Функция дает мне 21 рабочий день для марта, который является правильным. B
Он был назначен на проект с 01/03/2019 по 08/03/2019, он дает мне 5, чтоневерно, это должно дать мне 6.
Функция сравнения никогда не будет включать в себя последнюю дату.Если вы хотите включить эту последнюю дату, добавьте один день к последней дате перед вычислением:
? DateDiffWorkDays(#2019/03/01#, #2019/03/31#)
21
? DateDiffWorkDays(#2019/03/01#, #2019/04/01#)
21
? DateDiffWorkDays(#2019/03/01#, #2019/03/08#)
5
? DateDiffWorkDays(#2019/03/01#, #2019/03/09#)
6
Также, как уже отмечалось, укажите Понедельник в качестве первого дня недели.Кроме того, не используйте формат;День недели - это «прямой» метод.Таким образом:
If Weekday(DateCnt, vbMonday) < 6 Then
EndDays = EndDays + 1
End If
Для расширенного метода, который учитывает праздники, изучите мои функции:
Option Compare Database
Option Explicit
' Returns the count of full workdays between Date1 and Date2.
' The date difference can be positive, zero, or negative.
' Optionally, if WorkOnHolidays is True, holidays are regarded as workdays.
'
' Note that if one date is in a weekend and the other is not, the reverse
' count will differ by one, because the first date never is included in the count:
'
' Mo Tu We Th Fr Sa Su Su Sa Fr Th We Tu Mo
' 0 1 2 3 4 4 4 0 0 -1 -2 -3 -4 -5
'
' Su Mo Tu We Th Fr Sa Sa Fr Th We Tu Mo Su
' 0 1 2 3 4 5 5 0 -1 -2 -3 -4 -5 -5
'
' Sa Su Mo Tu We Th Fr Fr Th We Tu Mo Su Sa
' 0 0 1 2 3 4 5 0 -1 -2 -3 -4 -4 -4
'
' Fr Sa Su Mo Tu We Th Th We Tu Mo Su Sa Fr
' 0 0 0 1 2 3 4 0 -1 -2 -3 -3 -3 -4
'
' Execution time for finding working days of three years is about 4 ms.
'
' Requires table Holiday with list of holidays.
'
' 2015-12-19. Gustav Brock. Cactus Data ApS, CPH.
'
Public Function DateDiffWorkdays( _
ByVal Date1 As Date, _
ByVal Date2 As Date, _
Optional ByVal WorkOnHolidays As Boolean) _
As Long
Dim Holidays() As Date
Dim Diff As Long
Dim Sign As Long
Dim NextHoliday As Long
Dim LastHoliday As Long
Sign = Sgn(DateDiff("d", Date1, Date2))
If Sign <> 0 Then
If WorkOnHolidays = True Then
' Holidays are workdays.
Else
' Retrieve array with holidays between Date1 and Date2.
Holidays = GetHolidays(Date1, Date2, False) 'CBool(Sign < 0))
' Ignore error when using LBound and UBound on an unassigned array.
On Error Resume Next
NextHoliday = LBound(Holidays)
LastHoliday = UBound(Holidays)
' If Err.Number > 0 there are no holidays between Date1 and Date2.
If Err.Number > 0 Then
WorkOnHolidays = True
End If
On Error GoTo 0
End If
' Loop to sum up workdays.
Do Until DateDiff("d", Date1, Date2) = 0
Select Case Weekday(Date1)
Case vbSaturday, vbSunday
' Skip weekend.
Case Else
If WorkOnHolidays = False Then
' Check for holidays to skip.
If NextHoliday <= LastHoliday Then
' First, check if NextHoliday hasn't been advanced.
If NextHoliday < LastHoliday Then
If Sgn(DateDiff("d", Date1, Holidays(NextHoliday))) = -Sign Then
' Weekend hasn't advanced NextHoliday.
NextHoliday = NextHoliday + 1
End If
End If
' Then, check if Date1 has reached a holiday.
If DateDiff("d", Date1, Holidays(NextHoliday)) = 0 Then
' This Date1 hits a holiday.
' Subtract one day to neutralize the one
' being added at the end of the loop.
Diff = Diff - Sign
' Adjust to the next holiday to check.
NextHoliday = NextHoliday + 1
End If
End If
End If
Diff = Diff + Sign
End Select
' Advance Date1.
Date1 = DateAdd("d", Sign, Date1)
Loop
End If
DateDiffWorkdays = Diff
End Function
' Returns the holidays between Date1 and Date2.
' The holidays are returned as an array with the
' dates ordered ascending, optionally descending.
'
' The array is declared static to speed up
' repeated calls with identical date parameters.
'
' Requires table Holiday with list of holidays.
'
' 2015-12-18. Gustav Brock, Cactus Data ApS, CPH.
'
Public Function GetHolidays( _
ByVal Date1 As Date, _
ByVal Date2 As Date, _
Optional ByVal OrderDesc As Boolean) _
As Date()
' Constants for the arrays.
Const DimRecordCount As Long = 2
Const DimFieldOne As Long = 0
Static Date1Last As Date
Static Date2Last As Date
Static OrderLast As Boolean
Static DayRows As Variant
Static Days As Long
Dim rs As DAO.Recordset
' Cannot be declared Static.
Dim Holidays() As Date
If DateDiff("d", Date1, Date1Last) <> 0 Or _
DateDiff("d", Date2, Date2Last) <> 0 Or _
OrderDesc <> OrderLast Then
' Retrieve new range of holidays.
Set rs = DatesHoliday(Date1, Date2, OrderDesc)
' Save the current set of date parameters.
Date1Last = Date1
Date2Last = Date2
OrderLast = OrderDesc
Days = rs.RecordCount
If Days > 0 Then
' As repeated calls may happen, do a movefirst.
rs.MoveFirst
DayRows = rs.GetRows(Days)
' rs is now positioned at the last record.
End If
rs.Close
End If
If Days = 0 Then
' Leave Holidays() as an unassigned array.
Erase Holidays
Else
' Fill array to return.
ReDim Holidays(Days - 1)
For Days = LBound(DayRows, DimRecordCount) To UBound(DayRows, DimRecordCount)
Holidays(Days) = DayRows(DimFieldOne, Days)
Next
End If
Set rs = Nothing
GetHolidays = Holidays()
End Function
' Returns the holidays between Date1 and Date2.
' The holidays are returned as a recordset with the
' dates ordered ascending, optionally descending.
'
' Requires table Holiday with list of holidays.
'
' 2015-12-18. Gustav Brock, Cactus Data ApS, CPH.
'
Public Function DatesHoliday( _
ByVal Date1 As Date, _
ByVal Date2 As Date, _
Optional ByVal ReverseOrder As Boolean) _
As DAO.Recordset
' The table that holds the holidays.
Const Table As String = "Holiday"
' The field of the table that holds the dates of the holidays.
Const Field As String = "Date"
Dim rs As DAO.Recordset
Dim SQL As String
Dim SqlDate1 As String
Dim SqlDate2 As String
Dim Order As String
SqlDate1 = Format(Date1, "\#yyyy\/mm\/dd\#")
SqlDate2 = Format(Date2, "\#yyyy\/mm\/dd\#")
ReverseOrder = ReverseOrder Xor (DateDiff("d", Date1, Date2) < 0)
Order = IIf(ReverseOrder, "Desc", "Asc")
SQL = "Select " & Field & " From " & Table & " " & _
"Where " & Field & " Between " & SqlDate1 & " And " & SqlDate2 & " " & _
"Order By 1 " & Order
Set rs = CurrentDb.OpenRecordset(SQL, dbOpenSnapshot)
Set DatesHoliday = rs
End Function
Вы увидите, что по своей сути это всего лишь простой цикл,который настолько быстр, что попытки оптимизации не окупятся при обычном использовании.