> Imagine that you have an array:
>
> var a = [ 'a', 'b', 'c' ];
>
> Now you can call array methods on the array:
>
> var t = a.join(); // 'a,b,c'
>
> But what happens if you call an array method on one of the *elements* of the
> array?
>
> var x = a[0].join();
>
> That's an error, because a[0] is a *string*, not an array.
>
> In the same way, $(whatever) returns a jQuery object, with jQuery methods
> such as .val(), .css(), etc. The jQuery object is also an array-like object,
> so you can retrieve the array elements with [0], [1], etc. But those array
> elements are *DOM nodes*, not jQuery objects themselves.
>
> So, $(whatever)[0] gives you a DOM node, with properties like defaultValue,
> but this DOM node does not have the jQuery methods, because it's not a
> jQuery object.
>
> -Mike
>
> > for this form
> > <form name="f1" method="get" action="/search" id="searchform" >
> > <input type="text" name="q" value="Search...." size="70" />
> > <input type="submit" value="Search" />
> > </form>
>
> > why
> > $("#searchform input")[0]
> > is different than
> > $("#searchform input:first")
>
> > I can get values for $("#searchform input")[0].defaultValue
> > or .value but not with .fadeIn() or val() while the reverse
> > for the $ ("#searchform input:first")
Beautifully said, Mike. That response should be on the docs wiki
(maybe the FAQ page).
