三色鼠標(biāo)跟隨代碼特效

2024-09-05 12:22:22

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供稿：網(wǎng)友

一個(gè)鼠標(biāo)跟隨特效，鼠標(biāo)不管移上哪里，都會(huì)有一個(gè)三色的彩條相隨，鼠標(biāo)不再孤單。彩條基于Js代碼生成，是一個(gè)數(shù)組的完美應(yīng)用展示，對(duì)數(shù)組的學(xué)習(xí)是個(gè)很好的參考，鼠標(biāo)跟隨的效果也是許多朋友喜歡的，再些分享給大家。

<HTML><HEAD><TITLE>三色條跟隨鼠標(biāo)的效果</TITLE><META http-equiv=Content-Type content="text/html; charset=gb2312"></HEAD><BODY bgColor=#000000><SCRIPT language=JavaScript>var a_Colour='ff0000';var b_Colour='ffffff';var c_Colour='0000ff';var Size=50;var YDummy=new Array(),XDummy=new Array(),xpos=0,ypos=0,ThisStep=0;step=0.2;if (document.layers){window.captureEvents(Event.MOUSEMOVE);function nsMouse(evnt){xpos = window.pageYOffset+evnt.pageX+6;ypos = window.pageYOffset+evnt.pageY+16;}window.onMouseMove = nsMouse;}else if (document.all){function ieMouse(){xpos = document.body.scrollLeft+event.x+6;ypos = document.body.scrollTop+event.y+16;}document.onmousemove = ieMouse;}function swirl(){for (i = 0; i < 3; i++) {   YDummy[i]=ypos+Size*Math.cos(ThisStep+i*2)*Math.sin((ThisStep+i*25)/2);  XDummy[i]=xpos+Size*Math.sin(ThisStep+i*2)*Math.sin((ThisStep+i*25)/2)*Math.sin(ThisStep/4); }ThisStep+=step;setTimeout('swirl()',10);}var amount=10;if (document.layers){for (i = 0; i < amount; i++){document.write('<layer name=nsa'+i+' top=0 left=0 width='+i/2+' height='+i/2+' bgcolor='+a_Colour+'></layer>');document.write('<layer name=nsb'+i+' top=0 left=0 width='+i/2+' height='+i/2+' bgcolor='+b_Colour+'></layer>');document.write('<layer name=nsc'+i+' top=0 left=0 width='+i/2+' height='+i/2+' bgcolor='+c_Colour+'></layer>');}}else if (document.all){document.write('<div   class="a44otml"   id="ODiv" style="position:absolute;top:0px;left:0px">'+'<div   class="gom00e7"   id="IDiv" style="position:relative">');for (i = 0; i < amount; i++){document.write('<div   class="4xurpry"   id=x style="position:absolute;top:0px;left:0px;width:'+i/2+';height:'+i/2+';background:'+a_Colour+';font-size:'+i/2+'"></div>');document.write('<div   class="mbk2ekt"   id=y style="position:absolute;top:0px;left:0px;width:'+i/2+';height:'+i/2+';background:'+b_Colour+';font-size:'+i/2+'"></div>');document.write('<div   class="oovjffx"   id=z style="position:absolute;top:0px;left:0px;width:'+i/2+';height:'+i/2+';background:'+c_Colour+';font-size:'+i/2+'"></div>');}document.write('</div></div>');}function prepos(){var ntscp=document.layers;var msie=document.all;if (document.layers){for (i = 0; i < amount; i++){ if (i < amount-1)  { ntscp['nsa'+i].top=ntscp['nsa'+(i+1)].top;ntscp['nsa'+i].left=ntscp['nsa'+(i+1)].left; ntscp['nsb'+i].top=ntscp['nsb'+(i+1)].top;ntscp['nsb'+i].left=ntscp['nsb'+(i+1)].left; ntscp['nsc'+i].top=ntscp['nsc'+(i+1)].top;ntscp['nsc'+i].left=ntscp['nsc'+(i+1)].left; } else   { ntscp['nsa'+i].top=YDummy[0];ntscp['nsa'+i].left=XDummy[0]; ntscp['nsb'+i].top=YDummy[1];ntscp['nsb'+i].left=XDummy[1]; ntscp['nsc'+i].top=YDummy[2];ntscp['nsc'+i].left=XDummy[2]; }}}else if (document.all){for (i = 0; i <  amount; i++){ if (i < amount-1)  { msie.x[i].style.top=msie.x[i+1].style.top;msie.x[i].style.left=msie.x[i+1].style.left; msie.y[i].style.top=msie.y[i+1].style.top;msie.y[i].style.left=msie.y[i+1].style.left; msie.z[i].style.top=msie.z[i+1].style.top;msie.z[i].style.left=msie.z[i+1].style.left; } else  { msie.x[i].style.top=YDummy[0];msie.x[i].style.left=XDummy[0]; msie.y[i].style.top=YDummy[1];msie.y[i].style.left=XDummy[1]; msie.z[i].style.top=YDummy[2];msie.z[i].style.left=XDummy[2]; }}}setTimeout("prepos()",10);}function Start(){swirl(),prepos()}window.onload=Start;</SCRIPT></BODY></HTML>

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