javascript 如何从翻转开关或切换开关获取选定值
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时间：2020-10-28 09:00:21 来源:igfitidea点击:
How to get selected value from a Flip Switch or Toggle Switch
提问by Vijay Rajasekaran
<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>
I am not sure how to get the value from the flip switch (On or Off) and assign to a javascript variable.
我不确定如何从翻转开关（开或关）获取值并分配给 javascript 变量。
Edit: The above code is from App Designer (Intel XDK)
编辑：以上代码来自 App Designer (Intel XDK)
To get the value use,
为了获得价值使用，
if ($('#af-flipswitch-1').is(":checked"))
{
console.log("On");
} else {
console.log("Off");
}
回答by Angu
flip switch html code
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 code
查询代码
$(document).delegate("#submit", "tap", function() {
alert($("#flip-1").val());
});
