I have the following page using angular-dragdrop "ngDragDrop" (which uses the jQueryUI-Dragdrop)
the following code WORKS:
<div class="sourceContainer">
<div ng-repeat="oldLocation in oldLocations"
data-drag="true"
data-jqyoui-options="{revert: 'invalid', helper: 'clone'}"
ng-model="oldLocation"
jqyoui-draggable="{index: {{$index}}, animate: true, placeholder: 'keep'}"
class="oldLocation singleTag"
>{{oldLocation.name}}</div>
</div>
<div class="targetContainer" data-drop="true" ng-model='oldLocationsCopied' jqyoui-droppable="{multiple:false}">
<div ng-repeat="newLocation in oldLocationsCopied track by $index"
data-drag="true" data-jqyoui-options="{revert: 'invalid', helper: 'clone'}"
class="oldLocation singleTag"
ng-model="newLocation" jqyoui-draggable="{index: {{$index}},animate:true,onDrop:doDrop(newLocation.name)}">{{newLocation.name}}
</div>
</div>
So I can drag&drop DIVs from inside "sourceContainer" to "targetContainer" and the function "onDrop()" is called when I do that.
Now I want the same without ANY visible action. When a div is dragged, it should just disappear and the method onDrop() should be called.
I could just hide the targetContainer but there should be a better solution