加号。获得下一个日期的日期中的天数（不包括周末）

Question

我有一个约会，我需要添加没有。获得未来日期的天数，但应排除周末。 即

input date = "9-DEC-2011";
No. of days to add  = '13';

next date should be "28-Dec-2011"

这里不计算周末（周六/周日）。

Answer 1

试试这个

var startDate = "9-DEC-2011";
startDate = new Date(startDate.replace(/-/g, "/"));
var endDate = "", noOfDaysToAdd = 13, count = 0;
while(count < noOfDaysToAdd){
    endDate = new Date(startDate.setDate(startDate.getDate() + 1));
    if(endDate.getDay() != 0 && endDate.getDay() != 6){
       //Date.getDay() gives weekday starting from 0(Sunday) to 6(Saturday)
       count++;
    }
}
alert(endDate);//You can format this date as per your requirement

工作Demo

Answer 2

@ShankarSangoli

这是一个较新的版本，它避免在每个循环中重新创建 Date 对象，请注意它现在被包装在一个函数中。

function calcWorkingDays(fromDate, days) {
    var count = 0;
    while (count < days) {
        fromDate.setDate(fromDate.getDate() + 1);
        if (fromDate.getDay() != 0 && fromDate.getDay() != 6) // Skip weekends
            count++;
    }
    return fromDate;
}
alert(calcWorkingDays(new Date("9/DEC/2011"), 13));

Answer 3

这是一个没有任何循环或外部库的优雅解决方案：

function addBusinessDaysToDate(date, days) {
  var day = date.getDay();

  date = new Date(date.getTime());
  date.setDate(date.getDate() + days + (day === 6 ? 2 : +!day) + (Math.floor((days - 1 + (day % 6 || 1)) / 5) * 2));
  return date;
}

var date = "9-DEC-2011";
var newDate = addBusinessDaysToDate(new Date(date.replace(/-/g, "/")), 13);
console.log(newDate.toString().replace(/\S+\s(\S+)\s(\d+)\s(\d+)\s.*/, '$2-$1-$3')); // alerts "28-Dec-2011"

Answer 4

或者你也可以这样

function addWeekdays(date, weekdays) {
    var newDate = new Date(date.getTime());
    var i = 0;
    while (i < weekdays) {
        newDate.setDate(newDate.getDate() + 1);
        var day = newDate.getDay();
        if (day > 1 && day < 7) {
            i++;
        }
    }
    return newDate;
}
var currentDate = new Date('10/31/2014');
var targetDate = addWeekdays(currentDate, 45);
alert(targetDate);

Answer 5

使用moment.js：

const DATE_FORMAT = 'D-MMM-YYYY';
const SUNDAY = 0; // moment day index
const SATURDAY = 6; // moment day index
const WEEKENDS = [SATURDAY, SUNDAY];

function addBusinessDays(stringDate, numberOfDays, dateFormat = DATE_FORMAT) {
  const date = moment(stringDate, dateFormat);
    
  let count = 0;
        
  while (count < numberOfDays) {
    date.add(1, 'day');

    // Skip weekends
    if (WEEKENDS.includes(date.day())) {
      continue;
    }

    // Increment count
    count++;
  }
    
  return date.format(dateFormat);
}

// Test cases
console.log(addBusinessDays('3-Mar-2021', 1)); // 4-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 2)); // 5-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 3)); // 8-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 4)); // 9-Mar-2021
console.log(addBusinessDays('3-Mar-2021', 5)); // 10-Mar-2021
console.log(addBusinessDays('9-Dec-2011', 13)); // 28-Dec-2011
console.log(addBusinessDays('10-Dec-2011', 13)); // 28-Dec-2011 (Saturday, so remain on Friday)
console.log(addBusinessDays('11-Dec-2011', 13)); // 28-Dec-2011 (Sunday, so remain on Friday)
console.log(addBusinessDays('12-Dec-2011', 13)); // 29-Dec-2011
console.log(addBusinessDays('13-Dec-2011', 13)); // 30-Dec-2011
console.log(addBusinessDays('14-Dec-2011', 13)); // 2-Jan-2012

<script src="https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.24.0/moment.min.js"></script>

Answer 6

这个问题已经很老了，但是以前的所有答案都在一天一个地迭代。这可能会在很多天里效率低下。这对我有用，假设 days 是一个正整数并且 startDate 是一个工作日：

function addWorkingDates(startDate, days) {
  const current_day = startDate.getDay() - 1; // Week day, starting on Monday
  const weekend_days = 2 * parseInt((current_day + days) / 5);
  startDate.setDate(changed_to.getDate() + days + weekend_days);
}

addWorkingDates(new Date(),5)

Answer 7

出于某种原因，递归地尝试这个对我来说更直观。此版本不考虑节假日，但您可以更改 isValid 函数以检查任何内容。

function addWeekdaysToDate(date, numberToAdd) {
  var isValid = function(d) { return d.getDay() !== 0 && d.getDay() !== 6 }
  if(Math.abs(numberToAdd) > 1) {
    return addWeekdaysToDate(
      addWeekdaysToDate(date, Math.sign(numberToAdd)),
      numberToAdd - Math.sign(numberToAdd)
    )
  } else if(Math.abs(numberToAdd) === 1) {
    var result = new Date(date)
    result.setDate(result.getDate() + Math.sign(numberToAdd))
    if(isValid(result)) {
      return result
    } else {
      return addWeekdaysToDate(result, Math.sign(numberToAdd))
    }
  } else if(numberToAdd === 0) {
    return date
  }
  return false
}

console.log(addWeekdaysToDate(new Date(), 1))
console.log(addWeekdaysToDate(new Date(), 5))
console.log(addWeekdaysToDate(new Date(), -7))

在certain browsers 中，您可能需要为Math.sign 使用polyfill：

Math.sign = Math.sign || function(x) {
  x = +x; // convert to a number
  if (x === 0 || isNaN(x)) {
    return Number(x);
  }
  return x > 0 ? 1 : -1;
}

Answer 8

试试这个解决方案

<script language="javascript">

function getDateExcludeWeekends(startDay, startMonth, startYear, daysToAdd) {
var sdate = new Date();
var edate = new Date();
var dayMilliseconds = 1000 * 60 * 60 * 24;
sdate.setFullYear(startYear,startMonth,startDay);
edate.setFullYear(startYear,startMonth,startDay+daysToAdd);
var weekendDays = 0;
while (sdate <= edate) {
    var day = sdate.getDay()
    if (day == 0 || day == 6) {
        weekendDays++;
    }
    sdate = new Date(+sdate + dayMilliseconds);
} 
sdate.setFullYear(startYear,startMonth,startDay + weekendDays+daysToAdd);
return sdate;
}

</script>

Answer 9

如果要获取下一个工作日，从特定日期开始，请使用以下代码...

function getNextWorkingDay(originalDate) {
    var nextWorkingDayFound = false;
    var nextWorkingDate = new Date();
    var dateCounter = 1;

    while (!nextWorkingDayFound) {
        nextWorkingDate.setDate(originalDate.getDate() + dateCounter);
        dateCounter++;

        if (!isDateOnWeekend(nextWorkingDate)) {
            nextWorkingDayFound = true;
        } 
    }

    return nextWorkingDate;
}

function isDateOnWeekend(date) {
    if (date.getDay() === 6 || date.getDay() === 0)
        return true;
    else
        return false;
}

Answer 10

试试这个

function calculate() {
  var noOfDaysToAdd = 13;
  var startDate = "9-DEC-2011";
  startDate = new Date(startDate.replace(/-/g, "/"));
  var endDate = "";
  count = 0;
  while (count < noOfDaysToAdd) {
    endDate = new Date(startDate.setDate(startDate.getDate() + 1));
    if (endDate.getDay() != 0 && endDate.getDay() != 6) {
      count++;
    }
  }
  document.getElementById("result").innerHTML = endDate;
}

<div>
  Date of book delivery: <span id="result"></span><br /><br />
  <input type="button" onclick="calculate();" value="Calculate" />
  <br>
  <br>
</div>