日期:2014-05-17  浏览次数:20604 次

鼠标经过图片时变成另外一张
想实现鼠标经过图片时显示另外一张图片,鼠标离去后显示另一张图片,
如下代码

<head runat="server">

    <title>无标题页</title>    

   <style>
      .sy{background-image:url(Image/menusy.jpg);}

      .sy1{background-image:url(Image/menuOnSY.jpg);}

      .zj{background-image:url(Image/menuGSGK.jpg);}

      .zj1{background-image:url(Image/menuOnGSGK.jpg);}

      .xw{background-image:url(Image/menuNews.jpg);}

      .xw1{background-image:url(Image/menuOnNews.jpg);}

      .cp{background-image:url(Image/menuCPZS.jpg);}

      .cp1{background-image:url(Image/menuOnCPZS.jpg);}

      .zs{background-image:url(Image/menuJSZR.jpg);}

      .zs1{background-image:url(Image/menuOnJSZR.jpg);}

 

      .rc{background-image:url(Image/menuGYKF.jpg);}

      .rc1{background-image:url(Image/menuOnGYKF.jpg);}


      .lx{background-image:url(Image/menuZLXZ.jpg);}

      .lx1{background-image:url(Image/menuOnZLXZ.jpg);}
  

   </style>

  
</head>

<body>

    <form id="form1" runat="server">

    <div>     

       <ul>

            <li style="width:113px; height:30px;
background-image:url(Image/menuGSGK.jpg); list-style-type:none; "
onmouseup="this.className='zj'" onmouseout="this.className='zj1'"> 

            </li>

        </ul>

    </div>

    </form>

</body>

</html>




高手看看那儿错了,实现不了,鼠标经过时没有反应

------解决方案--------------------
onmouseover
------解决方案--------------------
引用:
onmouseover

+++
------解决方案--------------------
写一个方法,然后在onmouseover里面调用~~
------解决方案--------------------
把这个 去掉 background-image: url(Image/menuGSGK.jpg);

也换成  用 class表示的 。原因是 background-image: url(Image/menuGSGK.jpg); 是 行内样式,级别最高 ,回覆盖其他设定