fadeIn/fadeOut can take a second parameter, a callback. I would
assume in your case that the $('app') is hidden and you are fading it
in slowly, but you don't wait until that fade in is complete and have
it fade out, therefore appearing as its doing nothing at all.
$("#app").fadeIn("slow", function() { $('app').fadeOut("slow"); });
On Jul 15, 12:31 am, viewsonic712 <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I would like to fade 4 images with jQuery, 1.jpg, 2.jpg. 3.jpg. and
> 4.jpg
>
> I have the following code, but no luck yet :(
>
> <script type="text/javascript" src="jquery.js"></script>
> <script type="text/javascript">
>
> $(document).ready(function(){
>
> for(i=1; i<=4; i++) {
>
> $("#app").html("<img id='yass"+i+"' src='"+i+".jpg'>")
> $("#app").fadeIn("slow");
> $("#app").fadeOut("slow");
>
> }
>
> });
>
> </script>
