Question

javascript问题请教高手

<TABLE id=standard_table height=0 cellSpacing=0 cellPadding=0 width=0
align=center border=0>
</TABLE>
<DIV id=LayerFloatingBn
style="Z-INDEX: 1; VISIBILITY: visible; POSITION: absolute; left: 850px; top: 60px; height: 281px; width: 130px;">
<TABLE width=123 height="281" border=0 cellPadding=0 cellSpacing=0>
<TR>
<TD><img src="images/ss.jpg" width="127" height="461" border="1" /></TD>
</TR>
</TABLE>
</DIV>
<SCRIPT language=javascript>
LayerFloatingBn.style.left = standard_table.offsetLeft + 850;
LayerFloatingBn.style.top = standard_table.offsetTop +60;
window.onscroll = WindScroll;
window.onresize = WindReset;
window.onload = WindScroll;
function WindScroll()
{
LayerFloatingBn.style.top = document.body.scrollTop;
if(document.body.scrollTop <= standard_table.offsetTop)
{
LayerFloatingBn.style.top = standard_table.offsetTop +60;
}
else
{
LayerFloatingBn.style.top = document.body.scrollTop +60;
}
LayerFloatingBn.style.visibility = "visible";
}
function WindReset()
{
LayerFloatingBn.style.left = standard_table.offsetLeft + 850;
LayerFloatingBn.style.top = standard_table.offsetTop +60;
}
</SCRIPT>
如上代码，请教高手，如何实现在点击图片“ss.jpg”的时候链接道另外一个网页。

Answer 1

<img src="images/ss.jpg" width="127" height="461" border="1" onclick="javascript:window.location='page.html'"/></TD>

Answer 2

把图片放在超链接里不就行了
<a href="#"><img src="#"/></a>

Answer 3

这么做

<img src="图片路径" onclick="javascript:window.location='www.163.com'"/>

Answer 4

<img src="图片路径" onclick="javascript:window.open('www.163.com')"/>

Answer 5

LS正解~要用JS的话就是对图像上某个区域的点击制作超连~