分享程序员开发的那些事...
更新时间：2023-12-03 08:49:28
这是您解决问题的方法..！
Here is the solution for your problem..!
<HTML>
<HEAD>
<TITLE>Java Popup Window Example</TITLE>
<SCRIPT language="JavaScript" type="text/javascript">
var newwindow = ''
function popitup(url) {
if (newwindow.location && !newwindow.closed) {
newwindow.location.href = url;
newwindow.focus(); }
else {
newwindow=window.open(url,'htmlname','width=404,height=316,resizable=1');}
}
function tidy() {
if (newwindow.location && !newwindow.closed) {
newwindow.close(); }
}
</SCRIPT>
</HEAD>
<BODY onUnload="tidy()" >
<A HREF="javascript:popitup('gallery/templew.jpg')"><IMG SRC="gallery/templei.jpg" WIDTH="160" BORDER="0" HEIGHT="120" HSPACE="10" VSPACE="5" ALT="Decoration on the Sri Mariamman Temple" TITLE="Decoration on the Sri Mariamman Temple" ALIGN=left></A>
</BODY>
</HTML>
尝试这样可能会帮助您
使用夸脱字符串概念
在quary字符串中，您必须将image(asp控件)id传递到下一页，然后在下一页中检索该id并分配给image(asp控件)
try like this it may help you
use the quary string concept
in quary string u have to pass the image(asp control) id to the next page and in the next page retrive that id and assign to the image(asp control)