Hi, this proved to be a documentation bug.

The manual for .selectable[1] should state that you must provide an
array if the other selectables are more than one.

http://docs.jquery.com/UI/Sortable#options

The demo page says this:
"
Sort items from one list into another and vice versa, by passing an
array into the |connectWith| option. The simplest way to do this is to
group all related lists with a CSS class, and then pass that class into
the sortable function using array notation (i.e., |connectWith:
['.myclass']|).
"

This could have saved me 5 hours today :/. Apart from that, I like UI.
Thanks for making it.


Regards,
Tarjei


Tarjei Huse wrote:
> Hi,
>
> I got a menu I want to sort that has the following structure:
>
> <ul id="navbase">
> <li><div>Menu 1 </div>
> <ul class="submenu">
>    <li><a>Submenu 1</a></li>
>    ..
>    ..
> </ul>
> </li>
>
> <li><div>Menu 2 </div>
> <ul class="submenu">
>    <li><a>Submenu N</a></li>
>    ..
>    ..
> </ul>
> </li>
> ..
> </ul>
>
> Now, I want to be able to sort the top level menus and the bottom level
> menus separately and also be able to move submenu items between the
> different main menu items - but not place main menu items bellow each other.
>
> I tried looking at the UI Sortable plugin, but it seems not to do this.
> Am I wrong?
>
> I tried using the UI drag and drop plugin as well, but it seems just to
> place the element where I drop it (for example over one of the other
> menu elements) - not show how the element will fit in the collection
> like sortable does. Is there some way to make the drag and drop plugin
> be more like the sortable plugin?
>
> Kind regards,
> Tarjei
>
>
>
>   


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