我使用JQuery Datepicker爲客戶選擇交貨日期。我希望能夠檢查客戶是否在中午之前訂購,如果是,可以在第二天交貨。如果他們在中午之後訂購,第二天交貨不可用,所以這一天是不可選擇的。JQuery Datepicker - 根據當前時間設置MinDate
我有一些代碼來檢查當前時間,但如何將此值添加到MinDate中的頂部設置?
謝謝!
<div class="delivery-date">
<p>
<label for="date">Select a date for delivery below:</label>
<input id="date" type="text" name="properties[delivery-date]" readonly="readonly" style="background:white; width:30%" class="required" data-error="Please choose a delivery date." />
</p>
</div>
<script>
jQuery(function() {
jQuery("#date").datepicker({
// minDate: new Date(((new Date).getTime() + 49 * 60 * 60 * 1000)),
minDate: checkBeforeNoon,
maxDate: "+2M", // show up to 2 months
dateFormat: 'dd/mm/yy',
beforeShowDay: available_delivery_dates
});
});
/*========== check time ==========*/
// if time before 12pm, offer next day delivery
function checkBeforeNoon(nextDayDelivery){
var startTime = '12:00 AM';
var endTime = '12:00 PM';
var curr_time = getval();
if (get24Hr(curr_time) > get24Hr(startTime) && get24Hr(curr_time) < get24Hr(endTime)) {
// before 12pm - next day delivery available
var nextDayDelivery = '+1d';
} else {
// after 12pm - next day delivery unavailable
var nextDayDelivery = '+2d';
}
function get24Hr(time){
var hours = Number(time.match(/^(\d+)/)[1]);
var AMPM = time.match(/\s(.*)$/)[1];
if(AMPM == "PM" && hours<12) hours = hours+12;
if(AMPM == "AM" && hours==12) hours = hours-12;
var minutes = Number(time.match(/:(\d+)/)[1]);
hours = hours*100+minutes;
console.log(time +" - "+hours);
return hours;
}
function getval() {
var currentTime = new Date()
var hours = currentTime.getHours()
var minutes = currentTime.getMinutes()
if (minutes < 10) minutes = "0" + minutes;
var suffix = "AM";
if (hours >= 12) {
suffix = "PM";
hours = hours - 12;
}
if (hours == 0) {
hours = 12;
}
var current_time = hours + ":" + minutes + " " + suffix;
return current_time;
}
}
/*========== Make sundays always unavailable ==========*/
function available_delivery_dates(date) {
var sunday = 0; // unavailable for delivery
var mon = 1
var tue = 2;
var wed = 3;
var thu = 4;
var fri = 5;
var sat = 6;
var day_of_week = date.getDay();
var not_sun = day_of_week > 0;
if(not_sun){
var day = date.getDate();
return [true, ''];
}
else{
// all else - do not allow
return [false, ' ', 'Delivery is unavailable on this day'];
}
}
</script>
你設置'minDate'的功能,而不是返回值,你需要調用你的函數...'的minDate:checkBeforeNoon(),' –
謝謝,但並沒有工作 – casiokid
也'beforeShowDay: available_delivery_dates()' – kRiZ