服务器端button,用onmouseover、onmouseout来改变鼠标移动在button上时，改变背景的图片，从而改变button样式

Question

服务器端button,用onmouseover、onmouseout来改变鼠标移动在button上时，改变背景的图片，从而改变button样式 怎么写？？？？
onmouseover="this.Style.backgroundimage='url(images/button_hui1.gif)'" 这样写不对  应该怎写  望各位大侠指教啊.....

Answer 1

<input type=button style='background-image(url(a.gif))' onmouseover="this.style.backgroundImage='url(b.gif)'" onmouseout='this.style.backgroundImage='url(a.gif)''/>
 

Answer 2

<asp:Button ID="Button1" runat="server" Text="Button" 
style="background-image: url('b1.jpg')"  
onmouseover="this.style.backgroundImage='url(images/b3.jpg)'" onmouseout="this.style.backgroundImage='url(images/b1.jpg)'" 
/>

Answer 3

<style type="text/css"> 
.mouseOverStyle{
     background-image:url(images/bgbtn2.jpg);
 color:#cc0099;
 border:0px;
 margin:0px;
 padding:0px;
 height:23px;
 width:82px;
 font-size:14px;
  } 
  .mouseOutStyle{
     background-image:url(images/bgbtn1.jpg);
 color:#0000FF;
 border:0px;
 margin:0px;
 padding:0px;
 height:23px;
 width:82px;
 font-size:14px;
  }
</style>
<input name="btnLogin" type="button" class="mouseOutStyle" id="btnLogin" value=" 登 录" onMouseOver="this.className='mouseOverStyle'" onMouseOut="this.className='mouseOutStyle'">
</s