Uh, that's not what I'm asking at all.
In the documentation example you're passing .add() the "p" selector
which is a string. I want the end result to be the same except instead
of passing a selector string like "p" I want to pass an existing
jquery object. It doesn't seem to work this way and the only time more
elements get added to to jquery object is if you use a selector.
Sso I want to do more like :
$("div").css("border", "2px solid red")
.add($(this).children()) //see how this line is different
.css("background", "yellow");
On Dec 23, 12:50 pm, Charlie Griefer <charlie.grie...@gmail.com>
wrote:
> As per the docs (http://docs.jquery.com/Traversing/add),
>
> $("div").css("border", "2px solid red")
> .add("p")
> .css("background", "yellow");
>
> That will add a 2px red border around any 'div' element.
>
> Then add in a new selector and grab all of the <p> elements. Add a yellow
> background to the previously selected <div> elements, as well as the newly
> selected <p> elements.
>
> Does that clarify add()?
>
> I'm not sure I understand what it is you're trying to accomplish. Can you
> explain what you need to do?
>
>
>
> On Wed, Dec 23, 2009 at 10:45 AM, Scott <polyp...@gmail.com> wrote:
> > Maybe I'm not understanding the point of .add() but it doesn't really
> > do what I'd expect nor can I figure out how to do what I thought would
> > be simple.
>
> > I want to combine 2 jquery objects.
>
> > var $obj1 = $(".stuff");
> > var $obj2 = $(".morestuff");
> > $obj1.add($obj2);
> > alert($obj1.length);
>
> > This doesn't do anything, it looks like .add() only accepts selection
> > strings which in my example would work but in what I really need to do
> > I have no selection string which would find the specific jquery
> > objects.
>
> --
> Charlie Grieferhttp://charlie.griefer.com/
>
> I have failed as much as I have succeeded. But I love my life. I love my
> wife. And I wish you my kind of success.
