我目前正在使用此脚本
<script type="text/javascript">
var _next = new Date(new Date() * 1 + 24*60*60*1000*9);
document.write(" Your expected delivery date is " + (_next.getMonth() + 1) + "/" + _next.getDate() + "/" + _next.getFullYear());
</script>
我想知道如何尽可能排除周末。
如果愿意，只需增加几个小时。这是一个预计在下一个工作日交付的代码。
function myFunction() {
var d = new Date();
var weekday = new Array(7);
weekday[0] = "sunday";
weekday[1] = "monday";
weekday[2] = "tuesday";
weekday[3] = "wednesday";
weekday[4] = "thursday";
weekday[5] = "friday";
weekday[6] = "saturday";
var monthday = new Array(12);
monthday[0] = "01";
monthday[1] = "02";
monthday[2] = "03";
monthday[3] = "04";
monthday[4] = "05";
monthday[5] = "06";
monthday[6] = "07";
monthday[7] = "08";
monthday[8] = "09";
monthday[9] = "10";
monthday[10] = "11";
monthday[11] = "12";
if (d.getDay() > 0 && d.getDay() <= 4) {
d.setHours(d.getHours() + 24);
var deliver = weekday[d.getDay()];
var day = d.getDate();
var month = monthday[d.getMonth()];
var year = d .getFullYear();
document.getElementById("deliveryday").innerHTML = deliver + ' ' + day + '-' + month + '-' + year;
} else if (d.getDay() == 0) {
d.setHours(d.getHours() + 48);
var deliver = weekday[d.getDay()];
var day = d.getDate();
var month = monthday[d.getMonth()];
var year = d .getFullYear();
document.getElementById("deliveryday").innerHTML = deliver + ' ' + day + '-' + month + '-' + year;
} else {
d.setHours(d.getHours() + 72);
var deliver = weekday[d.getDay()];
var day = d.getDate();
var month = monthday[d.getMonth()];
var year = d .getFullYear();
document.getElementById("deliveryday").innerHTML = deliver + ' ' + day + '-' + month + '-' + year;
}
}
myFunction()
<script type="text/javascript">
<!--
var myDate=new Date();
if ( myDate.getHours() < 12 ) // less than 12pm
{
var daystodeliver = [7,5,5,5,5,8,7][myDate.getDay()];
}
else
{
var daystodeliver = [7,7,7,7,7,9,8][myDate.getDay()];
}
myDate.setDate(myDate.getDate()+daystodeliver);
document.write(['Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday'] [myDate.getDay()]);
var dayofmonth = myDate.getDate();
suffix = ((dayofmonth < 10)||(dayofmonth > 20)) ? ['th','st','nd','rd','th','th','th','th','th','th'][dayofmonth % 10] : 'th';
document.write(' ' + dayofmonth + suffix + ' ');
document.write(['January','February','March','April','May','June','July','August','Septembe r','October','November','December'][myDate.getMonth()]);
// -->
尝试一下：
function deliver(inDays, startingOn){
var s, f = 0, d;
if(!inDays)inDays = 0;
s = !startingOn ? new Date : new Date(startingOn);
for(var i=0,n,t=0,l=inDays; i<l; i++,t+=86400000){
n = new Date(s.getTime()+t).getDay();
if(n === 0 || n === 6)f++;
}
d = new Date(s.getTime()+86400000*(inDays+f));
return 'Your expected delivery date is '+d.toLocaleDateString();
}
// same day delivery
console.log(deliver());
// deliver in 9 days stating today
console.log(deliver(9));
// deliver in 9 days starting on October 12, 2013 - must be a valid Date String
console.log(deliver(9, 'October 12, 2013'));
/* Note that in the last example the Date starts on the weekend, therefore same
day becomes Monday, which if you don't work weekends is the first day you
would see the order anyways.
*/
这很高兴在交货时向您的客户展示。确保使用服务器端语言（如PHP）根据您所在的位置设置数据库的实际日期。
您应该为此付出代价，因为即使在on年也可以使用。哈！
我编写了一个例程，提供了一个日期对象，应用了方法，通过了一个单独的列表中的公共假日，并且跳过了包括周末在内的所有假日/公共假日。
其帖子
// array of ISO YYYY-MM-DD format dates
publicHolidays = {
uk:["2020-01-01","2020-04-10","2020-04-13","2020-05-08","2020-05-25",
"2020-08-03","2020-08-31","2020-12-25","2020-12-28"],
usa:["2020-01-01","2020-01-20","2020-02-14","2020-02-17","2020-04-10",
"2020-04-12","2020-05-10","2020-05-25","2020-06-21","2020-07-03",
"2020-07-04","2020-09-07","2020-10-12","2020-10-31","2020,11,11",
"2020-11-26","2020-12-25"]
}
// check if there is a match in the array
Date.prototype.isPublicHoliday = function( data ){// we check for a public holiday
if(!data) return 1;
return data.indexOf(this.toISOString().slice(0,10))>-1? 0:1;
}
// calculation of business days
Date.prototype.businessDays = function( d, holidays ){
var holidays = holidays || false, t = new Date( this ); // copy date.
while( d ){ // we loop while d is not zero...
t.setDate( t.getDate() + 1 ); // set a date and test it
switch( t.getDay() ){ // switch is used to allow easier addition of other days of the week
case 0: case 6: break;// sunday & saturday
default: // check if we are a public holiday or not
d -= t.isPublicHoliday( holidays );
}
}
return t.toISOString().slice(0,10); // just the YYY-MM-DD
}
// dummy var, could be a form field input
OrderDate = "2020-02-12";
// test with a UK holiday date
var deliveryDate = new Date(OrderDate).businessDays(7, publicHolidays.usa);
// expected output 2020-02-25
console.log("Order date: %s, Delivery date: %s",OrderDate,deliveryDate );