A cursory review of the code seems like it would only work for one layer. How would ROAllConnectedNodeDraggable work with the three layers of the following example[1]? For my project I came up with an alternative implementation 'LEKDraggableSubtree.' I'd be interested in what you think of the attached mcz.

cheers -ben

[1]------------------------------- | view toElement fromElement draggableCase | draggableCase := RODraggable. "draggableCase := ROAllConnectedNodeDraggable " "draggableCase := LEKDraggableSubtree withEdges: [ :selectedElement | selectedElement view onlyEdges ] . " view := ROView new. 1 to: 7 do: [ :n | toElement := ((ROLabel elementOn: n) extent: 50@50 ) + ROBorder @ draggableCase. view add: toElement. ((fromElement := view elementFromModel: n // 2) ) ifNotNil: [ view add: (ROEdge from: fromElement to: toElement ) + ROLine ] ]. ROHorizontalTreeLayout new on: view elements. view open. -------------------------------

moose-technology@googlecode.com wrote:

Updates: Status: Fixed

Comment #1 on issue 904 by alexandr...@gmail.com: Roassal dragging subtrees (feature contribution) http://code.google.com/p/moose-technology/issues/detail?id=904

Hi Ben,

Thanks for your suggestion. I wrote a contribution inspired from your example.

Fixed in 1.457 of Roassal. Try this:

-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= | view elements | view := ROView new.

0 to: 50 by: 10 do: [ :i | elements := (ROBox green size: 10) elementsOn: (i to: i + 9). elements do: [ :e | e @ RODraggable @ ROPopup ]. view addAll: elements.

ROLine buildEdgesFromElement: elements first from: #yourself 

toAll: [ :v | (v + 1 to: v + 9) ].

ROTreeLayout on: elements.

(view elementFromModel: i) - ROBox + ROBox red.
(view elementFromModel: i)
    @ ROAllConnectedNodeDraggable ;
    translateBy: i * 15 @ 10.

].

view open