[jQuery] jcarousel issue

[jQuery] jcarousel issue

I tried to pull up the image without the ";see|" at the end and I get a
404. That is probably why it is only showing the alt text. You probably
have your images in a subdirectory and you need to change the path to
reflect that.
If you want to use the code as is, you need to name your thumbnail
images with an "_s.jpg" and medium images to "_m.jpg". These are the
naming conventions used by flickr. That is where the images are coming
from.
The main hurdle you need to overcome is finding the right path to your
images.
http://www.liniverse.com/see.jpg
Is a 404, but
http://static.flickr.com/66/199481236_dc98b5abb3_s.jpg
pulls up an image.
-Marshall
-----Original Message-----
From: discuss-bounces@jquery.com [mailto:discuss-bounces@jquery.com] On
Behalf Of Indigo
Sent: Monday, March 05, 2007 3:11 PM
To: jQuery Discussion.
Subject: Re: [jQuery] jcarousel issue
The code is exactly what you'll find by viewing the source at
http://sorgalla.com/projects/jcarousel/example_dynamic_ajax.html . I
only
changed the image and ajax.txt file paths.
If you, however, look at the content of example_dynamic_ajax.html,
you'll find
lines like these (pay attention to the names of the files) :
http://static.flickr.com/66/199481236_dc98b5abb3_s.jpg;Flower1|
http://static.flickr.com/75/199481072_b4a0d09597_s.jpg;Flower2|
My file names are simpler and look like these:
http://www.liniverse.com/see.jpg;see|
http://www.liniverse.com/me.jpg;me|
I've checked and double-checked the file paths and I find no problems
there,
But looking at the js in the example, I find this section that seems to
be
performing a replacement trick on the file names:
<snip>
.....
function getItemHTML(data)
{
var split = data.split(";");
var url = jQuery.trim(split[0]);
var title = jQuery.trim(split[1]);
var url_m = url.replace(/_s.jpg/g, '_m.jpg');
return '<a href="' + url_m + '" title="' + title + '"
class="thickbox"><img src="' + url + '" width="' + 75 + '" height="' +
75 +
'" alt="' + title + '" /></a>';
};
....
--</snip>
I'm not a coder, so I do not know how to customize the above code
snippet for
my specific case. i think the offending lines are these:
var url_m = url.replace(/_s.jpg/g, '_m.jpg');
return '<a href="' + url_m + '" title="' + title + '"
'll appreciate any help.
TIA