如何从翻转开关或拨动开关获取选定值

Question

<div class="table-thing with-label widget uib_w_114 d-margins" data-uib="app_framework/flip_switch" data-ver="1">
    <label class="narrow-control label-inline">Notifications</label>
    <div class="wide-control">
        <input type="checkbox" class="toggle" id="af-flipswitch-1" name="af-flipswitch-1" checked="checked">
        <label for="af-flipswitch-1" data-off="Off" data-on="On">
            <span></span>
        </label>
    </div>
</div>

我不确定如何从翻转开关（打开或关闭）获取值并分配给 javascript 变量。

编辑：以上代码来自 App Designer (Intel XDK)

获取值使用，

    if ($('#af-flipswitch-1').is(":checked"))
            {
    console.log("On");
        } else {
             console.log("Off");
            }

Answer 1

翻转开关html代码

<label for="flip-1">Flip switch:</label>
<select name="flip-1" id="flip-1" data-role="slider">
    <option value="off">Off</option>
    <option value="on">On</option>
</select> 

<button id="submit">Submit</button>

Jquery代码

$(document).delegate("#submit", "tap", function() {
    alert($("#flip-1").val());
});

Click to Demo

如何从翻转开关或拨动开关获取选定值

How to get selected value from a Flip Switch or Toggle Switch

html

javascript

jquery

appframework

intel-xdk