我有三个部门(桌子、餐厅、服务员)
所以情况是:
1)创建表
2)拖到餐厅div
3)把服务员拖到桌子上
我使用HTML的拖放功能。
我想把受影响的服务员名单分类到一个特定的桌子上。我用了jquery插件。工作正常
现在我面临另一个问题,我不能删除一个服务生从一个表(把它放回服务生名单)刚刚添加插件…
这是密码..
<div class="col-md-6">
Restaurant
<div id="restaurant" class="restaurant" ondrop="drop(event,this)" ondragover="allowDrop(event)">
@for($i=1;$i<=count($tables);$i++)
<div id="table_{{$tables[$i-1]->id}}" style="height: {{count($tables[$i-1]->waitersAvailableTable)*80}}px;" class="table_restaurant sortable" draggable="true" ondragstart="drag(event)" ondragover="allowDrop(event)" ondrop="drop(event,this)">Table {{$tables[$i-1]->num_table}}
@foreach($tables[$i-1]->waitersAvailableTable as $waiterTable)
<div id="{{$waiterTable->waiter->id}}" class="waiter" draggable="true" ondragstart="drag(event)" table_id="table_{{$tables[$i-1]->id}}"><img src="{{asset($waiterTable->waiter->user->image)}}" height="40px;" />{{$waiterTable->waiter->user->name}} {{$waiterTable->waiter->user->surname}}</div>
@endforeach
</div>
@endfor
</div>
</div>
<div class="col-md-2">
<div>
<span>Serveurs </span>
</div>
<div id="waiters" class="waiters" ondrop="drop(event,this)" ondragover="allowDrop(event)">
@if(count($waiters)>0)
@foreach($waiters as $waiter)
<div id="{{$waiter->id}}" class="waiter" draggable="true" ondragstart="drag(event)" ><img src="{{asset($waiter->user->image)}}" height="40px;" />{{$waiter->user->name}} {{$waiter->user->surname}}</div>
@endforeach
@else
<li><a href="{{route('showAddWaiter')}}" title="Panel boxes"><span>Ajouter un serveur</span></a></li>
@endif
</div>
</div>
<script src="https://code.jquery.com/ui/1.12.1/jquery-ui.js"></script>
<script>
$( function() {
$( ".sortable" ).sortable();
$( ".sortable" ).disableSelection();
} );
</script>