这是我的解决方法，使列表可选和排序，并允许套索实现。这里是fiddle。
<html>
<meta charset="utf-8" />
<title>jQuery UI Sortable with Selectable</title>
<link rel="stylesheet" href="http://code.jquery.com/ui/1.10.3/themes/smoothness/jquery-ui.css" />
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://code.jquery.com/ui/1.10.3/jquery-ui.js"></script>
$(function() {
//
$('body').selectable({
filter: 'li'
//filter: '#album2 > li'
});
/*
Since the sortable seems unable to move more than one object at a
time, we'll do this:
The LIs should act only as wrappers for DIVs.
When sorting a LI, move all the DIVs that are children of selected
LIs to inside the sorting LI (this will make them move together);
but before doing that, save inside the DIVs a reference to their
respective original parents, so we can restore them later.
When the user drop the sorting, restore the DIVs to their original
parent LIs and place those LIs right after the just-dropped LI.
Voilá!
Tip: doesn't work so great if you try to stick the LIs inside the LI
*/
$('.connectedSortable').sortable({
connectWith: ".connectedSortable",
delay: 100,
start: function(e, ui) {
var topleft = 0;
// if the current sorting LI is not selected, select
$(ui.item).addClass('ui-selected');
$('.ui-selected div').each(function() {
// save reference to original parent
var originalParent = $(this).parent()[0];
$(this).data('origin', originalParent);
// position each DIV in cascade
$(this).css('position', 'absolute');
$(this).css('top', topleft);
$(this).css('left', topleft);
topleft += 20;
}).appendTo(ui.item); // glue them all inside current sorting LI
},
stop: function(e, ui) {
$(ui.item).children().each(function() {
// restore all the DIVs in the sorting LI to their original parents
var originalParent = $(this).data('origin');
$(this).appendTo(originalParent);
// remove the cascade positioning
$(this).css('position', '');
$(this).css('top', '');
$(this).css('left', '');
});
// put the selected LIs after the just-dropped sorting LI
$('#album .ui-selected').insertAfter(ui.item);
// put the selected LIs after the just-dropped sorting LI
$('#album2 .ui-selected').insertAfter(ui.item);
}
});
//
});
<style>
*,
*:before,
*:after {
-webkit-box-sizing: border-box;
-moz-box-sizing: border-box;
box-sizing: border-box;
}
#album {
list-style: none;
float: left;
width: 20%;
border: 1px solid blue;
}
#album2 {
list-style: none;
float: left;
width: 20%;
border: 1px solid blue;
}
#album li {
float: left;
margin: 5px;
}
#album2 li {
float: left;
margin: 5px;
}
#album div {
width: 100px;
height: 100px;
border: 1px solid #CCC;
background: #F6F6F6;
}
#album2 div {
width: 100px;
height: 100px;
border: 1px solid #CCC;
background: #F6F6F6;
}
#album .ui-sortable-placeholder {
border: 1px dashed #CCC;
width: 100px;
height: 100px;
background: none;
visibility: visible !important;
}
#album2 .ui-sortable-placeholder {
border: 1px dashed #CCC;
width: 100px;
height: 100px;
background: none;
visibility: visible !important;
}
#album .ui-selecting div,
#album .ui-selected div {
background-color: #3C6;
}
#album2 .ui-selecting div,
#album2 .ui-selected div {
background-color: #3C6;
}
#anotheralbum {
list-style: none;
float: left;
width: 20%;
height: 800px;
border: 1px solid green;
}
</style>
<body>
<ul id="album" class="connectedSortable">
<li id="li1"><div>1- First</div></li>
<li id="li2"><div>2- Second</div></li>
<li id="li3"><div>3- Third</div></li>
<li id="li4"><div>4- Fourth</div></li>
<li id="li5"><div>5- Fifth</div></li>
<li id="li6"><div>6- Sixth</div></li>
<li id="li7"><div>7- Seventh</div></li>
<li id="li8"><div>8- Eighth</div></li>
</ul>
<ul id="album2" class="connectedSortable">
<li id="li1"><div>1- 1</div></li>
<li id="li2"><div>2- 2</div></li>
<li id="li3"><div>3- 3</div></li>
<li id="li4"><div>4- 4</div></li>
<li id="li5"><div>5- 5</div></li>
<li id="li6"><div>6- 6</div></li>
<li id="li7"><div>7- 7</div></li>
<li id="li8"><div>8- 8</div></li>
</ul>
<div id="anotheralbum" class="connectedSortable">
another album - no style for the lists inside here
</div>
<br style="clear:both">
</body>
</html>
刚刚使用链接答案中的代码片段有什么问题？用这段代码，我很容易[得到这个工作]（http://jsfiddle.net/MattiasBuelens/vLM94/）。 –
当你放弃它时，只会拖动被拖动的物品而不是其他选择的物品，我已经到了我正在实施它的阶段。 –