Thanks All..
Great answers.. I learned two things.. that DOM objects are returned,
you have to understand the difference.
And duck!... thanks for the answer I was really hunting.. I guess I
need to read that part of the book again..
Moving on the event's now... going to click a next and prev arrow ..
On Jan 25, 7:33 pm, duck! <pooisb...@gmail.com> wrote:
> Exactly. $("div.image")[4] gives the raw dom element from the jQuery
> collection, and not a jQuery "wrapped" element like the OP was expecting
> (thus the .show() function isn't available).
>
> Re-wrapping the element like so: $($("div.image")[4]).show() would work. Of
> course you'd be better off using $("div.image:eq(4)") or
> $("div.image").eq(4) as the above poster mentioned.
>
>
>
> MorningZ wrote:
>
> > "But this does NOT
>
> > $("div.image")[4]."
>
> > Because that is not a jQuery object
>
> > On Jan 25, 4:52 pm, donb <falconwatc...@comcast.net> wrote:
> >> Try
>
> >> $("div.image").eq(4).show()
>
> >> I would only be guessing, but $("div.image")[4] must gets you
> >> something that's not an object possessing a show() method.
>
> >> On Jan 25, 4:41 pm,bartee<bar...@gmail.com> wrote:
>
> >> > bump..
>
> >> > Anyone have any ideas why this does not work.. ????
>
> >> > On Jan 25, 11:01 am,bartee<bar...@gmail.com> wrote:
>
> >> > > This works...
>
> >> > > $("div.image").slice(4,5).show()
>
> >> > > But this does NOT
>
> >> > > $("div.image")[4].show()
>
> >> > > Seems like the [4] should work !!
>
> --
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