I don't think I stated my problem fully.
What I'm trying to do it process the elements recursively, so using
the <ul>'s as an example:
1. Find the shallowest <ul>'s (Karl's solutions would work here).
2. For each <ul> found:
3. Add content to the <ul>.
4. Find the next shallowest <ul> within each <ul>.
5. Recurse from step 2 (until there are no more deeper <ul>'s).
var dostuff = function(elem) {
$(elem).findShallowest('ul').each(function() {
// do stuff, add content
dostuff(this);
});
}
It's important that the elements are processed in this manner,
as content is added which may include more <ul>'s which in turn
need processing.
Also the exact structure is not known, so I can't rely on child '>'
ops.
I think it could be done by stopping find() from going any deeper
once it finds match, but from looking at the jQuery src, find() is a
little baffling.
Thanks for help so far
- Mark.
On Nov 17, 7:28 pm, Karl Swedberg <[EMAIL PROTECTED]> wrote:
> Hi Mark,
>
> I took a look at your original post again, and from your description
> there it looks like this would work:
>
> $('ul').filter(function() {
> return !$(this).parents('ul').length;
>
> })
>
> --Karl
>
> ____________
> Karl Swedbergwww.englishrules.comwww.learningjquery.com
>
> On Nov 17, 2008, at 10:58 AM, Mark Gibson wrote:
>
>
>
> > The trouble is that the elements i'm looking for are not necessarily
> > direct children of the container,
> > so > isn't going to work. What I really need is to stop jQuery from
> > searching any deeper once it
> > finds the first <ul>, but continue the search at the same and
> > shallower levels.
> > Is there a way to do this with a $.expr[':'] plugin?
>
> > On Nov 11, 11:24 am, Klaus Hartl <[EMAIL PROTECTED]> wrote:
> >> That won't work. :first-child will give you *all* uls that are the
> >> first child of any element:
>
> >> <div id="container">
> >> <ul><!-- is first child -->
> >> <li>
> >> <ul> ... </ul><!-- is first child -->
> >> </li>
> >> </ul>
> >> </div>
>
> >> If you need the topmost ul use the ">" combinator:
>
> >> $('#container > u')
>
> >> Oh, jQuery has some documentation about
> >> selectors:http://docs.jquery.com/Selectors
>
> >> En plus knowing the spec doesn't hurt
> >> either:http://www.w3.org/TR/CSS21/selector.htmlhttp://www.w3.org/TR/css3-sel...
>
> >> --Klaus
>
> >> On 11 Nov., 02:07, "Hector Virgen" <[EMAIL PROTECTED]> wrote:
>
> >>> Oops, it's actually $('#container ul:first-child')
> >>> Here's a nifty page of CSS3
> >>> selectors:http://www.w3.org/TR/2001/CR-css3-selectors-20011113/#selectors
>
> >>> -Hector
>
> >>> On Mon, Nov 10, 2008 at 5:06 PM, Hector Virgen
> >>> <[EMAIL PROTECTED]> wrote:
> >>>> $('#container ul:first') should give you the first UL in the
> >>>> container
> >>>> -Hector
>
> >>>> On Mon, Nov 10, 2008 at 4:50 PM, Mark Gibson
> >>>> <[EMAIL PROTECTED]> wrote:
>
> >>>>> Anyone know how I can find the topmost elements of a certain
> >>>>> type, eg.
>
> >>>>> Say I have several <ul> lists on a page, which in turn contain
> >>>>> sub-
> >>>>> lists, and so on.
> >>>>> Is there a selector I can use to find all the topmost <ul>
> >>>>> elements.
>
> >>>>> I've been raking my brains over this, and can't see a way with css
> >>>>> style selectors,
> >>>>> have i overlooked something?
>
> >>>>> Ideally I'd like to do:
> >>>>> $('ul:topmost')
>
> >>>>> or:
> >>>>> $('#container ul:topmost')
>
> >>>>> Cheers
> >>>>> - Mark
