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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