onmousemove="if (表格id.style.background=='a.gif'){ 
表格id.style.background='b.gif';else 表格id.style.background='a.gif'}"

解决方案 »

  1.   

    更正onmousermove 为onmouseover
      

  2.   

    onmousemove="表格id.style.background='b.gif';“
    onmouseout = "id.style.background='a.gif';"
      

  3.   

    <style>
    .pic1{background:url(http://expert.csdn.net/images/csdn.gif)}
    .pic2{background:url(http://www.csdn.net/images/ad/vsnet_120.gif)}
    </style>
    <table width=100 height=100 border=1>
    <tr>
    <td onmouseover="this.className='pic2'" onmouseout="this.className='pic1'" class="pic1">&nbsp;</td>
    </tr>
    </table>
      

  4.   

    to bencalie(Bencalie)
    你的方法很不错:)
    但鼠标的移动,游览器每次都会去网站上下载该背景图,显得很慢。
      

  5.   

    用图片数组
    然后调用就调用数组内容
    以上是类似写法ImageArray=["loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif","loads.gif","up_d.gif","down_d.gif","down_o.gif","up_o.gif"];
    function preLoad(path,obj){
    for(i=0;i<=obj.length;i++){
    this[i]=new Image(); 
    this[i].src=path+obj[i];
    }
    return this;
    }
    preb=new preLoad("image/",ImageArray);
      

  6.   

    我现在已经预先加载了那几张背景图,可是每次触发onmouseover,onmouseout事件,背景还是要重新下载一遍?????为什么呀!!!!
    效果如下:
    http://www.21cantonfair.com/newgb/zdzn/index.asp