$("#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/
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Mário Alberto Chaves Moura
[EMAIL PROTECTED]
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