var businessDays = 10; // this will come from a form
var counter = 0; // I have a counter
var safety = 0; // I have a safety variable
var ship = today = new Date(); // I have the current date and an initialized shipping variable but the buy date will come from a form
while( ++safety <30 ){
ship.setDate( safety ); // add a number of days
switch( ship.getDay() ){
case 0: // Sunday
case 6: // Saturday
break;
default:
counter++;
}
if( counter >= businessDays ) break;
}
// the expected shipping date
console.log(">>> today = " + today);
console.log(">>> days = " + businessDays);
console.log(">>> ship = " + ship);
问题是：如果我在eBay上买东西，他们说发货时间是7到10个工作日，而且我在交货前的日期和工作日，我应该得到一个有效的日期，但我不是。我的日期是从今年1月开始，而不是从购买之日起的7到10个工作日
答案 0 :(得分：7)
我已经改编了Mark Giblin的修订代码，以更好地处理年终日期以及美国联邦假期。见下文......
function businessDaysFromDate(date,businessDays) {
var counter = 0, tmp = new Date(date);
while( businessDays>=0 ) {
tmp.setTime( date.getTime() + counter * 86400000 );
if(isBusinessDay (tmp)) {
--businessDays;
}
++counter;
}
return tmp;
}
function isBusinessDay (date) {
var dayOfWeek = date.getDay();
if(dayOfWeek === 0 || dayOfWeek === 6) {
// Weekend
return false;
}
holidays = [
'12/31+5', // New Year's Day on a saturday celebrated on previous friday
'1/1', // New Year's Day
'1/2+1', // New Year's Day on a sunday celebrated on next monday
'1-3/1', // Birthday of Martin Luther King, third Monday in January
'2-3/1', // Washington's Birthday, third Monday in February
'5~1/1', // Memorial Day, last Monday in May
'7/3+5', // Independence Day
'7/4', // Independence Day
'7/5+1', // Independence Day
'9-1/1', // Labor Day, first Monday in September
'10-2/1', // Columbus Day, second Monday in October
'11/10+5', // Veterans Day
'11/11', // Veterans Day
'11/12+1', // Veterans Day
'11-4/4', // Thanksgiving Day, fourth Thursday in November
'12/24+5', // Christmas Day
'12/25', // Christmas Day
'12/26+1', // Christmas Day
];
var dayOfMonth = date.getDate(),
month = date.getMonth() + 1,
monthDay = month + '/' + dayOfMonth;
if(holidays.indexOf(monthDay)>-1){
return false;
}
var monthDayDay = monthDay + '+' + dayOfWeek;
if(holidays.indexOf(monthDayDay)>-1){
return false;
}
var weekOfMonth = Math.floor((dayOfMonth - 1) / 7) + 1,
monthWeekDay = month + '-' + weekOfMonth + '/' + dayOfWeek;
if(holidays.indexOf(monthWeekDay)>-1){
return false;
}
var lastDayOfMonth = new Date(date);
lastDayOfMonth.setMonth(lastDayOfMonth.getMonth() + 1);
lastDayOfMonth.setDate(0);
var negWeekOfMonth = Math.floor((lastDayOfMonth.getDate() - dayOfMonth - 1) / 7) + 1,
monthNegWeekDay = month + '~' + negWeekOfMonth + '/' + dayOfWeek;
if(holidays.indexOf(monthNegWeekDay)>-1){
return false;
}
return true;
}
答案 1 :(得分：1)
感谢您的投入，我对我为此制作的方法进行了长时间的重新思考，并提出了这个小数字......
var businessDays = 7, counter = 0; // set to 1 to count from next business day
while( businessDays>0 ){
var tmp = new Date();
tmp.setDate( tmp.getDate() + counter++ );
switch( tmp.getDay() ){
case 0: case 6: break;// sunday & saturday
default:
businessDays--;
};
}
这个想法是从工作日开始，并且在遇到的每一天都倒数到零，这一天都落到了工作日的范围内。这种切换的使用将使一个人能够将一周中的一天宣布为非工作日，例如某人可能不在星期一工作，因此增加案例：1将包括星期一。
这是一个简单的脚本，并不考虑公共或银行假期，这将要求使用更复杂的脚本。
结果是设置为发货日期的日期，然后用户可以以他们喜欢的任何格式提取日期信息，例如
var shipDate = tmp.toUTCString().slice(1,15);
答案 2 :(得分：1)
我们的UI默认搜索输入为上一个工作日或一周前。 这是向前和向后都有效的东西。
// add (or subtract) business days to provided date
addBusinessDays = function (startingDate, daysToAdjust) {
var newDate = new Date(startingDate.valueOf()),
businessDaysLeft,
isWeekend,
direction;
// Timezones are scary, let's work with whole-days only
if (daysToAdjust !== parseInt(daysToAdjust, 10)) {
throw new TypeError('addBusinessDays can only adjust by whole days');
}
// short-circuit no work; make direction assignment simpler
if (daysToAdjust === 0) {
return startingDate;
}
direction = daysToAdjust > 0 ? 1 : -1;
// Move the date in the correct direction
// but only count business days toward movement
businessDaysLeft = Math.abs(daysToAdjust);
while (businessDaysLeft) {
newDate.setDate(newDate.getDate() + direction);
isWeekend = newDate.getDay() in {0: 'Sunday', 6: 'Saturday'};
if (!isWeekend) {
businessDaysLeft--;
}
}
return newDate;
};
传递一个可选的假日数据结构并调整它也很容易。
但是，生成假日数据结构，这将需要更多的努力，不仅针对每个国家和地区，而且针对每个组织。
答案 3 :(得分：0)
ship.setDate( safety ); // add a number of days
不添加天数。它设定了一天。更多信息：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Date/setDate
如果您想添加日期，请执行以下操作：
ship.setDate(ship.getDate()+1);
答案 4 :(得分：0)
从
更改第7行ship.setDate( safety ); // add a number of days
到
ship.setDate( ship.getDate() + safety );
问题在于您要添加天，而不是设置天数。
答案 5 :(得分：0)
你的主要问题是每次增加安全性意味着你每次循环都会增加多天，而不是1.所以第一次循环= 1，第二次= 1 + 2等等。
我相信这可以按你的意愿运作：
var businessDays = 10; // this will come from a form
var counter = 0; // I have a counter
var safety = 0; // I have a safety variable
var ship = today = new Date(); // I have the current date and an initialized shipping variable but the buy date will come from a form
console.log(">>> today = " + today);
// now the loop...
while( ++safety <30 ){
ship.setDate(ship.getDate()+1 );
switch( ship.getDay() ){
case 0: // Sunday
case 6: // Saturday
break;
default:
counter++;
}
if( counter >= businessDays ) break;
}
// add a number of days
// the expected shipping date
console.log(">>> days = " + businessDays);
console.log(">>> ship = " + ship);
答案 6 :(得分：0)
我需要类似但有所不同的东西，这就是我想出的。
对于每个有假期的月份，将假期添加到具有一个键的对象中。然后，该键在该月中有一系列被视为假期的天。
function getDueDate(date) {
var numBusinessDays = 20;
var saturday = 6;
var sunday = 0;
var holidays = {
/* Months are zero-based. 0 = Jan, 11 = Dec */
0: [1, 2],
1: [6],
3: [24],
11: [25, 26]
};
var dayOfWeek = null;
var dayOfMonth = null;
var month = null;
var isWeekday = null;
var monthHolidays = null;
var isHoliday = null;
while (numBusinessDays > 0) {
dayOfWeek = date.getDay();
dayOfMonth = date.getDate();
month = date.getMonth();
isWeekday = dayOfWeek !== saturday && dayOfWeek !== sunday;
monthHolidays = holidays[month];
isHoliday = monthHolidays
? monthHolidays.indexOf(dayOfMonth) > -1
: false;
if (isWeekday && !isHoliday) --numBusinessDays;
date.setDate(dayOfMonth + 1);
}
return date;
}
答案 7 :(得分：0)
我刚刚发现此脚本运行良好，可以为您所在国家/地区的假期提供可选数组
function addBusinessDays(date, days, holidays) {
var calendar = java.util.Calendar.getInstance();
calendar.setTimeInMillis(date.getTime());
var numberOfDaysToAdd = Math.abs(days);
var daysToAdd = days < 0 ? -1 : 1;
var businessDaysAdded = 0;
function isHoliday(dateToCheck) {
if (holidays && holidays.length > 0) {
for (var i = 0; i < holidays.length; i++) {
if (holidays[i].getFullYear() == dateToCheck.get(java.util.Calendar.YEAR) && holidays[i].getMonth() == dateToCheck.get(java.util.Calendar.MONTH) && holidays[i].getDate() == dateToCheck.get(java.util.Calendar.DAY_OF_MONTH)) {
return true;
}
}
}
return false;
}
while (businessDaysAdded < numberOfDaysToAdd) {
calendar.add(java.util.Calendar.DATE, daysToAdd);
if (calendar.get(java.util.Calendar.DAY_OF_WEEK) == java.util.Calendar.SATURDAY || calendar.get(java.util.Calendar.DAY_OF_WEEK) == java.util.Calendar.SUNDAY) {
// add another day
continue;
}
if (isHoliday(calendar)) {
// add another day
continue;
}
businessDaysAdded ++;
}
return new Date(calendar.getTimeInMillis());
}