Question

我想要恢复红色块的位置，这是绿色和灰色块上的可拖动div而不是蓝色块，它不能在绿色块上工作但是它在灰色块上工作请帮助我...为此目的我正在使用jquery还原。代码和lnk在下面请帮助我 http://galtech.org/testing/drag.php

<style type="text/css">
    #draggable { width: 100px; height: 70px; background: red; }
    #nodrag { width: 200px; height: 270px; background: #00CC66; }
  </style>

</head>
<body >
<div style="background:#CCCCCC;">
     <div id="droppble_blue" style="background:#99CCCC; height:500px; width:620px;"> 
        <div id="draggable" >Drag</div>
        <div id="nodrag" class="new">no Drag & drop here</div>
    </div>
</div>
</body>
</html>
  <script>
  $(document).ready(function() {
    $("#draggable").draggable({ revert: "invalid" });
    $("#droppble_blue").droppable({drop: function() { alert('dropped');}});
  });
  </script>

Answer 1

我没有使用可拖动的jquery但是

$("#nodrag").droppable({drop: function() { $('#draggable').css({top:'0px',left:'0px'});}});

工作？

进一步研究可拖动的东西没有明显的方法来恢复有效元素内的块，所以我会做以下

$("#draggable").draggable('option', 'start', function(){
   //when drag starts save the positions somewhere
   $(this).data('left',$(this).css('left')).data('top',$(this).css('top'))
});
$("#nodrag").droppable({drop: function(e,ui) {
    //when dropped on an invalid block move it back to where it was (you could change this to animate to mimic revert)
    $(ui.draggable)
   .css({left:$(ui.draggable).data('left'),top:$(ui.draggable).data('top')});
}});

jquery draggable有关于拖动的问题？

1 个答案: