jQuery Draggables和Droppables定位

Question

我正在使用jquery UI和jQuery draggable,我的所有draggable都使用jquery clone helper并将draggable追加到droppable.

这是我的代码

 $('#squeezePage #droppable').droppable({
  tolerance: 'fit',
  accept: '#squeezeWidgets .squeezeWidget',
  drop: function(event, ui) {
   var dropElem = ui.draggable.html();

   var clone = $(dropElem).clone();

   clone.css('position', 'absolute');
   clone.css('top', ui.absolutePosition.top);
   clone.css('left', ui.absolutePosition.left);

   $(this).append(clone);

   $(this).find('.top').remove();
   $(this).find('.widgetContent').slideDown('fast');

   $(this).find('.widgetContent').draggable({
    containment: '#squeezePage #droppable',
    cursor: 'crosshair',
    grid: [20, 20],
    scroll: true,
    snap: true,
    snapMode: 'outer',
    refreshPositions: true
   });

   $(this).find('.widgetContent').resizable({
    maxWidth: 560,
    minHeight: 60,
    minWidth: 180,
    grid: 20,
   });
  }
 });

我用.css('top', ui.absolutePosition.top);和设置克隆css('left', ui.absolutePosition.left);的位置,但位置相对于BODY.

该位置与droppable无关,这使得可拖动的下降到随机位置.总体而言,可投放和可拖动的集成并不紧张.我想让它更顺畅.

Answer 1

我正在获取正文的偏移量并从小部件克隆的偏移量中减去它.

        clone.css('top', ui.position.top - droppableOffset.top);
        clone.css('left', ui.position.left - droppableOffset.left);

有用!