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关于JAVASCRIPT的一个小题目 求解解决思路

热度:5153   发布时间:2013-02-25 21:21:22.0
关于JAVASCRIPT的一个小题目 求解
<html>
<title></title>
<head>
<script>
var i = 0;
var j = 0;
function move(){
var kuan = window.document.body.offsetWidth;
  var piao = document.getElementById('a');
setTimeout("move()",100);
i+=10;
piao.style.pixelLeft=i
</script>
</head>
<body onLoad="move();">
<div id="a" style="position:absolute;"><img src="1.jpg" height="50"></div>
</body>
</html>

做一个广告漂浮的。 要让一个图片从左到右边漂浮飘到浏览器边框的时候再往回飘 来回循环 想了很多办法 实在不行了。小弟初学,真心求解

------解决方案--------------------------------------------------------
<script language="JavaScript">
var pic="图片路径+文件名";
var alt="";
var speed=120;
var light=0;
var stop=0;
var num=0;
var brOK=false;
var mie=false;
var aver=parseInt(navigator.appVersion.substring(0,1));
var aname=navigator.appName;
var vmin=2;
var vmax=5;
var vr=2;
var timer1;
var allyes;
var ns = (document.layers) ? 1 : 0;
var ie = (document.all) ? 1 : 0;
var bFade=false;
function checkbrOK(){
if(aname.indexOf("Internet Explorer")!=-1){
if(aver>=4) brOK=navigator.javaEnabled();
mie=true;
 }
 if(aname.indexOf("Netscape")!=-1){
if(aver>=4) brOK=navigator.javaEnabled();
 }
}
function Chip(chipname,width,height){
 this.named=chipname;
 this.vx=vmin+vmax*Math.random();
 this.vy=vmin+vmax*Math.random();
 this.w=width;
 this.h=height;
 this.xx=10;
 this.yy=10;
 this.timer1=null;
}
function movechip(chipname){
if (stop!=1){
if(brOK){
 eval("chip="+chipname);
 if(!mie){
pageX=window.pageXOffset;
pageW=window.innerWidth;
pageY=window.pageYOffset;
pageH=window.innerHeight;
 }
 else{
pageX=window.document.body.scrollLeft;
pageW=window.document.body.offsetWidth;
pageY=window.document.body.scrollTop;
pageH=window.document.body.offsetHeight;
 } 
 chip.xx=chip.xx+chip.vx;
 chip.yy=chip.yy+chip.vy;
 chip.vx+=vr*(Math.random()-0.5);
 chip.vy+=vr*(Math.random()-0.5);
 if(chip.vx>(vmax+vmin)) chip.vx=(vmax+vmin)*2-chip.vx;
 if(chip.vx<(-vmax-vmin)) chip.vx=(-vmax-vmin)*2-chip.vx;
 if(chip.vy>(vmax+vmin)) chip.vy=(vmax+vmin)*2-chip.vy;
 if(chip.vy<(-vmax-vmin)) chip.vy=(-vmax-vmin)*2-chip.vy;
 if(chip.xx<=pageX){
chip.xx=pageX;
chip.vx=vmin+vmax*Math.random();
 }
 if(chip.xx>=pageX+pageW-chip.w){
chip.xx=pageX+pageW-chip.w;
chip.vx=-vmin-vmax*Math.random();
 }
 if(chip.yy<=pageY){
chip.yy=pageY;
chip.vy=vmin+vmax*Math.random();
 }
 if(chip.yy>=pageY+pageH-chip.h){
chip.yy=pageY+pageH-chip.h;
chip.vy=-vmin-vmax*Math.random();
 }
 if(!mie){
eval("document."+chip.named+".top ="+chip.yy);
eval("document."+chip.named+".left="+chip.xx);
 } 
 else{
eval("document.all."+chip.named+".style.pixelLeft="+chip.xx);
eval("document.all."+chip.named+".style.pixelTop ="+chip.yy); 
 }
 chip.timer1=setTimeout("movechip('"+chip.named+"')",speed);
}
}
else {chip.timer1=setTimeout("movechip('"+chip.named+"')",speed);}
if (ie){
 if (light==1){
var nOpacity=oImg.filters.alpha.opacity;
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