如何在拖放对象后保留其原始位置

Question

我有一个可拖放/可拖放的小型jquery应用程序，并且我在保留可拖动项目时遇到问题，并且在克隆被删除后仍保持其原始位置。

任何人都可以协助吗？

http://jsfiddle.net/franco13/vLSZf/1/

谢谢。

$(init); 

function init() { 

    $('.teamEmblem').draggable({ 
     // containment: $('.teamEmblem').parent(), // this does not work 
     cursor: 'move', 
     helper: 'clone', 
     obstacle: ".teamEmblem", // this does not work 
     preventCollision: true, // this does not work 
     revert: true 
    }); 

    $('.winner').droppable({ 
     hoverClass: 'hovered', 
     tolerance: 'touch', 
     drop: handleCardDrop1 
    }); 

} 

function handleCardDrop1(event, ui) { 

    if (true) { 
     ui.draggable.addClass('correct'); 
     ui.draggable.draggable('disable'); 
     $(this).droppable('disable'); 
     ui.draggable.position({ 
      of: $(this), 
      my: 'left top', 
      at: 'left top' 
     }); 
     ui.draggable.draggable('option', 'revert', false); 
    } 

} 

Answer 1

你可以克隆拖动元素和应用一点风度克隆的元素：

SEE DEMO

function handleCardDrop1(event, ui) { 
    if (true) { 

     ui.draggable.addClass('correct'); 
     ui.draggable.draggable('disable'); 
     $(this).droppable('disable'); 

     var dragged = ui.draggable.clone(true); 
     dragged.position({ 
      of: $(this), 
      my: 'left top', 
      at: 'left top' 
     }).css({ 
      position: 'absolute', 
      display: 'block', 
      margin: '0 auto' 
     }); 
     ui.draggable.draggable('option', 'revert', false); 
     $('body').append(dragged); 
    } 

}