I sketched up a plugin for this. It's at http://ff6600.org/j/jquery.dig.js
Live example at http://jsbin.com/ugena/ It just steps throught the children with the tag given and executes a function for each one, passing it it's 'hierarchy level' in relation to the container where the plug-in was called. Any new children added in the process will be taken into account on the next step. hope it helps! - ricardo On Nov 26, 3:52 pm, Mark Gibson <[EMAIL PROTECTED]> wrote: > 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
