I think it is because you are fading it back before the content has loaded
try this:
$('#list a').each(function(i){
$(this).click(function(event){
$("#list").fadeOut("slow").load(this.href, function () {
$(this).fadeIn("slow");
});
return false;
});
});
Haven't tested it, but what it should do is execute the fadeIn AFTER
the content has loaded.
//Kristinn
On 3/24/07, Sebastián V. Würtz <[EMAIL PROTECTED]> wrote:
> Well maybe isnt the return my problem, really is the behaviour, why when i
> clicl in the link it disapear- "appear with the same content" and then
> appear within the data loaded via ajax? I mean i want the container only
> hide and show whitin the content loaded nothing more.
>
> Maybe my english is very bad, so sorry.
> Sebastian
>
>
> ----- Original Message -----
> From: "Evan" <[EMAIL PROTECTED]>
> To: <discuss@jquery.com>
> Sent: Friday, March 23, 2007 9:32 PM
> Subject: Re: [jQuery] return false
>
>
> By putting return false, you're preventing the default behaviour of
> the anchor, otherwise it would still fire like a normal link.
>
> Alternatively, you could use another element (like a span, styled like
> an anchor) to achieve the same effect.
>
> On Mar 24, 8:30 am, Sebastián V. Würtz <[EMAIL PROTECTED]> wrote:
> > I have a html like this
> >
> > <div id="list">
> > <ul>
> > <li><a href="data/cat1.htm">cat1</a></li>
> > ....
> > ...
> > </ul>
> >
> > The html
> >
> > $('#list a').each(function(i){
> > $(this).click(function(event){
> > $("#list").fadeOut("slow").load(this.href);
> > $("#list").fadeIn("slow");return false;
> >
> > })
> >
> > When i click in 1 option the "list" hide (fadeOut); load the page; appear
> > again the list; and then the page appear. If i dont put the "return false"
> > i got the empty page not like a "content"
> >
> > How i can fix it?
> >
> > Thx
> > Sebastián
> >
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