Just bear in mind that that code will "fade in" as soon as the SRC attribute is set - it won't wait for the image to load...could result in an empty box fading in, and then the image suddenly appearing when it loads.
you'll need to trigger a ".load" function for the image too; this should be possible to do outside / independently of the click or fadein (i.e. an independent function within $(document).ready by itself).
James wrote:
You need to place your content inside the callback function for fadeOut. The callback is executed after the content is faded out. $(document).ready(function() { $("#lakeview-gallery a").click(function(event) { $("#lakeview-placeholder").fadeOut(800, function() { // this content will start once the content is faded out var image = $(this).attr("href"); $("img#lakeview-placeholder").attr("src", image); $("#lakeview-placeholder").fadeIn(800); return false; }); }); }); On Apr 2, 11:10 am, stinkysGTI <jesse.gr...@gmail.com> wrote:There's a pretty simple image replacement I'm trying to do, but can't get the current image to fadeOut, and the new one to fadeIn. There's only 3 images to choose from and always will be. With those 2 lines commented out, it works, just obviously without any kind of fade. I'm fairly new to jQuery, so any help would be greatly appreciated. $(document).ready(function() { $("#lakeview-gallery a").click(function(event) { // $("#lakeview-placeholder").fadeOut(800); // $("#lakeview-placeholder").fadeIn(800); event.preventDefault(); var image = $(this).attr("href"); $("img#lakeview-placeholder").attr("src", image); }); }); The page can be seen here:http://www.lakeshoremontana.com/dev/floorplans.html ------------------------------------------------------------------------ No virus found in this incoming message.Checked by AVG - www.avg.com Version: 8.0.238 / Virus Database: 270.11.40/2039 - Release Date: 04/03/09 06:19:00
