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我有一個timepicker從http://amsul.ca/pickadate.js/time/JS Timepicker設置和諧的最小和最大時間
有兩種情況,開始和結束:
<div align="center">
<span class="glyphicon glyphicon-time"></span>
<label for="start_time">Start</label>
<input name="start_time" id="start_time">
<p></p>
<span class="glyphicon glyphicon-time"></span>
<label for="finish_time">Finish</label>
<input name="finish_time" id="finish_time">
</div>
的JS設置最小和最大時間:
<script type="text/javascript">
$(document).ready(function(){
$('#start_time').pickatime({
//format: 'h:i A', // Displayed and application format
formatSubmit: 'HH:i:00',
hiddenName: true,
interval: 15, // Interval between values (in minutes)
min: '3:00 AM', // Starting value
max: '6:00 PM' // Ending value
//finish_time.set('min': start_time)
});
$('#finish_time').pickatime({
//format: 'h:i A', // Displayed and application format
formatSubmit: 'HH:i:00',
hiddenName: true,
interval: 15, // Interval between values (in minutes)
min: '3:00 AM', // Starting value
max: '6:00 PM' // Ending value
});
});
</script>
我想要做的是將start_time選擇器的最小開始時間設爲3:00 AM,但對於finish_time選擇器最小值無論w選爲start_time + 15min。你會如何去做這件事?
這個插件有一個方法名_onSet_。所以你可以在_start_time_元素上使用它來設置_finish_time_元素的_min_值。但我認爲你必須計算_start_time_值+ 15min是否在第二天! – EhsanT
所以在start_time,我添加set:updateMin()\t function updateMin(){ \t \t $('#finish_time')。setAttribute(min,$(「#start_time」)。value); \t} – Aaron
如果您沒有注意到,我昨天發佈了答案! – EhsanT