Looks like maybe you want
$(".label", this).show(1000)
or (nearly equivalent)
$(this).find(".label").show(1000)
- Richard
On Thu, Jan 15, 2009 at 4:58 AM, eXistent <[email protected]> wrote:
>
> Hi,
> is there some easy way to make something like this:
>
> $(document).ready(function() {
> $(".track").mouseover(function(){
> $(".track .label").show(1000);
> });
> });
>
> but only for each child element separately?
> I mean something like this:
>
> $(document).ready(function() {
> $(".track").mouseover(function(){
> $(this ".label").show(1000);
> });
> });
>
> Thanks for answers and tips.
>
