【发布时间】:2018-10-14 11:18:25
【问题描述】:
我正在一个网站上制作一个拖放界面,其中包含三个可拖放对象和四个可拖放对象。我希望每个可拖动的内容都显示在放置它的任何可放置内容中。
如您所见,我正在尝试从 droppable 创建一个名为 dropped1 的变量,但我是 jQuery 新手,迷路了。
<div class="demo">
<div id="snaptarget1" class="ui-widget-header">
<p>Div 1</p> Correct Answer is 1
</div>
<div id="snaptarget2" class="ui-widget-header">
<p>Div 2</p> Correct Answer is 2
</div>
<div id="snaptarget3" class="ui-widget-header">
<p>Div 3</p> Correct Answer is 3
</div>
<br clear="both" />
<div id="draggable1" class="draggable1 ui-widget-content">
<p>Answer 1</p>
</div>
<div id="draggable2" class="draggable1 ui-widget-content">
<p>Answer 2</p>
</div>
<div id="draggable3" class="draggable1 ui-widget-content">
<p>Answer 3</p>
</div>
</div>
$("#snaptarget1").droppable({
drop: function(event, ui) {
var dropped1 = ui.prop('innerHTML');
$('#drag1results').css('background-color', 'yellow')
$(this)
.droppable('disable')
.removeClass("ui-widget-header")
.addClass("ui-state-highlight")
.find("p")
.html("Dropped with " + dropped1 + "!")
ui.draggable("option", "disabled", true);
}
});
【问题讨论】:
标签: jquery html object jquery-ui-draggable jquery-ui-droppable