Hi
$($("E").children().get(1)).hide();
How can I hide all elements > 1
so I want hide all E except the first
Regards
PS : "eq" works fine too thanks Karl
2007/3/26, Karl Swedberg <[EMAIL PROTECTED]>:
Ah yes, of course. I'm glad you figured it out! I think part of the
problem with our solutions was that we didn't see your actual markup.
For future reference:
$('E:nth-child(n)') will select an E element that is the nth child of its
parent element.
$('E').children().get(n) will select an element that is the (n+1)th child
of the E element.
You should also be able to use .eq(n) if you want to keep the chain going
(returns the jQuery object rather than the DOM element):
$('E').children().eq(n)
Or you could do this:
$('E').children(':eq(n)')
Cheers,
--Karl
_________________
Karl Swedberg
www.englishrules.com
www.learningjquery.com
On Mar 26, 2007, at 8:28 AM, MARIO MOURA wrote:
Thanks Karl
I tried but didnt work (I made a lot of variations)
but this is working fine:
$("#imagefields").append( $(".group-image").children().get(3) );
Thanks for help
Regards
Mario
2007/3/24, Karl Swedberg <[EMAIL PROTECTED]>:
>
> $("#imagefields").append( $(".group-image":nth-child(2)) );
>
> Hi Mario, I know you were referring to the other Karl below, but if you
> don't mind, I'll jump in with an answer (I posted another response to your
> original post, because my email program didn't have it threaded with the the
> replies and I thought nobody had answered yet).
> You're almost there. You just need to put the quotation marks around the
> pseudo-class as well. So, this should work:
> $("#imagefields").append( $(".group-image:nth-child(2)") );
>
>
>
> --Karl
> _________________
> Karl Swedberg
> www.englishrules.com
> www.learningjquery.com
>
>
>
> On Mar 24, 2007, at 6:03 AM, MARIO MOURA wrote:
>
> Evan
>
> Thanks a lot your solution works for me.
>
> My real example is:
> $("#imagefields").append( $(".group-image").children().get(2) );
>
> But I am curious about what Karl said I tried a lot of things but I
> believe I have a error syntax.
>
> $("#imagefields").append( $(".group-image":2-child(n)) );
> or I tried too
> $("#imagefields").append( $(".group-image":nth-child(2)) );
>
> What is wrong?
> Regards
>
> macm
>
> 2007/3/24, Evan <[EMAIL PROTECTED] >:
> >
> > You could also use get(), though it's probably slower
> >
> > $('div').children().get(1);
> >
> > On Mar 24, 3:04 pm, "Karl Rudd" <[EMAIL PROTECTED]> wrote:
> > > Have read of this page:
> > >
> > > http://docs.jquery.com/Selectors#CSS_Selectors
> > >
> > > Especially E:nth-child(n)
> > >
> > > Karl Rudd
> > >
> > > On 3/24/07, MARIO MOURA <[EMAIL PROTECTED]> wrote:
> > >
> > >
> > >
> > > > Hi
> > >
> > > > I am newbie
> > >
> > > > How can I get a second children element?
> > >
> > > > In firebug I have
> > >
> > > > ---DIV
> > > > ------FIELD
> > > > ------FIELD
> > > > ------FIELD
> > > > ------FIELD
> > > > ------FIELD
> > >
> > > > My problem is all children have same name. How can I get second
> > and after
> > > > the third...
> > >
> > > > Regards
> > >
> > > > --
> > > > macm
> > > > _______________________________________________
> > > > jQuery mailing list
> > > > [EMAIL PROTECTED]
> > > >http://jquery.com/discuss/
> > >
> > > _______________________________________________
> > > jQuery mailing list
> > > [EMAIL PROTECTED]://jquery.com/discuss/
> >
> >
> > _______________________________________________
> > jQuery mailing list
> > discuss@jquery.com
> > http://jquery.com/discuss/
> >
>
>
>
> --
> Mário Alberto Chaves Moura
> [EMAIL PROTECTED]
> 31-9157-6000_______________________________________________
> jQuery mailing list
> discuss@jquery.com
> http://jquery.com/discuss/
>
>
>
> _______________________________________________
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> http://jquery.com/discuss/
>
>
--
Mário Alberto Chaves Moura
[EMAIL PROTECTED]
31-9157-6000_______________________________________________
jQuery mailing list
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[EMAIL PROTECTED]
31-9157-6000
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