thank you but it doesnt work, your code shows a row of 3 images that
fade out at the same time; what i need is one image in place then this
image fades out to the second one and the third one:
the document i need:
<body>
<image>
</body>
your code:
<body>
<image><image><image>
</body>
On Dec 31, 3:12 am, Karl Swedberg <[EMAIL PROTECTED]> wrote:
> On Dec 30, 2007, at 10:31 AM, Jeroen wrote:
>
>
>
> > On Dec 30, 2007 12:16 PM, tetrix <[EMAIL PROTECTED]> wrote:
>
> >> so no one knows how to do this simple thing?
>
> > If you want to perform the same function on more then one element,
> > just assign a
> > common class and use the appropriate jQuery selector.
>
> > <img class="fade" src="x.jpg">
> > <img class="fade" src="y.jpg">
> > <img class="fade" src="z.jpg">
>
> > $(window).load( function () {
> > $('.fade').fadeOut(4000);
> > })
>
> > This should work afaik.
>
> > --
> > Jeroen
>
> Or, enclose them in a containing div. Something like this, perhaps:
>
> <div id="fade">
> <img src="x.jpg">
> <img src="y.jpg">
> <img src="z.jpg">
> </div>
>
> and ...
>
> $(window).load( function () {
> $('#fade img').fadeOut(4000);
> })
>
> Lots of different ways. :-)
>
> --Karl
> _________________
> Karl Swedbergwww.englishrules.comwww.learningjquery.com
