嘿傢伙,我想我需要一組新的眼睛來幫助我看看我的代碼,這不符合我的預期。對不起,問題的相對容易,但我真的沒有用Javascript編碼。反正:代碼不按預期工作
基本上我想要做的是讓圖像更改爲鼠標上的倒轉版本。到目前爲止,我只用ID「crew1」測試圖像。下面的代碼:
<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>
和有關HTML代碼:
<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>
對不起,壞的格式,但由於某種原因,當我把它們放在一個不顯示腳本和DIV的結束標記新隊。
在此先感謝您的幫助。
什麼完全按預期不工作?你能詳細說明一下嗎? – 2010-07-14 18:03:39