Hello folks,
I'm looking for a script that will make a link list with onMouseOver show a thumbnail in the same spot on webpage. See example here. Thank you so much for taking the time to answer.
Best,
Dana O.
![]()
Hello folks,
I'm looking for a script that will make a link list with onMouseOver show a thumbnail in the same spot on webpage. See example here. Thank you so much for taking the time to answer.
Best,
Dana O.
![]()
Any rollover/mouseover image script should work.
Here is how to change it:
1. rollover images are applied to IMAGES-- apply it to the links instead. (<a> tag instead of <img> tag.)
2. There is no image at all of no link is hovered over. You can either leave a default image (or a "blank"-- white-- image), or you can use display:none when no link is active (onmouseout, and by default). Basically just add that action to the javascript function.
Daniel - Freelance Web Design | <?php?> | <html>| español | Deutsch | italiano | português | català | un peu de français | some knowledge of several other languages: I can sometimes help translate here on DD | Linguistics Forum
DanaOldfather (04-01-2010)
djr hit it on the head (as per usual) and I am a day late and a dollar short... again.
but, open forum right?
Have a look at Stu Nichol's CSS Play. Some GREAT menus there, and easily followed and adapted to your purpose.
For example the pop up image menu found here Instead of having the images appear with borders and such in different places, find the spot on your page and have them all appear in the same spot.
Or.. if you want a JS answer look at how the linked page you gave us does it:
The above goes in the <head> section... the following goes on the links:Code:<script language="javascript"> <!-- function showImage(image) { document.getElementById('imageC').src = image; } function hideImage() { document.getElementById('imageC').src = 'images/spacer30.gif'; } //--> </script>
Then have a div or such named id="imageC" where you want the images to appear.Code:<a href="gallery_of_artists/contemporary/bergholz.htm">david bergholz</a></span><br> <span id="choice3" style="cursor:hand" onmouseover="showImage('images/gallery_of_artists/tn/TN146_Biro.jpg');" onmouseout="hideImage();">
The page you referenced does it in table form, with the blank image as the last TD as such:
Code:<td width="146" valign="top" height="378"><span id="boxdescription"> <img src="images/spacer30.gif" id="imageC" width="146" height="146"></span> </td>
{CWoT - Riddle } {Freelance Copywriter} {Learn to Write}
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PHP Code:$result = mysql_query("SELECT finger FROM hand WHERE id=3");
echo $result;
DanaOldfather (04-01-2010)
Thanks so much! I'm really a beginner, and I really appreciate the time you took to break out all that code. I did view source on that page but wasn't sure if I was missing some scripting somewhere. One other question - the "images/spacer30.gif" is just a blank box that I will need to make, in say fireworks, and insert into my table right? If so, I think I get it. Thanks again!
Dana O.
That is correct. It is just a place holder so there is content there. That way the page doesn't jump and re-size when an image is inserted or removed via the hover.
{CWoT - Riddle } {Freelance Copywriter} {Learn to Write}
Follow Me on Twitter: @InkingHubris
PHP Code:$result = mysql_query("SELECT finger FROM hand WHERE id=3");
echo $result;
For example the pop up image menu found here Instead of having the images appear with borders and such in different places, find the spot on your page and have them all appear in the same spot.
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