C# 计算两个日期之间的工作日数？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1617049/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Calculate the number of business days between two dates?
提问by Martin
In C#, how can I calculate the number of business(or weekdays) days between two dates?
在C＃中，我怎么能计算的数量业务（或平日）天两个日期之间？
采纳答案by Alexander
I've had such a task before and I've got the solution. I would avoid enumerating all days in between when it's avoidable, which is the case here. I don't even mention creating a bunch of DateTime instances, as I saw in one of the answers above. This is really waste of processing power. Especially in the real world situation, when you have to examine time intervals of several months. See my code, with comments, below.
我以前有过这样的任务，我已经有了解决方案。我会避免在可以避免的情况下枚举其间的所有日子，这里就是这种情况。我什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。这真的是浪费处理能力。特别是在现实世界的情况下，当您必须检查几个月的时间间隔时。请参阅下面的我的代码和注释。
/// <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;
}
Edit by Slauma, August 2011
Slauma 编辑，2011 年 8 月
Great answer! There is little bug though. I take the freedom to edit this answer since the answerer is absent since 2009.
很好的答案！虽然有一点错误。我可以自由编辑这个答案，因为回答者自 2009 年以来就缺席了。
The code above assumes that DayOfWeek.Sundayhas the value 7which is not the case. The value is actually 0. It leads to a wrong calculation if for example firstDayand lastDayare both the same Sunday. The method returns 1in this case but it should be 0.
上面的代码假设DayOfWeek.Sunday具有7并非如此的值。值实际上是0. 例如firstDay，如果和lastDay都是同一个星期日，则会导致计算错误。该方法1在这种情况下返回，但它应该是0。
Easiest fix for this bug: Replace in the code above the lines where firstDayOfWeekand lastDayOfWeekare declared by the following:
最简单的修复此错误的方法：替换上面的代码，其中firstDayOfWeek和lastDayOfWeek由以下声明：
int firstDayOfWeek = firstDay.DayOfWeek == DayOfWeek.Sunday
? 7 : (int)firstDay.DayOfWeek;
int lastDayOfWeek = lastDay.DayOfWeek == DayOfWeek.Sunday
? 7 : (int)lastDay.DayOfWeek;
Now the result is:
现在的结果是：
- Friday to Friday -> 1
- Saturday to Saturday -> 0
- Sunday to Sunday -> 0
- Friday to Saturday -> 1
- Friday to Sunday -> 1
- Friday to Monday -> 2
- Saturday to Monday -> 1
- Sunday to Monday -> 1
- Monday to Monday -> 1
- 周五至周五 -> 1
- 周六至周六 -> 0
- 周日至周日 -> 0
- 周五至周六 -> 1
- 周五至周日 -> 1
- 周五至周一 -> 2
- 周六至周一 -> 1
- 周日至周一 -> 1
- 周一至周一 -> 1
回答by Qwerty
Define an Extension Method on DateTime like so:
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;
}
}
Then, use is within a Where clause to filter a broader list of dates:
然后，在 Where 子句中使用 is 来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() ;
回答by Homde
Here's some code for that purpose, with swedish holidays but you can adapt what holidays to count. Note that I added a limit you might want to remove, but it was for a web-based system and I didnt want anyone to enter some huge date to hog the process
这是用于此目的的一些代码，带有瑞典假期，但您可以调整要计算的假期。请注意，我添加了一个您可能想要删除的限制，但它是针对基于 Web 的系统的，我不希望任何人输入某个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);
}
// F?rsta 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);
}
// L?ngfredagen
public static DateTime GetGoodFriday(int year)
{
return GetEasterDay(year).AddDays(-3);
}
// Kristi himmelsf?rdsdag
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 p?sk
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;
}
回答by Alec Pojidaev
Ok. I think it's time to post the right answer:
好的。我认为是时候发布正确答案了：
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;
}
Original Source:
原始来源：
http://alecpojidaev.wordpress.com/2009/10/29/work-days-calculation-with-c/
http://alecpojidaev.wordpress.com/2009/10/29/work-days-calculation-with-c/
P.S. Solutions posted above making me sic for some reason.
由于某种原因，上面发布的 PS 解决方案让我感到很奇怪。
回答by Alpha
I know this question is already solved, but I thought I could provide a more straightforward-looking answer that may help other visitors in the future.
我知道这个问题已经解决了，但我想我可以提供一个更直截了当的答案，将来可能会帮助其他访问者。
Here's my take at it:
这是我的看法：
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);
}
This was my original submission:
这是我的原始提交：
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;
}
回答by manast
I think none of the above answers are actually correct. None of them solves all the special cases such as when the dates starts and ends on the middle of a weekend, when the date starts on a Friday and ends on next Monday, etc. On top of that, they all round the calculations to whole days, so if the start date is in the middle of a saturday for example, it will substract a whole day from the working days, giving wrong results...
我认为以上答案都不正确。他们都没有解决所有的特殊情况，例如日期何时开始和结束在周末的中间，日期从周五开始到下周一结束等等。 最重要的是，他们将计算全部四舍五入天，所以如果开始日期是在星期六的中间，例如，它会从工作日中减去一整天，给出错误的结果......
Anyway, here is my solution that is quite efficient and simple and works for all cases. The trick is just to find the previous Monday for start and end dates, and then do a small compensation when start and end happens during the weekend:
无论如何，这是我的解决方案，它非常有效且简单，适用于所有情况。诀窍是找到前一个星期一的开始和结束日期，然后在周末开始和结束时做一个小的补偿：
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;
}
回答by rbmeo
I'll just share my solution. It worked for me, maybe I just don't notice/know that theres a bug. I started by getting the first incomplete week if there's any. a complete week was from sunday for saturday, so if the (int)_now.DayOfWeek was not 0(Sunday), the first week was incomplete.
我只是分享我的解决方案。它对我有用，也许我只是没有注意到/知道存在错误。如果有的话，我从第一个不完整的星期开始。一个完整的星期是从星期日到星期六，所以如果 (int)_now.DayOfWeek 不是 0(Sunday)，那么第一周是不完整的。
I just subtract 1 to first weeks count for the first week's saturday then add it to new count;
我只是在第一周的周六减去第一周的计数，然后将其添加到新的计数中；
Then I get the last incomplete week, then subtract 1 for it's sunday then add to new count.
然后我得到最后一个不完整的一周，然后在星期日减去 1，然后添加到新计数。
Then finally, the number of complete weeks multiply by 5(weekdays) was added to new count.
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;
}
回答by Ajay Parmar
This is a generic solution.
这是一个通用的解决方案。
startdayvalue is day number of start date.
startdayvalue 是开始日期的天数。
weekendday_1 is day numner of week end.
weekendday_1 是周末的天数。
day number - MON - 1, TUE - 2, ... SAT - 6, SUN -7.
天数 - MON - 1，TUE - 2，... SAT - 6，SUN -7。
difference is difference between two dates..
差异是两个日期之间的差异..
Example : Start Date : 4 April, 2013, End Date : 14 April, 2013
示例：开始日期：2013 年 4 月 4 日，结束日期：2013 年 4 月 14 日
Difference : 10, startdayvalue : 4, weekendday_1 : 7 (if SUNDAY is a weekend for you.)
差 : 10, startdayvalue : 4,weekendday_1 : 7 (如果 SUNDAY 是你的周末。)
This will give you number of holidays.
这将为您提供假期数。
No of business day = (Difference + 1) - holiday1
工作日数 = (差值 + 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;
}
回答by paulslater19
I used the following code to also take in to account bank holidays:
我使用以下代码也考虑了银行假期：
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; }
}
And Unit Tests:
和单元测试：
[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");
}
}
回答by Greg Ogle
I was having trouble finding a solid TSQL version of this code. Below is essentially a conversion of the C# code herewith addition of the Holiday table which should be used to pre-calculate holidays.
我很难找到这段代码的可靠 TSQL 版本。下面基本上是C# 代码的转换，添加了用于预先计算假期的假日表。
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
