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/炫麗的倒計時效果Canvas繪圖與動畫基礎
不明白X軸Y軸里面的數值怎么弄
2017-10-17
源自:炫麗的倒計時效果Canvas繪圖與動畫基礎 2-2
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<!doctype?html> <html?lang="zh-CN"> <head> ????<meta?charset="UTF-8"> ????<meta?name="viewport" ?content="width=device-width,?user-scalable=no,?initial-scale=1.0,?maximum-scale=1.0,?minimum-scale=1.0"> ????<meta?http-equiv="X-UA-Compatible"?content="ie=edge"> ????<title>七巧板</title> ????<style?type="text/css"> ????????#canvas?{ ????????????border:?1px?solid?#ccc; ????????????display:?block; ????????????margin:?auto; ????????} ????</style> </head> <body> <canvas?id="canvas"?width="400"?height="400"></canvas> <script?type="text/javascript"> ????var?tangram?=?[ ????????{p:?[{x:?0,?y:?0},?{x:?400,?y:?0},?{x:?200,?y:?200}],?color:?"#ff0f09"}, ????????{p:?[{x:?0,?y:?0},?{x:?0,?y:?400},?{x:?200,?y:?200}],?color:?"#feff0a"}, ????????{p:?[{x:?0,?y:?400},?{x:?200,?y:?400},?{x:?100,?y:?300}],?color:?"#ff09f7"}, ????????{p:?[{x:?100,?y:?300},?{x:?200,?y:?400},?{x:?300,?y:?300},{x:?200,?y:?200}],?color:?"#ffdc52"}, ????????{p:?[{x:200,y:400},?{x:400,y:400},?{x:400,y:200}],?color:?"#0608ff"}, ????????{p:?[{x:200,y:200},?{x:300,y:300},?{x:300,y:100}],?color:?"#30ff10"}, ????????{p:?[{x:300,y:300},?{x:300,y:100},?{x:400,y:0},?{x:400,y:200}],?color:?"#ffa41a"}, ????]; ????var?canvas?=?document.getElementById("canvas");?//?獲取canvas ?var?context?=?canvas.getContext("2d");??//?獲取2d繪圖環境 ?for?(var?i?=?0;?i?<?tangram.length;?i++)?{ ????????draw(tangram[i],?context); ????} ????function?draw(piece,?context)?{ ????????context.beginPath(); ????????context.closePath(); ????????context.moveTo(piece.p[0].x,?piece.p[0].y); ????????for?(var?i?=?0;?i?<?piece.p.length;?i++)?{ ????????????context.lineTo(piece.p[i].x,?piece.p[i].y); ????????} ????????context.fillStyle?=?piece.color; ????????context.fill(); ????} </script> </body> </html>
照著老師那個坐標圖看,初始點(x,y),每個圓占用 2*(r+1)的寬高的正方形,現在求從左往右數第j個,從上往下數第i個圓的圓心:
customX = 2*(r+1)*j + (r+1);// j*每個圓占用的寬度 + 每個圓占用寬度/2
customY = 2*(r+1)*i + (r+1);//同理可證!
舉報
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2018-12-30
2017-10-18
照著老師那個坐標圖看,初始點(x,y),每個圓占用 2*(r+1)的寬高的正方形,現在求從左往右數第j個,從上往下數第i個圓的圓心:
customX = 2*(r+1)*j + (r+1);// j*每個圓占用的寬度 + 每個圓占用寬度/2
customY = 2*(r+1)*i + (r+1);//同理可證!