Closed juanbrujo closed 10 years ago
Unfortunately this would break ScrollMe if the .scrollme & .animateme aren't the same element as filter() would return an empty set.
That said, I see what you're getting at. This should to do the trick:
$( this ).find( '.animateme' ).addBack( '.animateme' ).each( function()
{
(...)
});
I'll include it in the next release.
Thanks!
using filter() instead of find() allows to reference the element itself and use it as .scrollme & .animateme