Calculate the number of weekdays between two dates in C#

O(1) solution:

// 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);
}

Examples for January 2014:

    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

N.B. The DateTime inputs should be at around the same time of the day. If you are creating DateTime objects based solely on year, month, and day as in the examples above, then you should be fine. As a counter example, 12:01am on Jan 1 to 11:59pm Jan 2 spans only 2 days, but the above function will count 3 if you use those times.

From this 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

Tests (each test returns 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);

eFloh's answer had an extra day if the last day was a Saturday or Sunday. This would fix it.

     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; 

    }