在C＃中，如何计算两个日期之间的业务（或工作日）天数？
答案 0 :(得分：110)
我之前有过这样的任务，而且我已经得到了解决方案。 我会避免在可以避免之间的所有日子进行枚举，这就是这种情况。我甚至没有提到创建一堆DateTime实例，正如我在上面的一个答案中看到的那样。这实际上是浪费处理能力。特别是在现实世界的情况下，你必须检查几个月的时间间隔。 请参阅下面的我的代码和评论。
/// <summary>
/// Calculates number of business days, taking into account:
/// - weekends (Saturdays and Sundays)
/// - bank holidays in the middle of the week
/// </summary>
/// <param name="firstDay">First day in the time interval</param>
/// <param name="lastDay">Last day in the time interval</param>
/// <param name="bankHolidays">List of bank holidays excluding weekends</param>
/// <returns>Number of business days during the 'span'</returns>
public static int BusinessDaysUntil(this DateTime firstDay, DateTime lastDay, params DateTime[] bankHolidays)
{
firstDay = firstDay.Date;
lastDay = lastDay.Date;
if (firstDay > lastDay)
throw new ArgumentException("Incorrect last day " + lastDay);
TimeSpan span = lastDay - firstDay;
int businessDays = span.Days + 1;
int fullWeekCount = businessDays / 7;
// find out if there are weekends during the time exceedng the full weeks
if (businessDays > fullWeekCount*7)
{
// we are here to find out if there is a 1-day or 2-days weekend
// in the time interval remaining after subtracting the complete weeks
int firstDayOfWeek = (int) firstDay.DayOfWeek;
int lastDayOfWeek = (int) lastDay.DayOfWeek;
if (lastDayOfWeek < firstDayOfWeek)
lastDayOfWeek += 7;
if (firstDayOfWeek <= 6)
{
if (lastDayOfWeek >= 7)// Both Saturday and Sunday are in the remaining time interval
businessDays -= 2;
else if (lastDayOfWeek >= 6)// Only Saturday is in the remaining time interval
businessDays -= 1;
}
else if (firstDayOfWeek <= 7 && lastDayOfWeek >= 7)// Only Sunday is in the remaining time interval
businessDays -= 1;
}
// subtract the weekends during the full weeks in the interval
businessDays -= fullWeekCount + fullWeekCount;
// subtract the number of bank holidays during the time interval
foreach (DateTime bankHoliday in bankHolidays)
{
DateTime bh = bankHoliday.Date;
if (firstDay <= bh && bh <= lastDay)
--businessDays;
}
return businessDays;
}
2011年8月由Slauma编辑
很棒的答案！虽然有一点虫子。自从2009年回答者缺席以来，我可以自由编辑这个答案。
上面的代码假定DayOfWeek.Sunday的值7不是这种情况。该值实际为0。如果例如firstDay和lastDay都是相同的星期日，则会导致错误的计算。在这种情况下，该方法返回1，但它应该是0。
最容易解决此错误：在以下代码中替换firstDayOfWeek和lastDayOfWeek的行代码：
int firstDayOfWeek = firstDay.DayOfWeek == DayOfWeek.Sunday
? 7 : (int)firstDay.DayOfWeek;
int lastDayOfWeek = lastDay.DayOfWeek == DayOfWeek.Sunday
? 7 : (int)lastDay.DayOfWeek;
现在的结果是：
答案 1 :(得分：87)
确定。我认为是时候发布正确的答案了：
public static double GetBusinessDays(DateTime startD, DateTime endD)
{
double calcBusinessDays =
1 + ((endD - startD).TotalDays * 5 -
(startD.DayOfWeek - endD.DayOfWeek) * 2) / 7;
if (endD.DayOfWeek == DayOfWeek.Saturday) calcBusinessDays--;
if (startD.DayOfWeek == DayOfWeek.Sunday) calcBusinessDays--;
return calcBusinessDays;
}
原始来源：
http://alecpojidaev.wordpress.com/2009/10/29/work-days-calculation-with-c/
P.S。上面发布的解决方案让我因某种原因而成为原因。
答案 2 :(得分：40)
我知道这个问题已经解决了，但我想我可以提供一个更直接的答案，可以帮助其他访客。
以下是我的看法：
public int GetWorkingDays(DateTime from, DateTime to)
{
var dayDifference = (int)to.Subtract(from).TotalDays;
return Enumerable
.Range(1, dayDifference)
.Select(x => from.AddDays(x))
.Count(x => x.DayOfWeek != DayOfWeek.Saturday && x.DayOfWeek != DayOfWeek.Sunday);
}
这是我最初提交的内容：
public int GetWorkingDays(DateTime from, DateTime to)
{
var totalDays = 0;
for (var date = from; date < to; date = date.AddDays(1))
{
if (date.DayOfWeek != DayOfWeek.Saturday
&& date.DayOfWeek != DayOfWeek.Sunday)
totalDays++;
}
return totalDays;
}
答案 3 :(得分：22)
在DateTime上定义一个扩展方法，如下所示：
public static class DateTimeExtensions
{
public static bool IsWorkingDay(this DateTime date)
{
return date.DayOfWeek != DayOfWeek.Saturday
&& date.DayOfWeek != DayOfWeek.Sunday;
}
}
然后，使用在Where子句中过滤更广泛的日期列表：
var allDates = GetDates(); // method which returns a list of dates
// filter dates by working day's
var countOfWorkDays = allDates
.Where(day => day.IsWorkingDay())
.Count() ;
答案 4 :(得分：10)
我使用以下代码来考虑银行假期：
public class WorkingDays
{
public List<DateTime> GetHolidays()
{
var client = new WebClient();
var json = client.DownloadString("https://www.gov.uk/bank-holidays.json");
var js = new JavaScriptSerializer();
var holidays = js.Deserialize <Dictionary<string, Holidays>>(json);
return holidays["england-and-wales"].events.Select(d => d.date).ToList();
}
public int GetWorkingDays(DateTime from, DateTime to)
{
var totalDays = 0;
var holidays = GetHolidays();
for (var date = from.AddDays(1); date <= to; date = date.AddDays(1))
{
if (date.DayOfWeek != DayOfWeek.Saturday
&& date.DayOfWeek != DayOfWeek.Sunday
&& !holidays.Contains(date))
totalDays++;
}
return totalDays;
}
}
public class Holidays
{
public string division { get; set; }
public List<Event> events { get; set; }
}
public class Event
{
public DateTime date { get; set; }
public string notes { get; set; }
public string title { get; set; }
}
单元测试：
[TestClass]
public class WorkingDays
{
[TestMethod]
public void SameDayIsZero()
{
var service = new WorkingDays();
var from = new DateTime(2013, 8, 12);
Assert.AreEqual(0, service.GetWorkingDays(from, from));
}
[TestMethod]
public void CalculateDaysInWorkingWeek()
{
var service = new WorkingDays();
var from = new DateTime(2013, 8, 12);
var to = new DateTime(2013, 8, 16);
Assert.AreEqual(4, service.GetWorkingDays(from, to), "Mon - Fri = 4");
Assert.AreEqual(1, service.GetWorkingDays(from, new DateTime(2013, 8, 13)), "Mon - Tues = 1");
}
[TestMethod]
public void NotIncludeWeekends()
{
var service = new WorkingDays();
var from = new DateTime(2013, 8, 9);
var to = new DateTime(2013, 8, 16);
Assert.AreEqual(5, service.GetWorkingDays(from, to), "Fri - Fri = 5");
Assert.AreEqual(2, service.GetWorkingDays(from, new DateTime(2013, 8, 13)), "Fri - Tues = 2");
Assert.AreEqual(1, service.GetWorkingDays(from, new DateTime(2013, 8, 12)), "Fri - Mon = 1");
}
[TestMethod]
public void AccountForHolidays()
{
var service = new WorkingDays();
var from = new DateTime(2013, 8, 23);
Assert.AreEqual(0, service.GetWorkingDays(from, new DateTime(2013, 8, 26)), "Fri - Mon = 0");
Assert.AreEqual(1, service.GetWorkingDays(from, new DateTime(2013, 8, 27)), "Fri - Tues = 1");
}
}
答案 5 :(得分：5)
这已经被打死了。 :)但是我仍然会提供另一个答案，因为我需要一些不同的东西。此解决方案的不同之处在于它在开始和结束之间返回Business TimeSpan ，您可以设置当天的营业时间，并添加假期。因此，您可以使用它来计算它是在一天，几天，周末甚至假日中发生的。只需从返回的TimeSpan对象中获取所需内容，您就可以获得工作日。它使用日期列表的方式，你可以看到，如果它不是典型的星期六和太阳，那么添加非工作日列表是多么容易。 我测试了一年，看起来超级快。
我只希望代码的粘贴准确无误。但我知道它有效。
public static TimeSpan GetBusinessTimespanBetween(
DateTime start, DateTime end,
TimeSpan workdayStartTime, TimeSpan workdayEndTime,
List<DateTime> holidays = null)
{
if (end < start)
throw new ArgumentException("start datetime must be before end datetime.");
// Just create an empty list for easier coding.
if (holidays == null) holidays = new List<DateTime>();
if (holidays.Where(x => x.TimeOfDay.Ticks > 0).Any())
throw new ArgumentException("holidays can not have a TimeOfDay, only the Date.");
var nonWorkDays = new List<DayOfWeek>() { DayOfWeek.Saturday, DayOfWeek.Sunday };
var startTime = start.TimeOfDay;
// If the start time is before the starting hours, set it to the starting hour.
if (startTime < workdayStartTime) startTime = workdayStartTime;
var timeBeforeEndOfWorkDay = workdayEndTime - startTime;
// If it's after the end of the day, then this time lapse doesn't count.
if (timeBeforeEndOfWorkDay.TotalSeconds < 0) timeBeforeEndOfWorkDay = new TimeSpan();
// If start is during a non work day, it doesn't count.
if (nonWorkDays.Contains(start.DayOfWeek)) timeBeforeEndOfWorkDay = new TimeSpan();
else if (holidays.Contains(start.Date)) timeBeforeEndOfWorkDay = new TimeSpan();
var endTime = end.TimeOfDay;
// If the end time is after the ending hours, set it to the ending hour.
if (endTime > workdayEndTime) endTime = workdayEndTime;
var timeAfterStartOfWorkDay = endTime - workdayStartTime;
// If it's before the start of the day, then this time lapse doesn't count.
if (timeAfterStartOfWorkDay.TotalSeconds < 0) timeAfterStartOfWorkDay = new TimeSpan();
// If end is during a non work day, it doesn't count.
if (nonWorkDays.Contains(end.DayOfWeek)) timeAfterStartOfWorkDay = new TimeSpan();
else if (holidays.Contains(end.Date)) timeAfterStartOfWorkDay = new TimeSpan();
// Easy scenario if the times are during the day day.
if (start.Date.CompareTo(end.Date) == 0)
{
if (nonWorkDays.Contains(start.DayOfWeek)) return new TimeSpan();
else if (holidays.Contains(start.Date)) return new TimeSpan();
return endTime - startTime;
}
else
{
var timeBetween = end - start;
var daysBetween = (int)Math.Floor(timeBetween.TotalDays);
var dailyWorkSeconds = (int)Math.Floor((workdayEndTime - workdayStartTime).TotalSeconds);
var businessDaysBetween = 0;
// Now the fun begins with calculating the actual Business days.
if (daysBetween > 0)
{
var nextStartDay = start.AddDays(1).Date;
var dayBeforeEnd = end.AddDays(-1).Date;
for (DateTime d = nextStartDay; d <= dayBeforeEnd; d = d.AddDays(1))
{
if (nonWorkDays.Contains(d.DayOfWeek)) continue;
else if (holidays.Contains(d.Date)) continue;
businessDaysBetween++;
}
}
var dailyWorkSecondsToAdd = dailyWorkSeconds * businessDaysBetween;
var output = timeBeforeEndOfWorkDay + timeAfterStartOfWorkDay;
output = output + new TimeSpan(0, 0, dailyWorkSecondsToAdd);
return output;
}
}
这是测试代码：请注意，您只需将此函数放在名为 DateHelper 的类中，即可使测试代码正常工作。
[TestMethod]
public void TestGetBusinessTimespanBetween()
{
var workdayStart = new TimeSpan(8, 0, 0);
var workdayEnd = new TimeSpan(17, 0, 0);
var holidays = new List<DateTime>()
{
new DateTime(2018, 1, 15), // a Monday
new DateTime(2018, 2, 15) // a Thursday
};
var testdata = new[]
{
new
{
expectedMinutes = 0,
start = new DateTime(2016, 10, 19, 9, 50, 0),
end = new DateTime(2016, 10, 19, 9, 50, 0)
},
new
{
expectedMinutes = 10,
start = new DateTime(2016, 10, 19, 9, 50, 0),
end = new DateTime(2016, 10, 19, 10, 0, 0)
},
new
{
expectedMinutes = 5,
start = new DateTime(2016, 10, 19, 7, 50, 0),
end = new DateTime(2016, 10, 19, 8, 5, 0)
},
new
{
expectedMinutes = 5,
start = new DateTime(2016, 10, 19, 16, 55, 0),
end = new DateTime(2016, 10, 19, 17, 5, 0)
},
new
{
expectedMinutes = 15,
start = new DateTime(2016, 10, 19, 16, 50, 0),
end = new DateTime(2016, 10, 20, 8, 5, 0)
},
new
{
expectedMinutes = 10,
start = new DateTime(2016, 10, 19, 16, 50, 0),
end = new DateTime(2016, 10, 20, 7, 55, 0)
},
new
{
expectedMinutes = 5,
start = new DateTime(2016, 10, 19, 17, 10, 0),
end = new DateTime(2016, 10, 20, 8, 5, 0)
},
new
{
expectedMinutes = 0,
start = new DateTime(2016, 10, 19, 17, 10, 0),
end = new DateTime(2016, 10, 20, 7, 5, 0)
},
new
{
expectedMinutes = 545,
start = new DateTime(2016, 10, 19, 12, 10, 0),
end = new DateTime(2016, 10, 20, 12, 15, 0)
},
// Spanning multiple weekdays
new
{
expectedMinutes = 835,
start = new DateTime(2016, 10, 19, 12, 10, 0),
end = new DateTime(2016, 10, 21, 8, 5, 0)
},
// Spanning multiple weekdays
new
{
expectedMinutes = 1375,
start = new DateTime(2016, 10, 18, 12, 10, 0),
end = new DateTime(2016, 10, 21, 8, 5, 0)
},
// Spanning from a Thursday to a Tuesday, 5 mins short of complete day.
new
{
expectedMinutes = 1615,
start = new DateTime(2016, 10, 20, 12, 10, 0),
end = new DateTime(2016, 10, 25, 12, 5, 0)
},
// Spanning from a Thursday to a Tuesday, 5 mins beyond complete day.
new
{
expectedMinutes = 1625,
start = new DateTime(2016, 10, 20, 12, 10, 0),
end = new DateTime(2016, 10, 25, 12, 15, 0)
},
// Spanning from a Friday to a Monday, 5 mins beyond complete day.
new
{
expectedMinutes = 545,
start = new DateTime(2016, 10, 21, 12, 10, 0),
end = new DateTime(2016, 10, 24, 12, 15, 0)
},
// Spanning from a Friday to a Monday, 5 mins short complete day.
new
{
expectedMinutes = 535,
start = new DateTime(2016, 10, 21, 12, 10, 0),
end = new DateTime(2016, 10, 24, 12, 5, 0)
},
// Spanning from a Saturday to a Monday, 5 mins short complete day.
new
{
expectedMinutes = 245,
start = new DateTime(2016, 10, 22, 12, 10, 0),
end = new DateTime(2016, 10, 24, 12, 5, 0)
},
// Spanning from a Saturday to a Sunday, 5 mins beyond complete day.
new
{
expectedMinutes = 0,
start = new DateTime(2016, 10, 22, 12, 10, 0),
end = new DateTime(2016, 10, 23, 12, 15, 0)
},
// Times within the same Saturday.
new
{
expectedMinutes = 0,
start = new DateTime(2016, 10, 22, 12, 10, 0),
end = new DateTime(2016, 10, 23, 12, 15, 0)
},
// Spanning from a Saturday to the Sunday next week.
new
{
expectedMinutes = 2700,
start = new DateTime(2016, 10, 22, 12, 10, 0),
end = new DateTime(2016, 10, 30, 12, 15, 0)
},
// Spanning a year.
new
{
expectedMinutes = 143355,
start = new DateTime(2016, 10, 22, 12, 10, 0),
end = new DateTime(2017, 10, 30, 12, 15, 0)
},
// Spanning a year with 2 holidays.
new
{
expectedMinutes = 142815,
start = new DateTime(2017, 10, 22, 12, 10, 0),
end = new DateTime(2018, 10, 30, 12, 15, 0)
},
};
foreach (var item in testdata)
{
Assert.AreEqual(item.expectedMinutes,
DateHelper.GetBusinessTimespanBetween(
item.start, item.end,
workdayStart, workdayEnd,
holidays)
.TotalMinutes);
}
}
答案 6 :(得分：4)
这里有一些代码用于瑞典假期，但您可以调整假期。请注意，我添加了一个您可能要删除的限制，但它是基于Web的系统，我不希望任何人输入一些巨大的日期来占用该过程
public static int GetWorkdays(DateTime from ,DateTime to)
{
int limit = 9999;
int counter = 0;
DateTime current = from;
int result = 0;
if (from > to)
{
DateTime temp = from;
from = to;
to = temp;
}
if (from >= to)
{
return 0;
}
while (current <= to && counter < limit)
{
if (IsSwedishWorkday(current))
{
result++;
}
current = current.AddDays(1);
counter++;
}
return result;
}
public static bool IsSwedishWorkday(DateTime date)
{
return (!IsSwedishHoliday(date) && date.DayOfWeek != DayOfWeek.Saturday && date.DayOfWeek != DayOfWeek.Sunday);
}
public static bool IsSwedishHoliday(DateTime date)
{
return (
IsSameDay(GetEpiphanyDay(date.Year), date) ||
IsSameDay(GetMayDay(date.Year), date) ||
IsSameDay(GetSwedishNationalDay(date.Year), date) ||
IsSameDay(GetChristmasDay(date.Year), date) ||
IsSameDay(GetBoxingDay(date.Year), date) ||
IsSameDay(GetGoodFriday(date.Year), date) ||
IsSameDay(GetAscensionDay(date.Year), date) ||
IsSameDay(GetAllSaintsDay(date.Year), date) ||
IsSameDay(GetMidsummersDay(date.Year), date) ||
IsSameDay(GetPentecostDay(date.Year), date) ||
IsSameDay(GetEasterMonday(date.Year), date) ||
IsSameDay(GetNewYearsDay(date.Year), date) ||
IsSameDay(GetEasterDay(date.Year), date)
);
}
// Trettondagen
public static DateTime GetEpiphanyDay(int year)
{
return new DateTime(year, 1, 6);
}
// Första maj
public static DateTime GetMayDay(int year)
{
return new DateTime(year,5,1);
}
// Juldagen
public static DateTime GetSwedishNationalDay(int year)
{
return new DateTime(year, 6, 6);
}
// Juldagen
public static DateTime GetNewYearsDay(int year)
{
return new DateTime(year,1,1);
}
// Juldagen
public static DateTime GetChristmasDay(int year)
{
return new DateTime(year,12,25);
}
// Annandag jul
public static DateTime GetBoxingDay(int year)
{
return new DateTime(year, 12, 26);
}
// Långfredagen
public static DateTime GetGoodFriday(int year)
{
return GetEasterDay(year).AddDays(-3);
}
// Kristi himmelsfärdsdag
public static DateTime GetAscensionDay(int year)
{
return GetEasterDay(year).AddDays(5*7+4);
}
// Midsommar
public static DateTime GetAllSaintsDay(int year)
{
DateTime result = new DateTime(year,10,31);
while (result.DayOfWeek != DayOfWeek.Saturday)
{
result = result.AddDays(1);
}
return result;
}
// Midsommar
public static DateTime GetMidsummersDay(int year)
{
DateTime result = new DateTime(year, 6, 20);
while (result.DayOfWeek != DayOfWeek.Saturday)
{
result = result.AddDays(1);
}
return result;
}
// Pingstdagen
public static DateTime GetPentecostDay(int year)
{
return GetEasterDay(year).AddDays(7 * 7);
}
// Annandag påsk
public static DateTime GetEasterMonday(int year)
{
return GetEasterDay(year).AddDays(1);
}
public static DateTime GetEasterDay(int y)
{
double c;
double n;
double k;
double i;
double j;
double l;
double m;
double d;
c = System.Math.Floor(y / 100.0);
n = y - 19 * System.Math.Floor(y / 19.0);
k = System.Math.Floor((c - 17) / 25.0);
i = c - System.Math.Floor(c / 4) - System.Math.Floor((c - k) / 3) + 19 * n + 15;
i = i - 30 * System.Math.Floor(i / 30);
i = i - System.Math.Floor(i / 28) * (1 - System.Math.Floor(i / 28) * System.Math.Floor(29 / (i + 1)) * System.Math.Floor((21 - n) / 11));
j = y + System.Math.Floor(y / 4.0) + i + 2 - c + System.Math.Floor(c / 4);
j = j - 7 * System.Math.Floor(j / 7);
l = i - j;
m = 3 + System.Math.Floor((l + 40) / 44);// month
d = l + 28 - 31 * System.Math.Floor(m / 4);// day
double days = ((m == 3) ? d : d + 31);
DateTime result = new DateTime(y, 3, 1).AddDays(days-1);
return result;
}
答案 7 :(得分：4)
这个解决方案避免了迭代，适用于+ ve和-ve工作日差异，并且包括一个单元测试套件，用于对较慢的工作日计数方法进行回归。我还添加了一个简洁的方法来添加工作日也可以以相同的非迭代方式工作。
单元测试涵盖几千个日期组合，以便详尽地测试所有开始/结束工作日组合，包括小日期范围和大日期范围。
重要：我们假设我们通过排除开始日期和包括结束日期来计算天数。这在计算工作日时非常重要，因为您包含/排除的特定开始/结束日期会影响结果。这也确保了两个相等天数之间的差异始终为零，并且我们只包括完整的工作日，因为您通常希望答案在当前开始日期（通常是今天）的任何时间都是正确的，并包括完整的结束日期（例如截止日期）。
注意：此代码需要对假期进行额外调整，但与上述假设一致，此代码必须在开始日期排除假期。
添加工作日：
private static readonly int[,] _addOffset =
{
// 0 1 2 3 4
{0, 1, 2, 3, 4}, // Su 0
{0, 1, 2, 3, 4}, // M 1
{0, 1, 2, 3, 6}, // Tu 2
{0, 1, 4, 5, 6}, // W 3
{0, 1, 4, 5, 6}, // Th 4
{0, 3, 4, 5, 6}, // F 5
{0, 2, 3, 4, 5}, // Sa 6
};
public static DateTime AddWeekdays(this DateTime date, int weekdays)
{
int extraDays = weekdays % 5;
int addDays = weekdays >= 0
? (weekdays / 5) * 7 + _addOffset[(int)date.DayOfWeek, extraDays]
: (weekdays / 5) * 7 - _addOffset[6 - (int)date.DayOfWeek, -extraDays];
return date.AddDays(addDays);
}
计算工作日差异：
static readonly int[,] _diffOffset =
{
// Su M Tu W Th F Sa
{0, 1, 2, 3, 4, 5, 5}, // Su
{4, 0, 1, 2, 3, 4, 4}, // M
{3, 4, 0, 1, 2, 3, 3}, // Tu
{2, 3, 4, 0, 1, 2, 2}, // W
{1, 2, 3, 4, 0, 1, 1}, // Th
{0, 1, 2, 3, 4, 0, 0}, // F
{0, 1, 2, 3, 4, 5, 0}, // Sa
};
public static int GetWeekdaysDiff(this DateTime dtStart, DateTime dtEnd)
{
int daysDiff = (int)(dtEnd - dtStart).TotalDays;
return daysDiff >= 0
? 5 * (daysDiff / 7) + _diffOffset[(int) dtStart.DayOfWeek, (int) dtEnd.DayOfWeek]
: 5 * (daysDiff / 7) - _diffOffset[6 - (int) dtStart.DayOfWeek, 6 - (int) dtEnd.DayOfWeek];
}
我发现堆栈溢出的大多数其他解决方案要么是缓慢的（迭代的）要么过于复杂而且很多都只是简单的错误。故事的道德是......不要相信它，除非<你已经详尽地测试了它！
基于NUnit Combinatorial testing和ShouldBe NUnit扩展的单元测试。
[TestFixture]
public class DateTimeExtensionsTests
{
/// <summary>
/// Exclude start date, Include end date
/// </summary>
/// <param name="dtStart"></param>
/// <param name="dtEnd"></param>
/// <returns></returns>
private IEnumerable<DateTime> GetDateRange(DateTime dtStart, DateTime dtEnd)
{
Console.WriteLine(@"dtStart={0:yy-MMM-dd ddd}, dtEnd={1:yy-MMM-dd ddd}", dtStart, dtEnd);
TimeSpan diff = dtEnd - dtStart;
Console.WriteLine(diff);
if (dtStart <= dtEnd)
{
for (DateTime dt = dtStart.AddDays(1); dt <= dtEnd; dt = dt.AddDays(1))
{
Console.WriteLine(@"dt={0:yy-MMM-dd ddd}", dt);
yield return dt;
}
}
else
{
for (DateTime dt = dtStart.AddDays(-1); dt >= dtEnd; dt = dt.AddDays(-1))
{
Console.WriteLine(@"dt={0:yy-MMM-dd ddd}", dt);
yield return dt;
}
}
}
[Test, Combinatorial]
public void TestGetWeekdaysDiff(
[Values(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 20, 30)]
int startDay,
[Values(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 20, 30)]
int endDay,
[Values(7)]
int startMonth,
[Values(7)]
int endMonth)
{
// Arrange
DateTime dtStart = new DateTime(2016, startMonth, startDay);
DateTime dtEnd = new DateTime(2016, endMonth, endDay);
int nDays = GetDateRange(dtStart, dtEnd)
.Count(dt => dt.DayOfWeek != DayOfWeek.Saturday && dt.DayOfWeek != DayOfWeek.Sunday);
if (dtEnd < dtStart) nDays = -nDays;
Console.WriteLine(@"countBusDays={0}", nDays);
// Act / Assert
dtStart.GetWeekdaysDiff(dtEnd).ShouldBe(nDays);
}
[Test, Combinatorial]
public void TestAddWeekdays(
[Values(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 20, 30)]
int startDay,
[Values(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 20, 30)]
int weekdays)
{
DateTime dtStart = new DateTime(2016, 7, startDay);
DateTime dtEnd1 = dtStart.AddWeekdays(weekdays); // ADD
dtStart.GetWeekdaysDiff(dtEnd1).ShouldBe(weekdays);
DateTime dtEnd2 = dtStart.AddWeekdays(-weekdays); // SUBTRACT
dtStart.GetWeekdaysDiff(dtEnd2).ShouldBe(-weekdays);
}
}
答案 8 :(得分：1)
我经常搜索一个易于消化的算法来计算两个日期之间的工作日，并且还排除了国定假日，最后我决定采用这种方法：
public static int NumberOfWorkingDaysBetween2Dates(DateTime start,DateTime due,IEnumerable<DateTime> holidays)
{
var dic = new Dictionary<DateTime, DayOfWeek>();
var totalDays = (due - start).Days;
for (int i = 0; i < totalDays + 1; i++)
{
if (!holidays.Any(x => x == start.AddDays(i)))
dic.Add(start.AddDays(i), start.AddDays(i).DayOfWeek);
}
return dic.Where(x => x.Value != DayOfWeek.Saturday && x.Value != DayOfWeek.Sunday).Count();
}
基本上我想跟每个日期一起评估我的条件：
但我也希望避免迭代日期。
通过运行和测量需要评估1年的时间，我得出以下结果：
static void Main(string[] args)
{
var start = new DateTime(2017, 1, 1);
var due = new DateTime(2017, 12, 31);
var sw = Stopwatch.StartNew();
var days = NumberOfWorkingDaysBetween2Dates(start, due,NationalHolidays());
sw.Stop();
Console.WriteLine($"Total working days = {days} --- time: {sw.Elapsed}");
Console.ReadLine();
// result is:
// Total working days = 249-- - time: 00:00:00.0269087
}
答案 9 :(得分：1)
这是一个快速示例代码。这是一个类方法，所以只能在你的类中工作。如果您希望它为static，请将签名更改为private static（或public static）。
private IEnumerable<DateTime> GetWorkingDays(DateTime sd, DateTime ed)
{
for (var d = sd; d <= ed; d = d.AddDays(1))
if (d.DayOfWeek != DayOfWeek.Saturday && d.DayOfWeek != DayOfWeek.Sunday)
yield return d;
}
此方法创建一个循环变量d，将其初始化为开始日sd，然后每次迭代增加一天（d = d.AddDays(1)）。
使用yield返回所需的值，这会创建iterator。关于迭代器的一个很酷的事情是它们没有在内存中保存IEnumerable的所有值，只是按顺序调用每个值。这意味着你可以从现在开始调用这种方法，而不必担心内存不足。
答案 10 :(得分：1)
此方法不使用任何循环，实际上非常简单。由于我们知道每个星期有5个工作日，因此它将日期范围扩展到整周。然后，它使用查找表查找要从开始和结束中减去以获得正确结果的工作日数。我已经扩展了计算以帮助显示正在发生的事情，但是如果需要的话，可以将整个事情压缩为一行。
无论如何，这对我有用，所以我认为我会在这里张贴它，以免对他人有所帮助。快乐的编码。
计算
文化
代码假定工作日为星期一至星期五。对于其他文化，例如星期日至星期四，您需要在计算之前抵消日期。
方法
public int Weekdays(DateTime min, DateTime max)
{
if (min.Date > max.Date) throw new Exception("Invalid date span");
var t = (max.AddDays(1).Date - min.Date).TotalDays;
var a = (int) min.DayOfWeek;
var b = 6 - (int) max.DayOfWeek;
var k = 1.4;
var m = new int[]{0, 0, 1, 2, 3, 4, 5};
var c = m[a] + m[b];
return (int)((t + a + b) / k) - c;
}
答案 11 :(得分：1)
这是我为该任务编写的帮助函数。
它还通过out参数返回周末计数。
如果希望的话，可以为使用不同周末的国家/地区自定义运行时间的“周末”天，或通过weekendDays[]可选参数包括假期：
public static int GetNetworkDays(DateTime startDate, DateTime endDate,out int totalWeekenDays, DayOfWeek[] weekendDays = null)
{
if (startDate >= endDate)
{
throw new Exception("start date can not be greater then or equel to end date");
}
DayOfWeek[] weekends = new DayOfWeek[] { DayOfWeek.Sunday, DayOfWeek.Saturday };
if (weekendDays != null)
{
weekends = weekendDays;
}
var totaldays = (endDate - startDate).TotalDays + 1; // add one to include the first day to
var counter = 0;
var workdaysCounter = 0;
var weekendsCounter = 0;
for (int i = 0; i < totaldays; i++)
{
if (weekends.Contains(startDate.AddDays(counter).DayOfWeek))
{
weekendsCounter++;
}
else
{
workdaysCounter++;
}
counter++;
}
totalWeekenDays = weekendsCounter;
return workdaysCounter;
}
答案 12 :(得分：1)
我认为上述答案中没有一个是正确的。它们都没有解决所有特殊情况，例如日期在周末中间开始和结束，日期从星期五开始到下周一结束等等。最重要的是，它们将计算结果全部变为整体天，所以如果开始日期是在星期六的中间，它将从工作日减去一整天，给出错误的结果......
无论如何，这是我的解决方案非常有效和简单，适用于所有情况。诀窍就是找到上周一的开始和结束日期，然后在周末开始和结束时做一个小补偿：
public double WorkDays(DateTime startDate, DateTime endDate){
double weekendDays;
double days = endDate.Subtract(startDate).TotalDays;
if(days<0) return 0;
DateTime startMonday = startDate.AddDays(DayOfWeek.Monday - startDate.DayOfWeek).Date;
DateTime endMonday = endDate.AddDays(DayOfWeek.Monday - endDate.DayOfWeek).Date;
weekendDays = ((endMonday.Subtract(startMonday).TotalDays) / 7) * 2;
// compute fractionary part of weekend days
double diffStart = startDate.Subtract(startMonday).TotalDays - 5;
double diffEnd = endDate.Subtract(endMonday).TotalDays - 5;
// compensate weekenddays
if(diffStart>0) weekendDays -= diffStart;
if(diffEnd>0) weekendDays += diffEnd;
return days - weekendDays;
}
答案 13 :(得分：0)
您只需要迭代时间范围内的每一天，并从计数器中减去一天（如果是星期六或星期日）。
private float SubtractWeekend(DateTime start, DateTime end) {
float totaldays = (end.Date - start.Date).Days;
var iterationVal = totalDays;
for (int i = 0; i <= iterationVal; i++) {
int dayVal = (int)start.Date.AddDays(i).DayOfWeek;
if(dayVal == 6 || dayVal == 0) {
// saturday or sunday
totalDays--;
}
}
return totalDays;
}
答案 14 :(得分：0)
我想出了以下解决方案
let arr = [
{Avobath_Count: "5"},
{DragonsEgg_Count: "3"},
{Intergalactic_Count: "2"},
{Twilight_Count: "9"},
{SexBomb_Count: "6"},
{TheExperimenter_Count: "6"},
{TurtleImmersion_Count: "5"},
{Butterball_Count: "3"},
{MarshmallowWorld_Count: "0"},
{ThinkPink_Count: "2"},
]
arr.sort((a, b) => Object.values(b)[0] - Object.values(a)[0]);
console.log(arr);答案 15 :(得分：0)
另一种计算工作日的方法是，不考虑假期，而是考虑一天中返回的部分天数：
public static double GetBusinessDays(DateTime startD, DateTime endD)
{
while (IsWeekend(startD))
startD = startD.Date.AddDays(1);
while (IsWeekend(endD))
endD = endD.Date.AddDays(-1);
var bussDays = (endD - startD).TotalDays -
(2 * ((int)(endD - startD).TotalDays / 7)) -
(startD.DayOfWeek > endD.DayOfWeek ? 2 : 0);
return bussDays;
}
public static bool IsWeekend(DateTime d)
{
return d.DayOfWeek == DayOfWeek.Saturday || d.DayOfWeek == DayOfWeek.Sunday;
}
您可以在这里摆弄它：https://rextester.com/ASHRS53997
答案 16 :(得分：0)
public static int CalculateBusinessDaysInRange(this DateTime startDate, DateTime endDate, params DateTime[] holidayDates)
{
endDate = endDate.Date;
if(startDate > endDate)
throw new ArgumentException("The end date can not be before the start date!", nameof(endDate));
int accumulator = 0;
DateTime itterator = startDate.Date;
do
{
if(itterator.DayOfWeek != DayOfWeek.Saturday && itterator.DayOfWeek != DayOfWeek.Sunday && !holidayDates.Any(hol => hol.Date == itterator))
{ accumulator++; }
}
while((itterator = itterator.AddDays(1)).Date <= endDate);
return accumulator
}
我之所以这样发布，是因为尽管给出了所有出色的答案，但数学对我来说都没有道理。这绝对是一种应该有效且可维护的KISS方法。当然，如果您计算的范围大于2-3个月，这将不是最有效的方法。我们只需确定是星期六还是星期日，或者该日期是给定的假日日期。如果不是，我们添加一个工作日。如果是这样，那一切都很好。
我敢肯定，使用LINQ可以使它更加简化，但是这种方式更容易理解。
答案 17 :(得分：0)
这是您也可以选择周末日期的另一个示例，
public int GetBuisnessDays(DateTime StartDate, DateTime EndDate)
{
int counter = 0;
if (StartDate.Date == EndDate.Date)
{
if (StartDate.DayOfWeek != DayOfWeek.Saturday && StartDate.DayOfWeek != DayOfWeek.Friday)
return 1;
return 0;
}
while (StartDate <= EndDate)
{
if (StartDate.DayOfWeek != DayOfWeek.Saturday && StartDate.DayOfWeek != DayOfWeek.Friday)
++counter;
StartDate = StartDate.AddDays(1);
}
return counter;
}
答案 18 :(得分：0)
如果您使用MVC，这是一种方法。 我还计算了国家法定假日或任何节日，通过从holidayscalendar中取出它来排除它，你需要制作它。
foreach (DateTime day in EachDay(model))
{
bool key = false;
foreach (LeaveModel ln in holidaycalendar)
{
if (day.Date == ln.Date && day.DayOfWeek != DayOfWeek.Saturday && day.DayOfWeek != DayOfWeek.Sunday)
{
key = true; break;
}
}
if (day.DayOfWeek == DayOfWeek.Saturday || day.DayOfWeek == DayOfWeek.Sunday)
{
key = true;
}
if (key != true)
{
leavecount++;
}
}
Leavemodel是一个列表
答案 19 :(得分：0)
因为我无法发表评论。接受的解决方案还有一个问题，即使它们位于周末，也会减去银行假期。看看如何检查其他输入，这也是合适的。
因此，foreach应该是：
// subtract the number of bank holidays during the time interval
foreach (DateTime bankHoliday in bankHolidays)
{
DateTime bh = bankHoliday.Date;
// Do not subtract bank holidays when they fall in the weekend to avoid double subtraction
if (bh.DayOfWeek == DayOfWeek.Saturday || bh.DayOfWeek == DayOfWeek.Sunday)
continue;
if (firstDay <= bh && bh <= lastDay)
--businessDays;
}
答案 20 :(得分：0)
所以我有一个类似的任务，除了我必须计算剩余的工作日（从日期不应超过至今），结束日期应跳到下一个工作日。
为了使其更易于理解/可读，我按照以下步骤进行了操作
更新至下一个工作日（如果是周末）。
找出日期之间完整的周数，并为每个完整的周考虑 5 天的总运行时间。
现在剩下的天数只与工作日不同（不会超过 6 天），所以写了一个小循环来获取它（跳过周六和周日）
{
"script": {
"source": "if (ctx._source._ignoredBy == null) {ctx._source._ignoredBy = []; } ctx._source._ignoredBy.removeIf(item -> item.user == params.by.user); ctx._source._ignoredBy.add(params.by)",
"params": {
...
}
}
}
答案 21 :(得分：0)
此方法返回两个日期之间的工作日数：
这里我使用 DayOfWeek 枚举来检查周末。
private static int BusinessDaysLeft(DateTime first, DateTime last)
{
var count = 0;
while (first.Date != last.Date)
{
if(first.DayOfWeek != DayOfWeek.Saturday && first.DayOfWeek != DayOfWeek.Sunday)
count++;
first = first.AddDays(1);
}
return count;
}
答案 22 :(得分：0)
这是另一个想法 - 此方法允许指定任何工作周和假期。
这里的想法是我们找到日期范围的核心，从一周的第一个工作日到一周的最后一个周末。这使我们能够轻松计算整周（没有迭代所有日期）。我们需要做的就是添加在核心范围开始和结束之前的工作日。
public static int CalculateWorkingDays(
DateTime startDate,
DateTime endDate,
IList<DateTime> holidays,
DayOfWeek firstDayOfWeek,
DayOfWeek lastDayOfWeek)
{
// Make sure the defined working days run contiguously
if (lastDayOfWeek < firstDayOfWeek)
{
throw new Exception("Last day of week cannot fall before first day of week!");
}
// Create a list of the days of the week that make-up the weekend by working back
// from the firstDayOfWeek and forward from lastDayOfWeek to get the start and end
// the weekend
var weekendStart = lastDayOfWeek == DayOfWeek.Saturday ? DayOfWeek.Sunday : lastDayOfWeek + 1;
var weekendEnd = firstDayOfWeek == DayOfWeek.Sunday ? DayOfWeek.Saturday : firstDayOfWeek - 1;
var weekendDays = new List<DayOfWeek>();
var w = weekendStart;
do {
weekendDays.Add(w);
if (w == weekendEnd) break;
w = (w == DayOfWeek.Saturday) ? DayOfWeek.Sunday : w + 1;
} while (true);
// Force simple dates - no time
startDate = startDate.Date;
endDate = endDate.Date;
// Ensure a progessive date range
if (endDate < startDate)
{
var t = startDate;
startDate = endDate;
endDate = t;
}
// setup some working variables and constants
const int daysInWeek = 7; // yeah - really!
var actualStartDate = startDate; // this will end up on startOfWeek boundary
var actualEndDate = endDate; // this will end up on weekendEnd boundary
int workingDaysInWeek = daysInWeek - weekendDays.Count;
int workingDays = 0; // the result we are trying to find
int leadingDays = 0; // the number of working days leading up to the firstDayOfWeek boundary
int trailingDays = 0; // the number of working days counting back to the weekendEnd boundary
// Calculate leading working days
// if we aren't on the firstDayOfWeek we need to step forward to the nearest
if (startDate.DayOfWeek != firstDayOfWeek)
{
var d = startDate;
do {
if (d.DayOfWeek == firstDayOfWeek || d >= endDate)
{
actualStartDate = d;
break;
}
if (!weekendDays.Contains(d.DayOfWeek))
{
leadingDays++;
}
d = d.AddDays(1);
} while(true);
}
// Calculate trailing working days
// if we aren't on the weekendEnd we step back to the nearest
if (endDate >= actualStartDate && endDate.DayOfWeek != weekendEnd)
{
var d = endDate;
do {
if (d.DayOfWeek == weekendEnd || d < actualStartDate)
{
actualEndDate = d;
break;
}
if (!weekendDays.Contains(d.DayOfWeek))
{
trailingDays++;
}
d = d.AddDays(-1);
} while(true);
}
// Calculate the inclusive number of days between the actualStartDate and the actualEndDate
var coreDays = (actualEndDate - actualStartDate).Days + 1;
var noWeeks = coreDays / daysInWeek;
// add together leading, core and trailing days
workingDays += noWeeks * workingDaysInWeek;
workingDays += leadingDays;
workingDays += trailingDays;
// Finally remove any holidays that fall within the range.
if (holidays != null)
{
workingDays -= holidays.Count(h => h >= startDate && (h <= endDate));
}
return workingDays;
}
答案 23 :(得分：0)
我相信这可能是一种更简单的方式：
public int BusinessDaysUntil(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;
}
答案 24 :(得分：0)
这是我们可以用来计算两个日期之间的工作日的功能。我没有使用假期列表，因为它可能因国家/地区而异。
如果我们想要使用它，我们可以将第三个参数作为假日列表，在递增计数之前我们应该检查列表不包含d
public static int GetBussinessDaysBetweenTwoDates(DateTime StartDate, DateTime EndDate)
{
if (StartDate > EndDate)
return -1;
int bd = 0;
for (DateTime d = StartDate; d < EndDate; d = d.AddDays(1))
{
if (d.DayOfWeek != DayOfWeek.Saturday && d.DayOfWeek != DayOfWeek.Sunday)
bd++;
}
return bd;
}
答案 25 :(得分：0)
我刚刚改进了@Alexander和@Slauma的答案，以支持商业周作为参数，对于星期六是工作日的情况，或者甚至是一周中只有几天被视为工作日的情况：
/// <summary>
/// Calculate the number of business days between two dates, considering:
/// - Days of the week that are not considered business days.
/// - Holidays between these two dates.
/// </summary>
/// <param name="fDay">First day of the desired 'span'.</param>
/// <param name="lDay">Last day of the desired 'span'.</param>
/// <param name="BusinessDaysOfWeek">Days of the week that are considered to be business days, if NULL considers monday, tuesday, wednesday, thursday and friday as business days of the week.</param>
/// <param name="Holidays">Holidays, if NULL, considers no holiday.</param>
/// <returns>Number of business days during the 'span'</returns>
public static int BusinessDaysUntil(this DateTime fDay, DateTime lDay, DayOfWeek[] BusinessDaysOfWeek = null, DateTime[] Holidays = null)
{
if (BusinessDaysOfWeek == null)
BusinessDaysOfWeek = new DayOfWeek[] { DayOfWeek.Monday, DayOfWeek.Tuesday, DayOfWeek.Wednesday, DayOfWeek.Thursday, DayOfWeek.Friday };
if (Holidays == null)
Holidays = new DateTime[] { };
fDay = fDay.Date;
lDay = lDay.Date;
if (fDay > lDay)
throw new ArgumentException("Incorrect last day " + lDay);
int bDays = (lDay - fDay).Days + 1;
int fullWeekCount = bDays / 7;
int fullWeekCountMult = 7 - WeekDays.Length;
// Find out if there are weekends during the time exceedng the full weeks
if (bDays > (fullWeekCount * 7))
{
int fDayOfWeek = (int)fDay.DayOfWeek;
int lDayOfWeek = (int)lDay.DayOfWeek;
if (fDayOfWeek > lDayOfWeek)
lDayOfWeek += 7;
// If they are the same, we already covered it right before the Holiday subtraction
if (lDayOfWeek != fDayOfWeek)
{
// Here we need to see if any of the days between are considered business days
for (int i = fDayOfWeek; i <= lDayOfWeek; i++)
if (!WeekDays.Contains((DayOfWeek)(i > 6 ? i - 7 : i)))
bDays -= 1;
}
}
// Subtract the days that are not in WeekDays[] during the full weeks in the interval
bDays -= (fullWeekCount * fullWeekCountMult);
// Subtract the number of bank holidays during the time interval
bDays = bDays - Holidays.Select(x => x.Date).Count(x => fDay <= x && x <= lDay);
return bDays;
}
答案 26 :(得分：0)
using System;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
DateTime start = new DateTime(2014, 1, 1);
DateTime stop = new DateTime(2014, 12, 31);
int totalWorkingDays = GetNumberOfWorkingDays(start, stop);
Console.WriteLine("There are {0} working days.", totalWorkingDays);
}
private static int GetNumberOfWorkingDays(DateTime start, DateTime stop)
{
TimeSpan interval = stop - start;
int totalWeek = interval.Days / 7;
int totalWorkingDays = 5 * totalWeek;
int remainingDays = interval.Days % 7;
for (int i = 0; i <= remainingDays; i++)
{
DayOfWeek test = (DayOfWeek)(((int)start.DayOfWeek + i) % 7);
if (test >= DayOfWeek.Monday && test <= DayOfWeek.Friday)
totalWorkingDays++;
}
return totalWorkingDays;
}
}
}
答案 27 :(得分：0)
基于标记为建议的答案和补丁的评论，以及 - ＆gt;此版本希望将天数转换为营业时间......也可以考虑相同的工作时间。
/// <summary>
/// Calculates number of business days, taking into account:
/// - weekends (Saturdays and Sundays)
/// - bank holidays in the middle of the week
/// </summary>
/// <param name="firstDay">First day in the time interval</param>
/// <param name="lastDay">Last day in the time interval</param>
/// <param name="bankHolidays">List of bank holidays excluding weekends</param>
/// <returns>Number of business hours during the 'span'</returns>
public static int BusinessHoursUntil(DateTime firstDay, DateTime lastDay, params DateTime[] bankHolidays)
{
var original_firstDay = firstDay;
var original_lastDay = lastDay;
firstDay = firstDay.Date;
lastDay = lastDay.Date;
if (firstDay > lastDay)
return -1; //// throw new ArgumentException("Incorrect last day " + lastDay);
TimeSpan span = lastDay - firstDay;
int businessDays = span.Days + 1;
int fullWeekCount = businessDays / 7;
// find out if there are weekends during the time exceedng the full weeks
if (businessDays > fullWeekCount * 7)
{
// we are here to find out if there is a 1-day or 2-days weekend
// in the time interval remaining after subtracting the complete weeks
int firstDayOfWeek = firstDay.DayOfWeek == DayOfWeek.Sunday ? 7 : (int)firstDay.DayOfWeek;
int lastDayOfWeek = lastDay.DayOfWeek == DayOfWeek.Sunday ? 7 : (int)lastDay.DayOfWeek;
if (lastDayOfWeek < firstDayOfWeek)
lastDayOfWeek += 7;
if (firstDayOfWeek <= 6)
{
if (lastDayOfWeek >= 7)// Both Saturday and Sunday are in the remaining time interval
businessDays -= 2;
else if (lastDayOfWeek >= 6)// Only Saturday is in the remaining time interval
businessDays -= 1;
}
else if (firstDayOfWeek <= 7 && lastDayOfWeek >= 7)// Only Sunday is in the remaining time interval
businessDays -= 1;
}
// subtract the weekends during the full weeks in the interval
businessDays -= fullWeekCount + fullWeekCount;
if (bankHolidays != null && bankHolidays.Any())
{
// subtract the number of bank holidays during the time interval
foreach (DateTime bankHoliday in bankHolidays)
{
DateTime bh = bankHoliday.Date;
if (firstDay <= bh && bh <= lastDay)
--businessDays;
}
}
int total_business_hours = 0;
if (firstDay.Date == lastDay.Date)
{//If on the same day, go granular with Hours from the Orginial_*Day values
total_business_hours = (int)(original_lastDay - original_firstDay).TotalHours;
}
else
{//Convert Business-Days to TotalHours
total_business_hours = (int)(firstDay.AddDays(businessDays).AddHours(firstDay.Hour) - firstDay).TotalHours;
}
return total_business_hours;
}
答案 28 :(得分：0)
以下是此问题的一个非常简单的解决方案。我们有开始日期，结束日期和“for循环”，用于增加日期并通过转换为字符串DayOfWeek来计算它是工作日还是周末。
class Program
{
static void Main(string[] args)
{
DateTime day = new DateTime();
Console.Write("Inser your end date (example: 01/30/2015): ");
DateTime endDate = DateTime.Parse(Console.ReadLine());
int numberOfDays = 0;
for (day = DateTime.Now.Date; day.Date < endDate.Date; day = day.Date.AddDays(1))
{
string dayToString = Convert.ToString(day.DayOfWeek);
if (dayToString != "Saturday" && dayToString != "Sunday") numberOfDays++;
}
Console.WriteLine("Number of working days (not including local holidays) between two dates is "+numberOfDays);
}
}
答案 29 :(得分：0)
int BusinessDayDifference(DateTime Date1, DateTime Date2)
{
int Sign = 1;
if (Date2 > Date1)
{
Sign = -1;
DateTime TempDate = Date1;
Date1 = Date2;
Date2 = TempDate;
}
int BusDayDiff = (int)(Date1.Date - Date2.Date).TotalDays;
if (Date1.DayOfWeek == DayOfWeek.Saturday)
BusDayDiff -= 1;
if (Date2.DayOfWeek == DayOfWeek.Sunday)
BusDayDiff -= 1;
int Week1 = GetWeekNum(Date1);
int Week2 = GetWeekNum(Date2);
int WeekDiff = Week1 - Week2;
BusDayDiff -= WeekDiff * 2;
foreach (DateTime Holiday in Holidays)
if (Date1 >= Holiday && Date2 <= Holiday)
BusDayDiff--;
BusDayDiff *= Sign;
return BusDayDiff;
}
private int GetWeekNum(DateTime Date)
{
return (int)(Date.AddDays(-(int)Date.DayOfWeek).Ticks / TimeSpan.TicksPerDay / 7);
}
答案 30 :(得分：0)
我无法找到此代码的可靠TSQL版本。以下基本上是C# code here的转换，加上Holiday表，应该用来预先计算假期。
CREATE TABLE dbo.Holiday
(
HolidayDt DATE NOT NULL,
Name NVARCHAR(50) NOT NULL,
IsWeekday BIT NOT NULL,
CONSTRAINT PK_Holiday PRIMARY KEY (HolidayDt)
)
GO
CREATE INDEX IDX_Holiday ON Holiday (HolidayDt, IsWeekday)
GO
CREATE function dbo.GetBusinessDays
(
@FirstDay datetime,
@LastDay datetime
)
RETURNS INT
AS
BEGIN
DECLARE @BusinessDays INT, @FullWeekCount INT
SELECT @FirstDay = CONVERT(DATETIME,CONVERT(DATE,@FirstDay))
, @LastDay = CONVERT(DATETIME,CONVERT(DATE,@LastDay))
IF @FirstDay > @LastDay
RETURN NULL;
SELECT @BusinessDays = DATEDIFF(DAY, @FirstDay, @LastDay) + 1
SELECT @FullWeekCount = @BusinessDays / 7;
-- find out if there are weekends during the time exceedng the full weeks
IF @BusinessDays > (@FullWeekCount * 7)
BEGIN
-- we are here to find out if there is a 1-day or 2-days weekend
-- in the time interval remaining after subtracting the complete weeks
DECLARE @firstDayOfWeek INT, @lastDayOfWeek INT;
SELECT @firstDayOfWeek = DATEPART(DW, @FirstDay), @lastDayOfWeek = DATEPART(DW, @LastDay);
IF @lastDayOfWeek < @firstDayOfWeek
SELECT @lastDayOfWeek = @lastDayOfWeek + 7;
IF @firstDayOfWeek <= 6
BEGIN
IF (@lastDayOfWeek >= 7) --Both Saturday and Sunday are in the remaining time interval
BEGIN
SELECT @BusinessDays = @BusinessDays - 2
END
ELSE IF @lastDayOfWeek>=6 --Only Saturday is in the remaining time interval
BEGIN
SELECT @BusinessDays = @BusinessDays - 1
END
END
ELSE IF @firstDayOfWeek <= 7 AND @lastDayOfWeek >=7 -- Only Sunday is in the remaining time interval
BEGIN
SELECT @BusinessDays = @BusinessDays - 1
END
END
-- subtract the weekends during the full weeks in the interval
DECLARE @Holidays INT;
SELECT @Holidays = COUNT(*)
FROM Holiday
WHERE HolidayDt BETWEEN @FirstDay AND @LastDay
AND IsWeekday = CAST(1 AS BIT)
SELECT @BusinessDays = @BusinessDays - (@FullWeekCount + @FullWeekCount) -- - @Holidays
RETURN @BusinessDays
END
答案 31 :(得分：0)
我将分享我的解决方案。它对我有用，也许我只是不注意/知道这是一个错误。 我开始的第一个不完整的一周，如果有的话。 完整的一周是星期六的星期日，所以如果（int）_now.DayOfWeek不是0（星期日），那么第一周就不完整了。
我只是在第一周的星期六减去1到第一周的数量，然后将其添加到新的计数中;
然后我得到最后一个不完整的一周，然后减去1，因为它是星期天，然后加入新的计数。
最后，将完整周数乘以5（工作日）添加到新计数中。
public int RemoveNonWorkingDays(int numberOfDays){
int workingDays = 0;
int firstWeek = 7 - (int)_now.DayOfWeek;
if(firstWeek < 7){
if(firstWeek > numberOfDays)
return numberOfDays;
workingDays += firstWeek-1;
numberOfDays -= firstWeek;
}
int lastWeek = numberOfDays % 7;
if(lastWeek > 0){
numberOfDays -= lastWeek;
workingDays += lastWeek - 1;
}
workingDays += (numberOfDays/7)*5;
return workingDays;
}
答案 32 :(得分：-1)
public enum NonWorkingDays { SaturdaySunday = 0, FridaySaturday = 1 };
public int getBusinessDates(DateTime dateSt, DateTime dateNd, NonWorkingDays nonWorkingDays = NonWorkingDays.SaturdaySunday)
{
List<DateTime> datelist = new List<DateTime>();
while (dateSt.Date < dateNd.Date)
{
datelist.Add((dateSt = dateSt.AddDays(1)));
}
if (nonWorkingDays == NonWorkingDays.SaturdaySunday)
{
return datelist.Count(d => d.DayOfWeek != DayOfWeek.Saturday &&
d.DayOfWeek != DayOfWeek.Friday);
}
else
{
return datelist.Count(d => d.DayOfWeek != DayOfWeek.Friday &&
d.DayOfWeek != DayOfWeek.Saturday);
}
}
答案 33 :(得分：-1)
检查此 1. https://github.com/yatishbalaji/moment-working-days#readme 2. https://www.npmjs.com/package/moment-working-days
它允许您考虑日期顺序来计算工作日。您可以自定义工作日，也可以声明假日（例如公共假日）的自定义日期，以将其排除在工作日之外
答案 34 :(得分：-1)
这是一个通用的解决方案。
startdayvalue是开始日期的天数。
weekendday_1是周末的日期数。
天数 - MON - 1，TUE - 2，...... SAT - 6，SUN -7。
差异是两个日期之间的差异..
示例：开始日期：2013年4月4日，结束日期：2013年4月14日
差异：10，startdayvalue：4，weekendday_1：7（如果SUNDAY是你的周末。）
这将为您提供假期数量。
营业日=（差异+ 1） - 假期1
if (startdayvalue > weekendday_1)
{
if (difference > ((7 - startdayvalue) + weekendday_1))
{
holiday1 = (difference - ((7 - startdayvalue) + weekendday_1)) / 7;
holiday1 = holiday1 + 1;
}
else
{
holiday1 = 0;
}
}
else if (startdayvalue < weekendday_1)
{
if (difference > (weekendday_1 - startdayvalue))
{
holiday1 = (difference - (weekendday_1 - startdayvalue)) / 7;
holiday1 = holiday1 + 1;
}
else if (difference == (weekendday_1 - startdayvalue))
{
holiday1 = 1;
}
else
{
holiday1 = 0;
}
}
else
{
holiday1 = difference / 7;
holiday1 = holiday1 + 1;
}