated/unrotated state
ctx.restore();
}
}); // end $(function(){});
</script>
</head>
<body>
<canvas id="canvas" width=300 height=300></canvas>
</body>
</html>
- @ user2479365 markE提供了一个非常详细的答案,这是一个很好的评论代码.只需根据您的需要调整`translate()`.你可能想要`-height`而不是`-height/2`或类似的东西. (7认同)
小智 10
您可以使用一个简单的函数来执行此操作,作为翻译和旋转的替代方法:
此功能可让您绘制一条从x和y开始的线,并以角度结束:
function lineToAngle(ctx, x1, y1, length, angle) {
angle *= Math.PI / 180;
var x2 = x1 + length * Math.cos(angle),
y2 = y1 + length * Math.sin(angle);
ctx.moveTo(x1, y1);
ctx.lineTo(x2, y2);
return {x: x2, y: y2};
}
用法:
lineToAngle(ctx, x, y, length, angle);
演示
var canvas = document.querySelector('canvas'),
ctx = canvas.getContext('2d'),
x = 100, y =50, length = 40, angle = 0, dlt = -2;
(function animate() {
//clear
ctx.clearRect(0, 0, canvas.width, canvas.height);
ctx.beginPath();
lineToAngle(ctx, x, y, length, angle);
ctx.lineWidth = 10;
ctx.stroke();
angle += dlt;
if (angle < -90 || angle > 0) dlt = -dlt;
requestAnimationFrame(animate);
})();
function lineToAngle(ctx, x1, y1, length, angle) {
angle *= Math.PI / 180;
var x2 = x1 + length * Math.cos(angle),
y2 = y1 + length * Math.sin(angle);
ctx.moveTo(x1, y1);
ctx.lineTo(x2, y2);
return {x: x2, y: y2};
}body {background-color:#555557}
canvas {background:#ddd;border:1px solid #000}<canvas></canvas>