I've put together a demo for you. Is this it? :) http://jquery-ui.googlecode.com/svn/trunk/demos/index.html#sortable|connect-lists-through-tabs
-- Bohdan On Feb 5, 4:12 pm, dirk w <[email protected]> wrote: > i am sorry but i can't figure out how the "connectWith" could help me. > i tried giving both lists the same class and add this as connectWith > value but it doesn't work at all (now it seems like it destoyed the > draggable option of those elements). > > remember, the two lists aren't visible at the same time (seperated > tabs) and to add the element of the first list the image needs to get > dropped on one Tab element. > now it appears on the second list and this one just should be > sortable. actually not too complicated but i am really struggeling :( > > dirk > > On 5 Feb., 14:02, Bohdan Ganicky <[email protected]> wrote: > > > Well, it seems that this is exactly what you > > want:http://ui.jquery.com/demos/sortable/#connect-lists:) > > > -- > > Bohdan > > > On Feb 5, 12:32 pm, dirk w <[email protected]> wrote: > > > > hi bohdan, > > > thanks for offering your help! i can't figure it out, even with the > > > examples. > > > > what works already: > > > i have a list of images which are draggable. they can get dropped on > > > various places on the site as well as on a Tab. > > > if dropped on a Tab they appear (append()) to an other list. in this > > > list they are still draggable (so you can still drop them on those > > > various places). > > > > the only thing i would like to do now is: > > > the second list (where the elements appear after they got dropped on > > > the tab) should be sortable. i mean people should drag those images > > > around and arrange them new (like in this > > > example:http://ui.jquery.com/demos/sortable/#display-grid) > > > > i tried to just make the second list droppable, but this leads to > > > problems with jquery (i get funny errors in firebug). i thought maybe > > > its because this list gets new elements all the time, so i declared > > > the second list as sortable every time a new item gets attached. but > > > this leads to a similar odd error. > > > > maybe i need to use the "connectToSortable" for my draggable elements, > > > but i am not sure ho to use it. > > > > your help is much appreciated! > > > dirk > > > > On 5 Feb., 12:10, Bohdan Ganicky <[email protected]> wrote: > > > > > Hi Dirk, > > > > check out the "Draggable + Sortable" > > > > demo:http://ui.jquery.com/demos/draggable/#sortable > > > > It could be a good starting point. Hope that helps. If not, let me > > > > know and I'll try to help you more. > > > > > -- > > > > Bohdan > > > > > On Feb 5, 11:23 am, dirk w <[email protected]> wrote: > > > > > > hi bohdan, > > > > > i changed my mind (since i just couldn't figure out how to do it like > > > > > i wanted). from now on the elements will not get copied, there's > > > > > always just one instance of an element (this is probably less > > > > > confusing to the user). > > > > > > but now i got a new problem, this time with sortable > > > > > > i have 2 ULs with LIs (which contain just 1 IMG) > > > > > it's possible to drag an LI from the first list on top of an TAB. if > > > > > dropped on the tab than the LI gets added to the second list. that > > > > > works fine. > > > > > but i would like to make this second list sortable. > > > > > so if you drop an element on the tab than this should get added to the > > > > > second list AND should be sortable in this second list (And still > > > > > drag&droppable). > > > > > the way i add items to the seconds list is this: > > > > > > // if result gets dropped on Playlist Tab than load it's Thumbnail and > > > > > Data to the Playlist > > > > > $(".playlist").droppable( > > > > > { > > > > > accept: ".searchResultEntry", > > > > > hoverClass: "videoDropHover", > > > > > drop: function(ev, ui) > > > > > { > > > > > if ($("#nextSongsList > > > > > > .searchResultEntry").length < 14) > > > > > { > > > > > > > > > > $("#nextSongsList").append(ui.draggable); > > > > > } > > > > > } > > > > > }); > > > > > > when i tried to make it sortable, than the items lost their draggable > > > > > functionality :( > > > > > > dirk > > > > > > On 4 Feb., 16:27, Bohdan Ganicky <[email protected]> wrote: > > > > > > > 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. 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