So you want the draggable to clone itself and still be draggable and clonable again? Can you briefly describe the whole desired functionality?
-- Bohdan On Feb 3, 6:26 pm, dirk w <[email protected]> wrote: > oh, that's odd. on the second sight it doesn't work. > by using .append(ui.draggable.clone()); the clone seems to be not > draggable anymore. > by making it draggable (directly after it got cloned) it seems to get > copied twice. > > have you heard about that? > > On 3 Feb., 18:09, dirk w <[email protected]> wrote: > > > works like a charm, thanks > > dirk > > > On 3 Feb., 17:21, Bohdan Ganicky <[email protected]> wrote: > > > > Hi, > > > what about this: > > > > $('.draggable').draggable({ > > > helper: 'clone'}); > > > > $('#droppable').droppable({ > > > accept: '.draggable', > > > drop: function(ev, ui) { > > > $(this).append(ui.draggable.clone()); > > > } > > > > }); > > > > -- > > > Bohdan > > > > On Feb 3, 3:51 pm, dirk w <[email protected]> wrote: > > > > > hello ui community! > > > > i would like to drop a draggable object and than append() it to an > > > > other list. > > > > > but i dislike that the draggable object than get's removed from it's > > > > original place (i mean the place where it was before dropping it) > > > > > is it possible to create a copy of the draggable object and than > > > > appending this one to the second place? so the real draggable object > > > > would still stay at it's original place. > > > > > your help is much appreciated > > > > dirk --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "jQuery UI" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/jquery-ui?hl=en -~----------~----~----~----~------~----~------~--~---
