> > $("div.image")[4]."
>
> > Because that is not a jQuery object
>
> > On Jan 25, 4:52 pm, donb wrote:
> >> Try
>
> >> $("div.image").eq(4).show()
>
> >> I would only be guessing, but $("div.image")[4] mus
bump..
Anyone have any ideas why this does not work..
On Jan 25, 11:01 am, bartee wrote:
> This works...
>
> $("div.image").slice(4,5).show()
>
> But this does NOT
>
> $("div.image")[4].show()
>
> Seems like the [4] should work !!
This works...
$("div.image").slice(4,5).show()
But this does NOT
$("div.image")[4].show()
Seems like the [4] should work !!
My alert always returns "1" as the divcount value ???
What's up with that.. I have read and tried bunches of things.
What am I missing
http://www.w3.org/1999/xhtml"; xml:lang="en" lang="en-AU">
Jquery Test
$(function() {
var divcount = $("div#div2").length;
alert(
THANKS to all
I had read about this and understand the problem. I am just new.
Great to find a place for answers.
Now I can start on my 2nd line of code :-)
I have this test code. My alert dialog always show zero as the div
count:
Help !!!
http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd";>
http://www.w3.org/1999/xhtml"; xml:lang="en" lang="en-AU">
Jquery Test
var divcount = $("div").length;
alert("start=" + divcount);
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