Maybe:
$("#container").children().not("#header, #content, #footer").empty();
--John
On Sun, Jan 4, 2009 at 2:05 PM, Dirceu Barquette
<dirceu.barque...@gmail.com> wrote:
> Hi all!
>
> Is there better way?
>
> var elem = $('#container')[0];
> var arr = ["header","content","footer"];
> jQuery.each(elem.childNodes,function(k,v) {
> if (jQuery.inArray(v.id, arr) < 0) {
> $(v).empty();
> }
> })
>
> thanks
>
> Dirceu Barquette
>
