嘿家伙,我想我需要一组新的眼睛来帮助我看看我的代码,这不符合我的预期。对不起,问题的相对容易,但我真的没有用Javascript编码。反正:代码不按预期工作
基本上我想要做的是让图像更改为鼠标上的倒转版本。到目前为止,我只用ID“crew1”测试图像。下面的代码:
<script type="text/javascript">
function imageChange(oldID) {
var a = "http://www.affiliateskeptic.com/wp-content/uploads/2010/06/face_placeholder_new.jpg";
var b = "http://www.affiliateskeptic.com/wp-content/uploads/2010/06/face_placeholder_new.jpg";
var c = "http://www.affiliateskeptic.com/wp-content/uploads/2010/06/face_placeholder_new.jpg";
var d = "http://www.affiliateskeptic.com/wp-content/uploads/2010/06/face_placeholder_new.jpg";
switch(oldID) {
case "crew1":
document.getElementById(oldID).src=a;
break;
case "crew2":
document.getElementById(oldID).src=b;
break;
case "crew3":
document.getElementById(oldID).src=c;
break;
case "crew4":
document.getElementById(oldID).src=d;
break;
}
}</script>
和有关HTML代码:
<div>
<img onmouseover="imageChange("crew1")" id="crew1" src="http://www.affiliateskeptic.com/wp-content/uploads/2010/06/face_placeholder.jpg" alt="Picture of Crew 1" width="224" height="235">
<p>Crew 1 character description.</p></div>
对不起,坏的格式,但由于某种原因,当我把它们放在一个不显示脚本和DIV的结束标记新队。
在此先感谢您的帮助。
什么完全按预期不工作?你能详细说明一下吗? – 2010-07-14 18:03:39