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 Swedberg
www.englishrules.com
www.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-selectors/
--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
