[ACCEPTED]-Count number of Mondays in a given date range-datediff

Try this:

static int CountDays(DayOfWeek day, DateTime start, DateTime end)
{
    TimeSpan ts = end - start;                       // Total duration
    int count = (int)Math.Floor(ts.TotalDays / 7);   // Number of whole weeks
    int remainder = (int)(ts.TotalDays % 7);         // Number of remaining days
    int sinceLastDay = (int)(end.DayOfWeek - day);   // Number of days since last [day]
    if (sinceLastDay < 0) sinceLastDay += 7;         // Adjust for negative days since last [day]

    // If the days in excess of an even week are greater than or equal to the number days since the last [day], then count this one, too.
    if (remainder >= sinceLastDay) count++;          

    return count;
}

0

Since you're using C#, if you're using C#3.0, you 5 can use LINQ.

Assuming you have an Array/List/IQueryable 4 etc that contains your dates as DateTime 3 types:

DateTime[] dates = { new DateTime(2008,10,6), new DateTime(2008,10,7)}; //etc....

var mondays = dates.Where(d => d.DayOfWeek == DayOfWeek.Monday); // = {10/6/2008}

Added:

Not sure if you meant grouping 2 them and counting them, but here's how to 1 do that in LINQ as well:

var datesgrouped = from d in dates
                   group d by d.DayOfWeek into grouped
                   select new { WeekDay = grouped.Key, Days = grouped };

foreach (var g in datesgrouped)
{
    Console.Write (String.Format("{0} : {1}", g.WeekDay,g.Days.Count());
}

It's fun to look at different algorithms 9 for calculating day of week, and @Gabe Hollombe's 8 pointing to WP on the subject was a great 7 idea (and I remember implementing Zeller's Congruence in COBOL 6 about twenty years ago), but it was rather 5 along the line of handing someone a blueprint 4 of a clock when all they asked what time 3 it was.

In C#:

    private int CountMondays(DateTime startDate, DateTime endDate)
    {
        int mondayCount = 0;

        for (DateTime dt = startDate; dt < endDate; dt = dt.AddDays(1.0))
        {
            if (dt.DayOfWeek == DayOfWeek.Monday)
            {
                mondayCount++;
            }
        }

        return mondayCount;
    }

This of course does not evaluate 2 the end date for "Mondayness", so 1 if this was desired, make the for loop evaluate

dt < endDate.AddDays(1.0)

Here's some pseudocode:

DifferenceInDays(Start, End) / 7   // Integer division discarding remainder
+ 1 if DayOfWeek(Start) <= DayImLookingFor
+ 1 if DayOfWeek(End)   >= DayImLookingFor
- 1

Where DifferenceInDays returns End - Start in 8 days, and DayOfWeek returns the day of the week as 7 an integer. It doesn't really matter what 6 mapping DayOfWeek uses, as long as it is increasing 5 and matches up with DayImLookingFor.

Note that this algorithm 4 assumes the date range is inclusive. If 3 End should not be part of the range, you'll 2 have to adjust the algorithm slightly.

Translating 1 to C# is left as an exercise for the reader.

Any particular language and therefore date 7 format?

If dates are represented as a count 6 of days, then the difference between two 5 values plus one (day), and divide by 7, is 4 most of the answer. If both end dates are 3 the day in question, add one.

Edited: corrected 2 'modulo 7' to 'divide by 7' - thanks. And 1 that is integer division.

public List<DateTime> GetSelectedDaysInPeriod(DateTime startDate, DateTime endDate, List<DayOfWeek> daysToCheck)
{
    var selectedDates = new List<DateTime>();

    if (startDate >= endDate)
        return selectedDates; //No days to return

    if (daysToCheck == null || daysToCheck.Count == 0)
        return selectedDates; //No days to select

    try
    {
        //Get the total number of days between the two dates
        var totalDays = (int)endDate.Subtract(startDate).TotalDays;

        //So.. we're creating a list of all dates between the two dates:
        var allDatesQry = from d in Enumerable.Range(1, totalDays)
                             select new DateTime(
                                                  startDate.AddDays(d).Year,
                                                  startDate.AddDays(d).Month,
                                                  startDate.AddDays(d).Day);

        //And extracting those weekdays we explicitly wanted to return
        var selectedDatesQry = from d in allDatesQry
                                  where daysToCheck.Contains(d.DayOfWeek)
                                  select d;

        //Copying the IEnumerable to a List
        selectedDates = selectedDatesQry.ToList();
    }
    catch (Exception ex)
    {
        //Log error
        //...

        //And re-throw
        throw;
    }
    return selectedDates;
}

0

Add the smallest possible number to make 3 the first day a Monday. Subtract the smallest 2 possible number to make the last day a Monday. Calculate 1 the difference in days and divide by 7.

Convert the dates to Julian Day Number, then 3 do a little bit of math. Since Mondays are 2 zero mod 7, you could do the calculation 1 like this:

JD1=JulianDayOf(the_first_date)
JD2=JulianDayOf(the_second_date)
Round JD1 up to nearest multiple of 7
Round JD2 up to nearest multiple of 7
d = JD2-JD1
nMondays = (JD2-JD1+7)/7    # integer divide

I have had the same need today. I started 8 with the cjm function since I don't understand 7 the JonB function and since the Cyberherbalist function is 6 not linear.

I had have to correct

DifferenceInDays(Start, End) / 7   // Integer division discarding remainder
+ 1 if DayOfWeek(Start) <= DayImLookingFor
+ 1 if DayOfWeek(End)   >= DayImLookingFor
- 1

to

DifferenceInDays(Start, End) / 7   // Integer division discarding remainder
+ 1 if DayImLookingFor is between Start.Day and End.Day 

With the 5 between function that return true if, starting 4 from the start day, we meet first the dayImLookingFor 3 before the endDay.

I have done the between 2 function by computing the number of day 1 from startDay to the other two days:

private int CountDays(DateTime start, DateTime end, DayOfWeek selectedDay)
{
    if (start.Date > end.Date)
    {
        return 0;
    }
    int totalDays = (int)end.Date.Subtract(start.Date).TotalDays;
    DayOfWeek startDay = start.DayOfWeek;
    DayOfWeek endDay = end.DayOfWeek;
    ///look if endDay appears before or after the selectedDay when we start from startDay.
    int startToEnd = (int)endDay - (int)startDay;
    if (startToEnd < 0)
    {
        startToEnd += 7;
    }
    int startToSelected = (int)selectedDay - (int)startDay;
    if (startToSelected < 0)
    {
        startToSelected += 7;
    }
    bool isSelectedBetweenStartAndEnd = startToEnd >= startToSelected;
    if (isSelectedBetweenStartAndEnd)
    {
        return totalDays / 7 + 1;
    }
    else
    {
        return totalDays / 7;
    }
}

This will return a collection of integers 4 showing how many times each day of the week 3 occurs within a date range

    int[] CountDays(DateTime firstDate, DateTime lastDate)
    {
        var totalDays = lastDate.Date.Subtract(firstDate.Date).TotalDays + 1;
        var weeks = (int)Math.Floor(totalDays / 7);

        var result = Enumerable.Repeat<int>(weeks, 7).ToArray();
        if (totalDays % 7 != 0)
        {
            int firstDayOfWeek = (int)firstDate.DayOfWeek;
            int lastDayOfWeek = (int)lastDate.DayOfWeek;
            if (lastDayOfWeek < firstDayOfWeek)
                lastDayOfWeek += 7;
            for (int dayOfWeek = firstDayOfWeek; dayOfWeek <= lastDayOfWeek; dayOfWeek++)
                result[dayOfWeek % 7]++;
        }
        return result;
    }

Or a slight variation 2 which lets you do FirstDate.TotalDaysOfWeeks(SecondDate) and 1 returns a Dictionary

    public static Dictionary<DayOfWeek, int> TotalDaysOfWeeks(this DateTime firstDate, DateTime lastDate)
    {
        var totalDays = lastDate.Date.Subtract(firstDate.Date).TotalDays + 1;
        var weeks = (int)Math.Floor(totalDays / 7);

        var resultArray = Enumerable.Repeat<int>(weeks, 7).ToArray();
        if (totalDays % 7 != 0)
        {
            int firstDayOfWeek = (int)firstDate.DayOfWeek;
            int lastDayOfWeek = (int)lastDate.DayOfWeek;
            if (lastDayOfWeek < firstDayOfWeek)
                lastDayOfWeek += 7;
            for (int dayOfWeek = firstDayOfWeek; dayOfWeek <= lastDayOfWeek; dayOfWeek++)
                resultArray[dayOfWeek % 7]++;
        }
        var result = new Dictionary<DayOfWeek, int>();
        for (int dayOfWeek = 0; dayOfWeek < 7; dayOfWeek++)
            result[(DayOfWeek)dayOfWeek] = resultArray[dayOfWeek];
        return result;
    }

A bit Modified Code is here works and Tested 1 by me

        private int CountDays(DayOfWeek day, DateTime startDate, DateTime endDate)
        {
            int dayCount = 0;

            for (DateTime dt = startDate; dt < endDate; dt = dt.AddDays(1.0))
            {
                if (dt.DayOfWeek == day)
                {
                    dayCount++;
                }
            }

            return dayCount;
        }

Example:

int Days = CountDays(DayOfWeek.Friday, Convert.ToDateTime("2019-07-04"), 
             Convert.ToDateTime("2019-07-27")).ToString();

I had a similar problem for a report. I 7 needed the number of workdays between two 6 dates. I could have cycled through the dates 5 and counted but my discrete math training 4 wouldn't let me. Here is a function I wrote 3 in VBA to get the number of workdays between 2 two dates. I'm sure .net has a similar 1 WeekDay function.

   1  
   2  ' WorkDays
   3  ' returns the number of working days between two dates
   4  Public Function WorkDays(ByVal dtBegin As Date, ByVal dtEnd As Date) As Long
   5  
   6     Dim dtFirstSunday As Date
   7     Dim dtLastSaturday As Date
   8     Dim lngWorkDays As Long
   9  
  10     ' get first sunday in range
  11     dtFirstSunday = dtBegin + ((8 - Weekday(dtBegin)) Mod 7)
  12  
  13     ' get last saturday in range
  14     dtLastSaturday = dtEnd - (Weekday(dtEnd) Mod 7)
  15  
  16     ' get work days between first sunday and last saturday
  17     lngWorkDays = (((dtLastSaturday - dtFirstSunday) + 1) / 7) * 5
  18  
  19     ' if first sunday is not begin date
  20     If dtFirstSunday <> dtBegin Then
  21  
  22        ' assume first sunday is after begin date
  23        ' add workdays from begin date to first sunday
  24        lngWorkDays = lngWorkDays + (7 - Weekday(dtBegin))
  25  
  26     End If
  27  
  28     ' if last saturday is not end date
  29     If dtLastSaturday <> dtEnd Then
  30  
  31        ' assume last saturday is before end date
  32        ' add workdays from last saturday to end date
  33        lngWorkDays = lngWorkDays + (Weekday(dtEnd) - 1)
  34  
  35     End If
  36  
  37     ' return working days
  38     WorkDays = lngWorkDays
  39  
  40  End Function

private System.Int32 CountDaysOfWeek(System.DayOfWeek dayOfWeek, System.DateTime date1, System.DateTime date2)
{
  System.DateTime EndDate;
  System.DateTime StartDate;

  if (date1 > date2)
  {
    StartDate = date2;
    EndDate = date1;
  }
  else
  {
    StartDate = date1;
    EndDate = date2;
  }

  while (StartDate.DayOfWeek != dayOfWeek)
    StartDate = StartDate.AddDays(1);

  return EndDate.Subtract(StartDate).Days / 7 + 1;
}

0

Four years later, I thought I'd run a test:

[TestMethod]
public void ShouldFindFridaysInTimeSpan()
{
    //reference: http://stackoverflow.com/questions/248273/count-number-of-mondays-in-a-given-date-range

    var spanOfSixtyDays = new TimeSpan(60, 0, 0, 0);
    var setOfDates = new List<DateTime>(spanOfSixtyDays.Days);
    var now = DateTime.Now;

    for(int i = 0; i < spanOfSixtyDays.Days; i++)
    {
        setOfDates.Add(now.AddDays(i));
    }

    Assert.IsTrue(setOfDates.Count == 60,
        "The expected number of days is not here.");

    var fridays = setOfDates.Where(i => i.DayOfWeek == DayOfWeek.Friday);

    Assert.IsTrue(fridays.Count() > 0,
        "The expected Friday days are not here.");
    Assert.IsTrue(fridays.First() == setOfDates.First(i => i.DayOfWeek == DayOfWeek.Friday),
        "The expected first Friday day is not here.");
    Assert.IsTrue(fridays.Last() == setOfDates.Last(i => i.DayOfWeek == DayOfWeek.Friday),
        "The expected last Friday day is not here.");
}

My 2 use of TimeSpan is a bit of overkill---actually 1 I wanted to query TimeSpan directly.