jQuery.draggable()-单击按钮时恢复 [英] jQuery.draggable() - Revert on button click

问题描述

我有几个可拖动&我页面上的可投放元素，具有accept属性.

I have several draggable & droppable elements on my page, which have accept properties.

目前，我的代码设置如下:

Currently my code is set up like:

$(".answer." + answer).draggable({
    revert: "invalid"
    , snap: ".graph"
});
$(".graph." + graph).droppable({
    accept: ".answer." + answer
});

因此，如果答案不正确，它将恢复为原始位置.

Therefore if the answer isn't correct, its reverted to its original position.

用户还需要能够重置页面上的所有内容.我已经尝试了以下方法，但是不起作用:

The user also needs the ability to reset all on the page. I've tried the following, but it doesn't work:

$("btnReset").click(function(){
   $(".graph.A").draggable({revert:true});
});

推荐答案

由于没有内置方法可以执行所需的操作，因此您必须自己模拟还原行为.您可以存储每个可拖动元素的原始位置，然后在单击重置按钮时将其重新设置为这些位置的动画.

Since there's no built-in method to do what you need, you'd have to simulate the revert behavior yourself. You can store the original position of each draggable element and then when clicking a reset button, animate them back to those positions.

这是一个简化的 jsFiddle示例 .

Here's a simplified jsFiddle example.

jQuery:

$("#draggable").draggable({
    revert: "invalid"
});
$("#draggable2").draggable({
    revert: "invalid"
});
$("#droppable").droppable({
});
$("#btnReset").click(function() {
    $("#draggable, #draggable2").animate({
        "left": $("#draggable").data("left"),
        "top": $("#draggable").data("top")
    });
});
$(".ui-widget-content").data("left", $(".ui-widget-content").position().left).data("top", $(".ui-widget-content").position().top);

HTML:

<div id="draggable" class="ui-widget-content">
    <p>I revert when I'm dropped</p>
</div>
<div id="draggable2" class="ui-widget-content">
    <p>I revert when I'm dropped</p>
</div>
<button id="btnReset">Reset</button>
<div id="droppable" class="ui-widget-header">
    <p>Drop me here</p>
</div>​

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