AddBusinessDays和GetBusinessDays

public static DateTime AddBusinessDays(DateTime date, int days)
{
    if (days < 0)
    {
        throw new ArgumentException("days cannot be negative", "days");
    }

    if (days == 0) return date;

    if (date.DayOfWeek == DayOfWeek.Saturday)
    {
        date = date.AddDays(2);
        days -= 1;
    }
    else if (date.DayOfWeek == DayOfWeek.Sunday)
    {
        date = date.AddDays(1);
        days -= 1;
    }

    date = date.AddDays(days / 5 * 7);
    int extraDays = days % 5;

    if ((int)date.DayOfWeek + extraDays > 5)
    {
        extraDays += 2;
    }

    return date.AddDays(extraDays);

}

第二个功能GetBusinessDays可以按如下方式实现:

public static int GetBusinessDays(DateTime start, DateTime end)
{
    if (start.DayOfWeek == DayOfWeek.Saturday)
    {
        start = start.AddDays(2);
    }
    else if (start.DayOfWeek == DayOfWeek.Sunday)
    {
        start = start.AddDays(1);
    }

    if (end.DayOfWeek == DayOfWeek.Saturday)
    {
        end = end.AddDays(-1);
    }
    else if (end.DayOfWeek == DayOfWeek.Sunday)
    {
        end = end.AddDays(-2);
    }

    int diff = (int)end.Subtract(start).TotalDays;

    int result = diff / 7 * 5 + diff % 7;

    if (end.DayOfWeek < start.DayOfWeek)
    {
        return result - 2;
    }
    else{
        return result;
    }
}

使用 Fluent DateTime

var now = DateTime.Now;
var dateTime1 = now.AddBusinessDays(3);
var dateTime2 = now.SubtractBusinessDays(5);

内部代码如下

    /// <summary>
    /// Adds the given number of business days to the <see cref="DateTime"/>.
    /// </summary>
    /// <param name="current">The date to be changed.</param>
    /// <param name="days">Number of business days to be added.</param>
    /// <returns>A <see cref="DateTime"/> increased by a given number of business days.</returns>
    public static DateTime AddBusinessDays(this DateTime current, int days)
    {
        var sign = Math.Sign(days);
        var unsignedDays = Math.Abs(days);
        for (var i = 0; i < unsignedDays; i++)
        {
            do
            {
                current = current.AddDays(sign);
            }
            while (current.DayOfWeek == DayOfWeek.Saturday ||
                current.DayOfWeek == DayOfWeek.Sunday);
        }
        return current;
    }

    /// <summary>
    /// Subtracts the given number of business days to the <see cref="DateTime"/>.
    /// </summary>
    /// <param name="current">The date to be changed.</param>
    /// <param name="days">Number of business days to be subtracted.</param>
    /// <returns>A <see cref="DateTime"/> increased by a given number of business days.</returns>
    public static DateTime SubtractBusinessDays(this DateTime current, int days)
    {
        return AddBusinessDays(current, -days);
    }

使用负数的工作日进行减去。我认为这是一个相当优雅的解决方案。它似乎在所有情况下都有效。

namespace Extensions.DateTime
{
    public static class BusinessDays
    {
        public static System.DateTime AddBusinessDays(this System.DateTime source, int businessDays)
        {
            var dayOfWeek = businessDays < 0
                                ? ((int)source.DayOfWeek - 12) % 7
                                : ((int)source.DayOfWeek + 6) % 7;

            switch (dayOfWeek)
            {
                case 6:
                    businessDays--;
                    break;
                case -6:
                    businessDays++;
                    break;
            }

            return source.AddDays(businessDays + ((businessDays + dayOfWeek) / 5) * 2);
        }
    }
}

例子:

using System;
using System.Windows.Forms;
using Extensions.DateTime;

namespace AddBusinessDaysTest
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
            label1.Text = DateTime.Now.AddBusinessDays(5).ToString();
            label2.Text = DateTime.Now.AddBusinessDays(-36).ToString();
        }
    }
}

例如:

• 周三-3个工作日=上周五
• 周五-7个工作日=上周三
• 周二-5个工作日=上周二

好吧,你明白了;)

最后我写了这个扩展类

public static partial class MyExtensions
{
    public static DateTime AddBusinessDays(this DateTime date, int addDays)
    {
        while (addDays != 0)
        {
            date = date.AddDays(Math.Sign(addDays));
            if (MyClass.IsBusinessDay(date))
            {
                addDays = addDays - Math.Sign(addDays);
            }
        }
        return date;
    }
}

它使用了我认为在其他地方有用的方法。。。

public class MyClass
{
    public static bool IsBusinessDay(DateTime date)
    {
        switch (date.DayOfWeek)
        {
            case DayOfWeek.Monday:
            case DayOfWeek.Tuesday:
            case DayOfWeek.Wednesday:
            case DayOfWeek.Thursday:
            case DayOfWeek.Friday:
                return true;
            default:
                return false;
        }
    }
}

如果你不想为此烦恼,你可以把它换掉 if (MyClass.IsBusinessDay(date)) 如果 if ((date.DayOfWeek != DayOfWeek.Saturday) && (date.DayOfWeek != DayOfWeek.Sunday))

所以现在你可以做了

var myDate = DateTime.Now.AddBusinessDays(-3);

或

var myDate = DateTime.Now.AddBusinessDays(5);

以下是一些测试的结果:

Test                         Expected   Result
Wednesday -4 business days   Thursday   Thursday
Wednesday -3 business days   Friday     Friday
Wednesday +3 business days   Monday     Monday
Friday -7 business days      Wednesday  Wednesday
Tuesday -5 business days     Tuesday    Tuesday
Friday +1 business days      Monday     Monday
Saturday +1 business days    Monday     Monday
Sunday -1 business days      Friday     Friday
Monday -1 business days      Friday     Friday
Monday +1 business days      Tuesday    Tuesday
Monday +0 business days      Monday     Monday

public static DateTime AddBusinessDays(this DateTime date, int days)
{
    date = date.AddDays((days / 5) * 7);

    int remainder = days % 5;

    switch (date.DayOfWeek)
    {
        case DayOfWeek.Tuesday:
            if (remainder > 3) date = date.AddDays(2);
            break;
        case DayOfWeek.Wednesday:
            if (remainder > 2) date = date.AddDays(2);
            break;
        case DayOfWeek.Thursday:
            if (remainder > 1) date = date.AddDays(2);
            break;
        case DayOfWeek.Friday:
            if (remainder > 0) date = date.AddDays(2);
            break;
        case DayOfWeek.Saturday:
            if (days > 0) date = date.AddDays((remainder == 0) ? 2 : 1);
            break;
        case DayOfWeek.Sunday:
            if (days > 0) date = date.AddDays((remainder == 0) ? 1 : 0);
            break;
        default:  // monday
            break;
    }

    return date.AddDays(remainder);
}

我来晚了,但我做了一个小图书馆,里面有在工作日进行简单操作所需的所有定制。。。我把它留在这里: Working Days Management

唯一真正的解决方案是让这些调用访问定义业务日历的数据库表。您可以将其编码为周一到周五的工作周,而不会有太多困难,但处理假期将是一项挑战。

编辑以添加非优雅和未经测试的部分解决方案:

public static DateTime AddBusinessDays(this DateTime date, int days)
{
    for (int index = 0; index < days; index++)
    {
        switch (date.DayOfWeek)
        {
            case DayOfWeek.Friday:
                date = date.AddDays(3);
                break;
            case DayOfWeek.Saturday:
                date = date.AddDays(2);
                break;
            default:
                date = date.AddDays(1);
                break;
         }
    }
    return date;
}

我还违反了无循环要求。

我之所以重新发布这篇文章,是因为今天我必须找到一种方法来排除 和 星期日

• 国家固定假日(至少对于西方国家,如2001年1月)。
• 计算假期(如复活节和复活节星期一)。
• 国家特定假日(如意大利解放日或美国ID4)。
• 特定于城镇的假日(如罗马圣守护神日)。
• 任何其他定制假日(如“明天我们的办公室将关闭”)。

最后,我提出了以下一组帮助器/扩展类:尽管它们并不十分优雅,因为它们大量使用了非有效循环,但它们足够好,可以永远解决我的问题。我在这篇文章中删除了全部源代码,希望它对其他人也有用。

源代码

/// <summary>
/// Helper/extension class for manipulating date and time values.
/// </summary>
public static class DateTimeExtensions
{
    /// <summary>
    /// Calculates the absolute year difference between two dates.
    /// </summary>
    /// <param name="dt1"></param>
    /// <param name="dt2"></param>
    /// <returns>A whole number representing the number of full years between the specified dates.</returns>
    public static int Years(DateTime dt1,DateTime dt2)
    {
        return Months(dt1,dt2)/12;
        //if (dt2<dt1)
        //{
        //    DateTime dt0=dt1;
        //    dt1=dt2;
        //    dt2=dt0;
        //}

        //int diff=dt2.Year-dt1.Year;
        //int m1=dt1.Month;
        //int m2=dt2.Month;
        //if (m2>m1) return diff;
        //if (m2==m1 && dt2.Day>=dt1.Day) return diff;
        //return (diff-1);
    }

    /// <summary>
    /// Calculates the absolute year difference between two dates.
    /// Alternative, stand-alone version (without other DateTimeUtil dependency nesting required)
    /// </summary>
    /// <param name="start"></param>
    /// <param name="end"></param>
    /// <returns></returns>
    public static int Years2(DateTime start, DateTime end)
    {
        return (end.Year - start.Year - 1) +
            (((end.Month > start.Month) ||
            ((end.Month == start.Month) && (end.Day >= start.Day))) ? 1 : 0);
    }

    /// <summary>
    /// Calculates the absolute month difference between two dates.
    /// </summary>
    /// <param name="dt1"></param>
    /// <param name="dt2"></param>
    /// <returns>A whole number representing the number of full months between the specified dates.</returns>
    public static int Months(DateTime dt1,DateTime dt2)
    {
        if (dt2<dt1)
        {
            DateTime dt0=dt1;
            dt1=dt2;
            dt2=dt0;
        }

        dt2=dt2.AddDays(-(dt1.Day-1));
        return (dt2.Year-dt1.Year)*12+(dt2.Month-dt1.Month);
    }

    /// <summary>
    /// Returns the higher of the two date time values.
    /// </summary>
    /// <param name="dt1">The first of the two <c>DateTime</c> values to compare.</param>
    /// <param name="dt2">The second of the two <c>DateTime</c> values to compare.</param>
    /// <returns><c>dt1</c> or <c>dt2</c>, whichever is higher.</returns>
    public static DateTime Max(DateTime dt1,DateTime dt2)
    {
        return (dt2>dt1?dt2:dt1);
    }

    /// <summary>
    /// Returns the lower of the two date time values.
    /// </summary>
    /// <param name="dt1">The first of the two <c>DateTime</c> values to compare.</param>
    /// <param name="dt2">The second of the two <c>DateTime</c> values to compare.</param>
    /// <returns><c>dt1</c> or <c>dt2</c>, whichever is lower.</returns>
    public static DateTime Min(DateTime dt1,DateTime dt2)
    {
        return (dt2<dt1?dt2:dt1);
    }

    /// <summary>
    /// Adds the given number of business days to the <see cref="DateTime"/>.
    /// </summary>
    /// <param name="current">The date to be changed.</param>
    /// <param name="days">Number of business days to be added.</param>
    /// <param name="holidays">An optional list of holiday (non-business) days to consider.</param>
    /// <returns>A <see cref="DateTime"/> increased by a given number of business days.</returns>
    public static DateTime AddBusinessDays(
        this DateTime current, 
        int days, 
        IEnumerable<DateTime> holidays = null)
    {
        var sign = Math.Sign(days);
        var unsignedDays = Math.Abs(days);
        for (var i = 0; i < unsignedDays; i++)
        {
            do
            {
                current = current.AddDays(sign);
            }
            while (current.DayOfWeek == DayOfWeek.Saturday
                || current.DayOfWeek == DayOfWeek.Sunday
                || (holidays != null && holidays.Contains(current.Date))
                );
        }
        return current;
    }

    /// <summary>
    /// Subtracts the given number of business days to the <see cref="DateTime"/>.
    /// </summary>
    /// <param name="current">The date to be changed.</param>
    /// <param name="days">Number of business days to be subtracted.</param>
    /// <param name="holidays">An optional list of holiday (non-business) days to consider.</param>
    /// <returns>A <see cref="DateTime"/> increased by a given number of business days.</returns>
    public static DateTime SubtractBusinessDays(
        this DateTime current, 
        int days,
        IEnumerable<DateTime> holidays)
    {
        return AddBusinessDays(current, -days, holidays);
    }

    /// <summary>
    /// Retrieves the number of business days from two dates
    /// </summary>
    /// <param name="startDate">The inclusive start date</param>
    /// <param name="endDate">The inclusive end date</param>
    /// <param name="holidays">An optional list of holiday (non-business) days to consider.</param>
    /// <returns></returns>
    public static int GetBusinessDays(
        this DateTime startDate, 
        DateTime endDate,
        IEnumerable<DateTime> holidays)
    {
        if (startDate > endDate)
            throw new NotSupportedException("ERROR: [startDate] cannot be greater than [endDate].");

        int cnt = 0;
        for (var current = startDate; current < endDate; current = current.AddDays(1))
        {
            if (current.DayOfWeek == DayOfWeek.Saturday
                || current.DayOfWeek == DayOfWeek.Sunday
                || (holidays != null && holidays.Contains(current.Date))
                )
            {
                // skip holiday
            }
            else cnt++;
        }
        return cnt;
    }

    /// <summary>
    /// Calculate Easter Sunday for any given year.
    /// src.: https://stackoverflow.com/a/2510411/1233379
    /// </summary>
    /// <param name="year">The year to calcolate Easter against.</param>
    /// <returns>a DateTime object containing the Easter month and day for the given year</returns>
    public static DateTime GetEasterSunday(int year)
    {
        int day = 0;
        int month = 0;

        int g = year % 19;
        int c = year / 100;
        int h = (c - (int)(c / 4) - (int)((8 * c + 13) / 25) + 19 * g + 15) % 30;
        int i = h - (int)(h / 28) * (1 - (int)(h / 28) * (int)(29 / (h + 1)) * (int)((21 - g) / 11));

        day = i - ((year + (int)(year / 4) + i + 2 - c + (int)(c / 4)) % 7) + 28;
        month = 3;

        if (day > 31)
        {
            month++;
            day -= 31;
        }

        return new DateTime(year, month, day);
    }

    /// <summary>
    /// Retrieve holidays for given years
    /// </summary>
    /// <param name="years">an array of years to retrieve the holidays</param>
    /// <param name="countryCode">a country two letter ISO (ex.: "IT") to add the holidays specific for that country</param>
    /// <param name="cityName">a city name to add the holidays specific for that city</param>
    /// <returns></returns>
    public static IEnumerable<DateTime> GetHolidays(IEnumerable<int> years, string countryCode = null, string cityName = null)
    {
        var lst = new List<DateTime>();

        foreach (var year in years.Distinct())
        {
            lst.AddRange(new[] {
                new DateTime(year, 1, 1),       // 1 gennaio (capodanno)
                new DateTime(year, 1, 6),       // 6 gennaio (epifania)
                new DateTime(year, 5, 1),       // 1 maggio (lavoro)
                new DateTime(year, 8, 15),      // 15 agosto (ferragosto)
                new DateTime(year, 11, 1),      // 1 novembre (ognissanti)
                new DateTime(year, 12, 8),      // 8 dicembre (immacolata concezione)
                new DateTime(year, 12, 25),     // 25 dicembre (natale)
                new DateTime(year, 12, 26)      // 26 dicembre (s. stefano)
            });

            // add easter sunday (pasqua) and monday (pasquetta)
            var easterDate = GetEasterSunday(year);
            lst.Add(easterDate);
            lst.Add(easterDate.AddDays(1));

            // country-specific holidays
            if (!String.IsNullOrEmpty(countryCode))
            {
                switch (countryCode.ToUpper())
                {
                    case "IT":
                        lst.Add(new DateTime(year, 4, 25));     // 25 aprile (liberazione)
                        break;
                    case "US":
                        lst.Add(new DateTime(year, 7, 4));     // 4 luglio (Independence Day)
                        break;

                    // todo: add other countries

                    case default:
                        // unsupported country: do nothing
                        break;
                }
            }

            // city-specific holidays
            if (!String.IsNullOrEmpty(cityName))
            {
                switch (cityName)
                {
                    case "Rome":
                    case "Roma":
                        lst.Add(new DateTime(year, 6, 29));  // 29 giugno (s. pietro e paolo)
                        break;
                    case "Milano":
                    case "Milan":
                        lst.Add(new DateTime(year, 12, 7));  // 7 dicembre (s. ambrogio)
                        break;

                    // todo: add other cities

                    default:
                        // unsupported city: do nothing
                        break;

                }
            }
        }
        return lst;
    }
}

这段代码很容易解释,但是这里有几个例子来解释如何使用它。

增加10个工作日(仅跳过周六和周日工作日)

var dtResult = DateTimeUtil.AddBusinessDays(srcDate, 10);

增加10个工作日(跳过2019年的周六、周日和所有国家/地区固定假日)

var dtResult = DateTimeUtil.AddBusinessDays(srcDate, 10, GetHolidays(2019));

var dtResult = DateTimeUtil.AddBusinessDays(srcDate, 10, GetHolidays(2019, "IT"));

增加10个工作日(跳过2019年的周六、周日、所有意大利假日和罗马特定假日)

var dtResult = DateTimeUtil.AddBusinessDays(srcDate, 10, GetHolidays(2019, "IT", "Rome"));

进一步解释了上述函数和代码示例 in this post 我的博客。

好的,这个解决方案略有不同(有一些优点和缺点):

2. https://www.nuget.org/packages/Nager.Date/
3. 只涉及增加营业日
4. 也跳过假期(可以删除)
5. 使用递归

public static class DateTimeExtensions
{
    public static DateTime AddBusinessDays(this DateTime date, int days, CountryCode countryCode)
    {
        if (days < 0)
        {
            throw new ArgumentException("days cannot be negative", "days");
        }

        if (days == 0)
        {
            return date;
        }

        date = date.AddDays(1);

        if (DateSystem.IsWeekend(date, countryCode) || DateSystem.IsPublicHoliday(date, countryCode))
        {
            return date.AddBusinessDays(days, countryCode);
        }

        days -= 1;

        return date.AddBusinessDays(days, countryCode);
    }
}

用法:

[TestFixture]
public class BusinessDaysTests
{
    [TestCase("2021-06-04", 5, "2021-06-11", Nager.Date.CountryCode.GB)]
    [TestCase("2021-06-04", 6, "2021-06-14", Nager.Date.CountryCode.GB)]
    [TestCase("2021-05-28", 6, "2021-06-08", Nager.Date.CountryCode.GB)] // UK holidays 2021-05-31
    [TestCase("2021-06-01", 3, "2021-06-06", Nager.Date.CountryCode.KW)] // Friday-Saturday weekend in Kuwait
    public void AddTests(DateTime initDate, int delayDays, DateTime expectedDate, Nager.Date.CountryCode countryCode)
    {
        var outputDate = initDate.AddBusinessDays(delayDays, countryCode);
        Assert.AreEqual(expectedDate, outputDate);
    }
}

    public static DateTime AddBusinessDays(DateTime date, int days)
    {
        if (days == 0) return date;
        int i = 0;
        while (i < days)
        {
            if (!(date.DayOfWeek == DayOfWeek.Saturday ||  date.DayOfWeek == DayOfWeek.Sunday)) i++;  
            date = date.AddDays(1);
        }
        return date;
    }

我想要一个“AddBusinessDays”,支持负数的天数添加,结果是:

// 0 == Monday, 6 == Sunday
private static int epochDayToDayOfWeek0Based(long epochDay) {
    return (int)Math.floorMod(epochDay + 3, 7);
}

public static int daysBetween(long fromEpochDay, long toEpochDay) {
    // http://stackoverflow.com/questions/1617049/calculate-the-number-of-business-days-between-two-dates
    final int fromDOW = epochDayToDayOfWeek0Based(fromEpochDay);
    final int toDOW = epochDayToDayOfWeek0Based(toEpochDay);
    long calcBusinessDays = ((toEpochDay - fromEpochDay) * 5 + (toDOW - fromDOW) * 2) / 7;

    if (toDOW   == 6) calcBusinessDays -= 1;
    if (fromDOW == 6) calcBusinessDays += 1;
    return (int)calcBusinessDays;
}

public static long addDays(long epochDay, int n) {
    // https://alecpojidaev.wordpress.com/2009/10/29/work-days-calculation-with-c/
    // NB: in .NET, Sunday == 0, but in our code Monday == 0
    final int dow = (epochDayToDayOfWeek0Based(epochDay) + 1) % 7;
    final int wds = n + (dow == 0 ? 1 : dow); // Adjusted number of working days to add, given that we now start from the immediately preceding Sunday
    final int wends = n < 0 ? ((wds - 5) / 5) * 2
                            : (wds / 5) * 2 - (wds % 5 == 0 ? 2 : 0);
    return epochDay - dow + // Find the immediately preceding Sunday
           wds +            // Add computed working days
           wends;           // Add weekends that occur within each complete working week
}

它适用于自纪元以来以日历天数表示的天数,因为这是由新的JDK8 LocalDate类公开的,而我是在Java中工作的。但是应该很容易适应其他设置。

其基本性质是 addDays 总是返回一个工作日,这对所有 d 和 n , daysBetween(d, addDays(d, n)) == n

请注意,从理论上讲,加0天和减0天应该是不同的操作(如果您的日期是星期天,则加0天应带您到星期一,减0天应带您到星期五)。因为不存在负0(浮点之外!),所以我选择将参数n=0解释为意义 零天。

我相信这可能是一种更简单的获取工作日的方式:

    public int GetBusinessDays(DateTime start, DateTime end, params DateTime[] bankHolidays)
    {
        int tld = (int)((end - start).TotalDays) + 1; //including end day
        int not_buss_day = 2 * (tld / 7); //Saturday and Sunday
        int rest = tld % 7; //rest.

        if (rest > 0)
        {
            int tmp = (int)start.DayOfWeek - 1 + rest;
            if (tmp == 6 || start.DayOfWeek == DayOfWeek.Sunday) not_buss_day++; else if (tmp > 6) not_buss_day += 2;
        }

        foreach (DateTime bankHoliday in bankHolidays)
        {
            DateTime bh = bankHoliday.Date;
            if (!(bh.DayOfWeek == DayOfWeek.Saturday || bh.DayOfWeek == DayOfWeek.Sunday) && (start <= bh && bh <= end))
            {
                not_buss_day++;
            }
        }
        return tld - not_buss_day;
    }

这是我的代码,包括出发日期和客户的交货日期。

            // Calculate departure date
            TimeSpan DeliveryTime = new TimeSpan(14, 30, 0); 
            TimeSpan now = DateTime.Now.TimeOfDay;
            DateTime dt = DateTime.Now;
            if (dt.TimeOfDay > DeliveryTime) dt = dt.AddDays(1);
            if (dt.DayOfWeek == DayOfWeek.Saturday) dt = dt.AddDays(1);
            if (dt.DayOfWeek == DayOfWeek.Sunday) dt = dt.AddDays(1);
            dt = dt.Date + DeliveryTime;
            string DepartureDay = "today at "+dt.ToString("HH:mm");
            if (dt.Day!=DateTime.Now.Day)
            {
                DepartureDay = dt.ToString("dddd at HH:mm", new CultureInfo(WebContextState.CurrentUICulture));
            }
            Return DepartureDay;

            // Caclulate delivery date
            dt = dt.AddDays(1);
            if (dt.DayOfWeek == DayOfWeek.Saturday) dt = dt.AddDays(1);
            if (dt.DayOfWeek == DayOfWeek.Sunday) dt = dt.AddDays(1);
            string DeliveryDay = dt.ToString("dddd", new CultureInfo(WebContextState.CurrentUICulture));
            return DeliveryDay;

快乐编码。

public static DateTime AddWorkingDays(this DateTime date, int daysToAdd)
{
    while (daysToAdd > 0)
    {
        date = date.AddDays(1);

        if (date.DayOfWeek != DayOfWeek.Saturday && date.DayOfWeek != DayOfWeek.Sunday)
        {
            daysToAdd -= 1;
        }
    }

    return date;
}

public static int GetBusinessDays(this DateTime start, DateTime end)
            {
                return Enumerable.Range(0, (end- start).Days)
                                .Select(a => start.AddDays(a))
                                .Where(a => a.DayOfWeek != DayOfWeek.Sunday)
                                .Where(a => a.DayOfWeek != DayOfWeek.Saturday)
                                .Count();
    
            }

希望这对别人有帮助。

private DateTime AddWorkingDays(DateTime addToDate, int numberofDays)
    {
        addToDate= addToDate.AddDays(numberofDays);
        while (addToDate.DayOfWeek == DayOfWeek.Saturday || addToDate.DayOfWeek == DayOfWeek.Sunday)
        {
            addToDate= addToDate.AddDays(1);
        }
        return addToDate;
    }