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Javascript HTML5 canvas implements Chinese Valentine's Day 3D rose effect code_javascript skills

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Release: 2016-05-16 15:47:33
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The example in this article describes how to use javascript HTML5 canvas to achieve the 3D rose effect on Chinese Valentine’s Day. Share it with everyone for your reference. The details are as follows:

The rose drawing below uses HTML 5 canvas, so your browser needs to support HTML 5. Personally, I still recommend chrome. This effect will also be slightly stuck under Firefox.

Rendering:

Demo address: http://demo.jb51.net/js/2015/js-flower-canvas.html

The specific code is as follows:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd">
<html>
 <head>
 <title>3D玫瑰花</title>
 <meta name="Generator" content="EditPlus">
 <meta name="Author" content="">
 <meta name="Keywords" content="">
 <meta name="Description" content="">
 </head>
 <body>
 情人节快到了,这里送大家一枝玫瑰,无论是有对象还是没对象的朋友们,情人节快乐~
下面的玫瑰绘制用到了HTML 5的canvas,所以你的浏览器需要支持HTML 5。个人还是比较推荐chrome,这个效果在Firefox下也会稍卡。
<div id="demo" style="width:520; height:500px;"><canvas id="c" height="500" width="500"></canvas></div>
<script> 
  var b = document.body;
  var c = document.getElementsByTagName('canvas')[0];
  var a = c.getContext('2d');
  var canvas = document.getElementsByTagName('canvas')[0];
  var ctx = canvas.getContext('2d');
  document.body.clientWidth; 
  with(m=Math)C=cos,S=sin,P=pow,R=random;
  c.width=c.height=f=500;h=-250;
  function p(a,b,c){
    if(c>60)
      return[S(a*7)*(13+5/(.2+P(b*4,4)))-S(b)*50,b*f+50,625+C(a*7)*(13+5/(.2+P(b*4,4)))+b*400,a*1-b/2,a];
    A=a*2-1;
    B=b*2-1;
    if(A*A+B*B<1)
    {
      if(c>37)
      {
        n=(j=c&1)&#63;6:4;o=.5/(a+.01)+C(b*125)*3-a*300;
        w=b*h;
        return[o*C(n)+w*S(n)+j*610-390,o*S(n)-w*C(n)+550-j*350,1180+C(B+A)*99-j*300,.4-a*.1+P(1-B*B,-h*6)*.15-a*b*.4+C(a+b)/5+P(C((o*(a+1)+(B>0&#63;w:-w))/25),30)*.1*(1-B*B),o/1e3+.7-o*w*3e-6]
      }
      if(c>32)
      {
        c=c*1.16-.15;o=a*45-20;w=b*b*h;z=o*S(c)+w*C(c)+620;
        return[o*C(c)-w*S(c),28+C(B*.5)*99-b*b*b*60-z/2-h,z,(b*b*.3+P((1-(A*A)),7)*.15+.3)*b,b*.7]
      }
      o=A*(2-b)*(80-c*2);
      w=99-C(A)*120-C(b)*(-h-c*4.9)+C(P(1-b,7))*50+c*2;z=o*S(c)+w*C(c)+700;
      return[o*C(c)-w*S(c),B*99-C(P(b, 7))*50-c/3-z/1.35+450,z,(1-b/1.2)*.9+a*.1, P((1-b),20)/4+.05]
    }
  }
  var draw = setInterval('for(i=0;i<1e4;i++)if(s=p(R(),R(),i%46/.74)){z=s[2];x=~~(s[0]*f/z-h);y=~~(s[1]*f/z-h);if(!m[q=y*f+x]|m[q]>z)m[q]=z,a.fillStyle="rgb("+~(s[3]*h)+","+~(s[4]*h)+","+~(s[3]*s[3]*-80)+")",a.fillRect(x,y,1,1)}',0);
  var demo = document.getElementById('demo');
  function redraw(){
    /*
    var d_c = document.createElement("canvas");
    d_c.setAttribute("id","c");
    d_c.setAttribute("width","520");
    d_c.setAttribute("height","500");
    demo.appendChild(d_c);
    */
    draw = setInterval('for(i=0;i<1e4;i++)if(s=p(R(),R(),i%46/.74)){z=s[2];x=~~(s[0]*f/z-h);y=~~(s[1]*f/z-h);if(!m[q=y*f+x]|m[q]>z)m[q]=z,a.fillStyle="rgb("+~(s[3]*h)+","+~(s[4]*h)+","+~(s[3]*s[3]*-80)+")",a.fillRect(x,y,1,1)}',0);
    //alert(d_c);
  }
  function clear_canvas()
  {
    ctx.clearRect(0,0,520,500);
    //canvas.parentNode.removeChild(canvas);  //删除
  }
  function stop_draw(obj){
    clearInterval(obj);
  }
</script>
 </body>
</html>

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I hope this article will be helpful to everyone’s JavaScript programming design.

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