html5 canvas 画圆弧有锯齿怎么解决
2个回答
展开全部
Html5 Canvas 画椭圆有锯齿：因为在Canvas中整数坐标值对应的位置恰巧是屏幕象素点中间的夹缝，那么当按这样的坐标进行线条渲染时所要用到的就是夹缝两边的象素点，这样即便设置了lineWidth为1也将看到两个象素效果的线条，解决方法原象素点+0.5进行偏移。
下面是处理前后的效果比较：
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>canvasTest</title>
<script type="text/javascript" src="http://www.pyzy.net/Demo/html5_cancas_js/excanvas.js"></script>
<script type="text/javascript">
var MyCanvas = function(boxObj, width, height) {
//序号、计数
this.index = arguments.callee.prototype.Count = (arguments.callee.prototype.Count || 0) + 1;
var cvs = document.createElement("canvas");
cvs.id = "myCanvas" + this.index;
cvs.width = width || 800;
cvs.height = height || 600;
(boxObj || document.body).appendChild(cvs);
//excanvas框架中针对ie加载canvas延时问题手动初始化对象
if (typeof G_vmlCanvasManager != "undefined") G_vmlCanvasManager.initElement(cvs);
//2D画布对象
this.ctx = cvs.getContext("2d");
/* * 绘制线条
* @ops JSON对象，可按实际支持属性扩展，示例： { lineWidth:1,strokeStyle:'rgb(255,255,255)' }
* @dotXY:{ x:0, y:0 } ||[{ x:0, y:0 },{ x:0, y:0 }]
*/
this.drawLine = function(dotXY, ops) {
this.ctx.beginPath();
for (var att in ops) this.ctx[att] = ops[att];
dotXY = dotXY.constructor == Object ? [dotXY || { x: 0, y: 0}] : dotXY;
this.ctx.moveTo(dotXY[0].x, dotXY[0].y);
for (var i = 1, len = dotXY.length; i < len; i++) this.ctx.lineTo(dotXY[i].x, dotXY[i].y);
this.ctx.stroke();
};
};
window.onload=function(){
var c1 = new MyCanvas();
c1.drawLine([{ x: 10, y: 10 }, { x: 10, y: 200 }],{lineWidth:2,strokeStyle:'rgb(0,0,0)'});
c1.drawLine([{ x: 11, y: 10 }, { x: 11, y: 200 }],{lineWidth:2,strokeStyle:'rgb(255,255,255)'});
c1.drawLine([{ x: 100, y: 10 }, { x: 100, y: 200 }],{lineWidth:1,strokeStyle:'rgb(0,0,0)'}); //普通线
c1.drawLine([{ x: 200.5, y: 10 }, { x: 200.5, y: 200 }],{lineWidth:1,strokeStyle:'rgb(0,0,0)'}); //+0.5偏移
}
</script>
</head>
<body>
↓ 处理的 ↓ 普通的 ↓ +0.5偏移的<br />
</body>
</html>
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
