我怎样才能得到两个给定日期之间的工作日数，而不只是遍历并计算工作日之间的日期？
似乎相当简单，但我似乎无法找到符合以下规定的结论性正确答案：
我发现similar question 和answer 接近但不正确
【问题讨论】：
标签： c#
O(1) 解：
// Count days from d0 to d1 inclusive, excluding weekends
public static int countWeekDays(DateTime d0, DateTime d1)
{
int ndays = 1 + Convert.ToInt32((d1 - d0).TotalDays);
int nsaturdays = (ndays + Convert.ToInt32(d0.DayOfWeek)) / 7;
return ndays - 2 * nsaturdays
- (d0.DayOfWeek == DayOfWeek.Sunday ? 1 : 0)
+ (d1.DayOfWeek == DayOfWeek.Saturday ? 1 : 0);
}
2014 年 1 月的示例：
January 2014
Su Mo Tu We Th Fr Sa
1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31
countWeekDays(new DateTime(2014, 1, 1), new DateTime(2014, 1, 1)); // 1
countWeekDays(new DateTime(2014, 1, 1), new DateTime(2014, 1, 2)); // 2
countWeekDays(new DateTime(2014, 1, 1), new DateTime(2014, 1, 3)); // 3
countWeekDays(new DateTime(2014, 1, 1), new DateTime(2014, 1, 4)); // 3
countWeekDays(new DateTime(2014, 1, 1), new DateTime(2014, 1, 5)); // 3
countWeekDays(new DateTime(2014, 1, 1), new DateTime(2014, 1, 6)); // 4
注意DateTime 输入应该在一天中的同一时间。如果您在上面的示例中仅基于年、月和日创建 DateTime 对象，那么您应该没问题。举个反例，1 月 1 日凌晨 12:01 到 1 月 2 日晚上 11:59 仅持续 2 天，但如果您使用这些时间，上述函数将计为 3。
【讨论】：
来自link：
public static int Weekdays(DateTime dtmStart, DateTime dtmEnd)
{
// This function includes the start and end date in the count if they fall on a weekday
int dowStart = ((int)dtmStart.DayOfWeek == 0 ? 7 : (int)dtmStart.DayOfWeek);
int dowEnd = ((int)dtmEnd.DayOfWeek == 0 ? 7 : (int)dtmEnd.DayOfWeek);
TimeSpan tSpan = dtmEnd - dtmStart;
if (dowStart <= dowEnd)
{
return (((tSpan.Days / 7) * 5) + Math.Max((Math.Min((dowEnd + 1), 6) - dowStart), 0));
}
return (((tSpan.Days / 7) * 5) + Math.Min((dowEnd + 6) - Math.Min(dowStart, 6), 5));
}
[1]: http://www.eggheadcafe.com/community/aspnet/2/44982/how-to-calculate-num-of-w.aspx
测试（每个测试返回 5）：
int ndays = Weekdays(new DateTime(2009, 11, 30), new DateTime(2009, 12, 4));
System.Console.WriteLine(ndays);
// leap year test
ndays = Weekdays(new DateTime(2000, 2,27), new DateTime(2000, 3, 5));
System.Console.WriteLine(ndays);
// non leap year test
ndays = Weekdays(new DateTime(2007, 2, 25), new DateTime(2007, 3, 4));
System.Console.WriteLine(ndays);
【讨论】：
如果最后一天是周六或周日，eFloh 的答案会多出一天。这将解决它。
public static int Weekdays(DateTime dtmStart, DateTime dtmEnd)
{
if (dtmStart > dtmEnd)
{
DateTime temp = dtmStart;
dtmStart = dtmEnd;
dtmEnd = temp;
}
/* Move border dates to the monday of the first full week and sunday of the last week */
DateTime startMonday = dtmStart;
int startDays = 1;
while (startMonday.DayOfWeek != DayOfWeek.Monday)
{
if (startMonday.DayOfWeek != DayOfWeek.Saturday && startMonday.DayOfWeek != DayOfWeek.Sunday)
{
startDays++;
}
startMonday = startMonday.AddDays(1);
}
DateTime endSunday = dtmEnd;
int endDays = 0;
while (endSunday.DayOfWeek != DayOfWeek.Sunday)
{
if (endSunday.DayOfWeek != DayOfWeek.Saturday && endSunday.DayOfWeek != DayOfWeek.Sunday)
{
endDays++;
}
endSunday = endSunday.AddDays(1);
}
int weekDays;
/* calculate weeks between full week border dates and fix the offset created by moving the border dates */
weekDays = (Math.Max(0, (int)Math.Ceiling((endSunday - startMonday).TotalDays + 1)) / 7 * 5) + startDays - endDays;
if (dtmEnd.DayOfWeek == DayOfWeek.Saturday || dtmEnd.DayOfWeek == DayOfWeek.Sunday)
{
weekDays -= 1;
}
return weekDays;
}
【讨论】：
这应该比 dcp 的解决方案做得更好：
/// <summary>
/// Count Weekdays between two dates
/// </summary>
/// <param name="dtmStart">first date</param>
/// <param name="dtmEnd">second date</param>
/// <returns>weekdays between the two dates, including the start and end day</returns>
internal static int getWeekdaysBetween(DateTime dtmStart, DateTime dtmEnd)
{
if (dtmStart > dtmEnd)
{
DateTime temp = dtmStart;
dtmStart = dtmEnd;
dtmEnd = temp;
}
/* Move border dates to the monday of the first full week and sunday of the last week */
DateTime startMonday = dtmStart;
int startDays = 1;
while (startMonday.DayOfWeek != DayOfWeek.Monday)
{
if (startMonday.DayOfWeek != DayOfWeek.Saturday && startMonday.DayOfWeek != DayOfWeek.Sunday)
{
startDays++;
}
startMonday = startMonday.AddDays(1);
}
DateTime endSunday = dtmEnd;
int endDays = 0;
while (endSunday.DayOfWeek != DayOfWeek.Sunday)
{
if (endSunday.DayOfWeek != DayOfWeek.Saturday && endSunday.DayOfWeek != DayOfWeek.Sunday)
{
endDays++;
}
endSunday = endSunday.AddDays(1);
}
int weekDays;
/* calculate weeks between full week border dates and fix the offset created by moving the border dates */
weekDays = (Math.Max(0, (int)Math.Ceiling((endSunday - startMonday).TotalDays + 1)) / 7 * 5) + startDays - endDays;
return weekDays;
}
【讨论】：
我需要正数/负数（不是绝对值），所以我是这样解决的：
public static int WeekdayDifference(DateTime StartDate, DateTime EndDate)
{
DateTime thisDate = StartDate;
int weekDays = 0;
while (thisDate != EndDate)
{
if (thisDate.DayOfWeek != DayOfWeek.Saturday && thisDate.DayOfWeek != DayOfWeek.Sunday) { weekDays++; }
if (EndDate > StartDate) { thisDate = thisDate.AddDays(1); } else { thisDate = thisDate.AddDays(-1); }
}
/* Determine if value is positive or negative */
if (EndDate > StartDate) {
return weekDays;
}
else
{
return weekDays * -1;
}
}
【讨论】：
Math.Abs 获得绝对值怎么样！
public static int GetWeekDays(DateTime startDay, DateTime endDate, bool countEndDate = true)
{
var daysBetween = (int)(endDate - startDay).TotalDays;
daysBetween = countEndDate ? daysBetween += 1 : daysBetween;
return Enumerable.Range(0, daysBetween).Count(d => !startDay.AddDays(d).DayOfWeek.In(DayOfWeek.Saturday, DayOfWeek.Sunday));
}
public static bool In<T>(this T source, params T[] list)
{
if (null == source)
{
throw new ArgumentNullException("source");
}
return list.Contains(source);
}
【讨论】：
public static List<DateTime> Weekdays(DateTime startDate, DateTime endDate)
{
if (startDate > endDate)
{
Swap(ref startDate, ref endDate);
}
List<DateTime> days = new List<DateTime>();
var ts = endDate - startDate;
for (int i = 0; i < ts.TotalDays; i++)
{
var cur = startDate.AddDays(i);
if (cur.DayOfWeek != DayOfWeek.Saturday && cur.DayOfWeek != DayOfWeek.Sunday)
days.Add(cur);
//if (startingDate.AddDays(i).DayOfWeek != DayOfWeek.Saturday || startingDate.AddDays(i).DayOfWeek != DayOfWeek.Sunday)
//yield return startingDate.AddDays(i);
}
return days;
}
和交换日期
private static void Swap(ref DateTime startDate, ref DateTime endDate)
{
object a = startDate;
startDate = endDate;
endDate = (DateTime)a;
}
【讨论】：
获取日期范围的实用函数：
public IEnumerable<DateTime> GetDates(DateTime begin, int count)
{
var first = new DateTime(begin.Year, begin.Month, begin.Day);
var maxYield = Math.Abs(count);
for (int i = 0; i < maxYield; i++)
{
if(count < 0)
yield return first - TimeSpan.FromDays(i);
else
yield return first + TimeSpan.FromDays(i);
}
yield break;
}
public IEnumerable<DateTime> GetDates(DateTime begin, DateTime end)
{
var days = (int)Math.Ceiling((end - begin).TotalDays);
return GetDates(begin, days);
}
LINQPad 演示代码：
var begin = DateTime.Now;
var end = begin + TimeSpan.FromDays(14);
var dates = GetDates(begin, end);
var weekdays = dates.Count(x => x.DayOfWeek != DayOfWeek.Saturday && x.DayOfWeek != DayOfWeek.Sunday);
var mondays = dates.Count(x => x.DayOfWeek == DayOfWeek.Monday);
var firstThursday = dates
.OrderBy(d => d)
.FirstOrDefault(d => d.DayOfWeek == DayOfWeek.Thursday);
dates.Dump("Dates in Range");
weekdays.Dump("Count of Weekdays");
mondays.Dump("Count of Mondays");
firstThursday.Dump("First Thursday");
【讨论】：
var dates = new List<DateTime>();
for (var dt = YourStartDate; dt <= YourEndDate; dt = dt.AddDays(1))
{
if (dt.DayOfWeek != DayOfWeek.Sunday && dt.DayOfWeek != DayOfWeek.Saturday)
{ dates.Add(dt); }
}
在此代码中，您可以列出两个日期之间的所有营业日。
如果你想要这些日期的计数，你可以得到dates.Count 作为一个整数。
或者如果你想得到每一天，你可以将列表加入一个字符串。
【讨论】：
这里的函数计算两个日期之间的 DayOfWeek 计数。小心它包含计算它（包括计算中的开始日期和结束日期）：
private int GetWeekdayCount(DayOfWeek dayOfWeek, DateTime begin, DateTime end)
{
if (begin < end)
{
var timeSpan = end.Subtract(begin);
var fullDays = timeSpan.Days;
var count = fullDays / 7; // количество дней равно как минимум количеству полных недель попавших в диапазон
var remain = fullDays % 7; // остаток от деления
// и вычислим попал ли еще один день в те куски недели, что остаются от полной
if (remain > 0)
{
var dowBegin = (int)begin.DayOfWeek;
var dowEnd = (int)end.DayOfWeek;
var dowDay = (int)dayOfWeek;
if (dowBegin < dowEnd)
{
// когда день недели начала меньше дня недели конца, например:
// начало конец
// \/ \/
// -- -- -- -- --
// Вс Пн Вт Ср Чт Пт Сб
if (dowDay >= dowBegin && dowDay <= dowEnd)
count++;
}
else
{
// когда день недели начала больше дня недели конца, например:
// конец начало
// \/ \/
// -- -- -- --
// Вс Пн Вт Ср Чт Пт Сб
if (dowDay <= dowEnd || dowDay >= dowBegin)
count++;
}
}
else if (begin.DayOfWeek == dayOfWeek)
count++;
return count;
}
return 0;
}
这里是之前函数的另一个简单模拟：
private int GetWeekdayCountStupid(DayOfWeek dayOfWeek, DateTime begin, DateTime end)
{
if (begin < end)
{
var count = 0;
var day = begin;
while (day <= end)
{
if (day.DayOfWeek == dayOfWeek)
count++;
day = day.AddDays(1);
}
return count;
}
return 0;
}
并测试这两个功能：
[TestMethod()]
public void TestWeekdayCount()
{
var init = new DateTime(2000, 01, 01);
for (int day = 0; day < 7; day++)
{
var dayOfWeek = (DayOfWeek)day;
for (int shift = 0; shift < 8; shift++)
{
for (int i = 0; i < 365; i++)
{
var begin = init.AddDays(shift);
var end = init.AddDays(shift + i);
var count1 = GetWeekdayCount(dayOfWeek, begin, end);
var count2 = GetWeekdayCountStupid(dayOfWeek, begin, end);
Assert.AreEqual(count1, count2);
}
}
}
}
【讨论】：
这是一个老问题，但我想我会分享一个更灵活的答案，允许删除一周中的任何一天。
我试图通过最多循环 6 天来保持代码简洁易读，同时保持高效。
所以对于 OP，你可以像这样使用它：
myDate.DaysUntill(DateTime.UtcNow, new List<DayOfWeek> { DayOfWeek.Saturday, DayOfWeek.Sunday });
/// <summary>
/// For better accuracy make sure <paramref name="startDate"/> and <paramref name="endDate"/> have the same time zone.
/// This is only really useful if we use <paramref name="daysOfWeekToExclude"/> - otherwise the computation is trivial.
/// </summary>
/// <param name="startDate"></param>
/// <param name="endDate"></param>
/// <param name="daysOfWeekToExclude"></param>
/// <returns></returns>
public static int DaysUntill(this DateTime startDate, DateTime endDate, IEnumerable<DayOfWeek> daysOfWeekToExclude = null)
{
if (startDate >= endDate) return 0;
daysOfWeekToExclude = daysOfWeekToExclude?.Distinct() ?? new List<DayOfWeek>();
int nbOfWeeks = (endDate - startDate).Days / 7;
int nbOfExtraDays = (endDate - startDate).Days % 7;
int result = nbOfWeeks * (7 - daysOfWeekToExclude.Count());
for (int i = 0; i < nbOfExtraDays; i++)
{
if (!daysOfWeekToExclude.Contains(startDate.AddDays(i + 1).DayOfWeek)) result++;
}
return result;
}
【讨论】：