$D.fn = $D.prototype;
Cheers
--
Ariel Flesler
http://flesler.blogspot.com
On 17 jun, 03:54, "R. Rajesh Jeba Anbiah"
<[EMAIL PROTECTED]> wrote:
> On May 25, 2:48 am, "R. Rajesh Jeba Anbiah"
>
>
>
>
>
> <[EMAIL PROTECTED]> wrote:
> > I want to clone the jQuery object $ to $D; but it actually seems to be
> > overwriting. Can anyone please suggest any solution?
>
> > Here is the code:
>
> > $D = $.extend($); // same result for $D = $;
> > $D.fn.extend({
> > click: function() {
> > alert('overridden click');
> > return this;
> > }
> > });
>
> > $(document).ready(function(){
> > $('body').click(); // this also triggers the 'overridden click' --
> > which is what I want to avoid.
> > $D('body').click();
>
> > });
>
> > As you see, $('body').click() triggers the 'overridden click', but I
> > wanted it to trigger the original jQuery object's click.
>
> Anyone? Or is it impossible (can someone please confirm)?
>
> --
> <?php echo 'Just another PHP saint'; ?>
> Email: rrjanbiah-at-Y!com Blog:http://rajeshanbiah.blogspot.com/- Ocultar
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