hmahonen — Sat, 17 Feb 2001 10:34:04 +0000

Since the basic JavaScript method for displaying images is rather simple, it doesn't need any modification at all.

As the mouse goes over a link (a picture) you want to display another picture at another place.

Hard to explain, but let's give it a shot. As you have something like:

<a href="#"><img src="link.gif" width="100" height="20" alt="The link which changes another picture while onmouseover" border="0" onmouseover="change('<strong>wombat</strong>', 'down')" onmouseout="change('<strong>wombat</strong>', 'up')"></a>

and the image, which needs to be changed..

<img src="imagetobechanged.gif" <strong>name="wombat"</strong> width="200" height="50" alt="" border="0">

As you can see, normally we just would name the picture between A-tags and the onmouseover and -out event's would change it's image respectivily while onmouseover, but now we change the another picture.

I don't take any responsibility whatsoever about does this even work or do I have any clue about the matter, but as far as I can think this is the way it goes. Never tested it though

Update: it does work after all, just tested.