:O
I'm not sure about it. But I thought it was faster to clone and then remove,
than move it all.
Thanks for taking your time to explain this.
On 9/12/07, Pyrolupus <[EMAIL PROTECTED]> wrote:
>
>
> On Sep 11, 7:06 pm, "Joan Piedra" <[EMAIL PROTECTED]> wrote:
> > Hmm.. Can't you clone the nodes and then just remove the _old_ parent
> div?
>
> Sure, but if I'm not mistaken, using clone() is more expensive: it
> creates copies of everything just to subsequently remove the
> originals. By using children(), we just work with the same number of
> DOM elements, just change their root. (This is based on an untested
> assumption that the browser is re-linking existing items, not doing
> its own internal clone-then-delete type behavior.)
>
> In either case, though, the old parent div does get removed.
>
> Pyro
>
> >
> > On 9/11/07, Pyrolupus <[EMAIL PROTECTED]> wrote:
> >
> > > On Sep 11, 3:03 pm, "Brook Davies" <[EMAIL PROTECTED]> wrote:
> > > > Easy Question I think. If I use (from Jquery 1.2):
> >
> > > > $("#myElem").wrapAll("<div id='myDiv'></div>");
> >
> > > > To wrap the div 'myDiv' around 'myElem', how can I later remove that
> div
> > > > without removing the pre-existing myElem and any other contents of
> that
> > > div?
> >
> > > I'm newish to jQuery, so my syntax is not the uberest, but something
> > > like the following would work:
> >
> > > $('div#myDiv').after($('div#myDiv').children()).remove()
> >
> > > In fact, it does work--I just tested it in FireBug. (I just don't
> > > know the correct syntax to avoid referring to div#myDiv twice.)
> >
> > > Pyro
> >
> > --
> > Joan Piedra || Frontend web developerhttp://www.justaquit.com/ ||
> http://www.joanpiedra.com/
>
>
--
Joan Piedra || Frontend web developer
http://www.justaquit.com/ || http://www.joanpiedra.com/
