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
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 !!
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
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
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
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