这是有效的:
创建表:
newRows += "<table name ='activityTable'>";
newRows += "<tbody id='activity2Tablebody' name='test1'>";
newRows += "<tr>";
newRows += "<td class='dropValue' name='test2'>";
newRows += "<div class='dragabbleRemove'>";
newRows += "<input class='droppableItem activityWidth' name='test3' disabled></input>";//Droppable Activity Class
newRows += "</div>";
newRows += "</td>";
newRows += "</tr>";
newRows += "</tbody>";
newRows += "</table>";
在拖放时,将name属性的值设置为拖动项的id属性的值:
draggableId = (draggable.prop('id'));
droppable.prop('name', draggableId);
alert ("$(this).find('input').attr('name'): " + $(this).find('input').attr('name'));//id value
然后我在inputs元素上放置一个透明div,这样可以拖动div来移除它:
newRows += "<div class='dragabbleRemove'><div> </div>";
newRows += "<input class='droppableItem activityWidth' name='test3' disabled></input>";//Droppable Activity Class
newRows += "</div>";
现在返回空白:
警报($(this).find('input').attr('name'):”+$(this).find('input').attr('name');//id值
我怀疑需要更改以下行以设置输入元素的名称值;但是,我无法确定需要将其更改为什么(正确的代码):
droppable.prop('name', draggableId);