Количество дней между датами

[root]

Function Betwindate(Byval d1 As Integer , Byval m1 As Integer , Byval y1 As Integer , Byval d2 As Integer , Byval m2 As Integer, Byval y2 As Integer ) As Long 
Dim Result As Long
Dim n1 As Long
Dim n2 As Long

If m1>2 Then
m1 = m1+1
Else
m1 = m1+13
y1 = y1-1
End If
n1 = 36525*y1\100+306*m1\10+d1
If m2>2 Then
m2 = m2+1
Else
m2 = m2+13
y2 = y2-1
End If
n2 = 36525*y2\100+306*m2\10+d2
Result = n2-n1

Betwindate= Result


End Function

 

[morpheus]

хм.... захожу в справку к собаке @Adjust
и виже такою себе штуку -

Tip

To find the difference between two dates, subtract them. The result is returned in seconds. To adjust the result to days, divide the result by 86,400 - which is the number of seconds in a day. For example, if you have two date fields, date1, which contains [07/01/01] and date2, which contains [07/05/01], use the following formula to return the number of days between the two dates:
(date2-date1)/86400
This code returns 4.

Вапрос: чем Вам нее подходит вышеприведенный способ?

 

[Domino6]

DateTimeDif=cdate(cdbl(DateTimeEnd)-cdbl(DateTimeBegin))

Сразу вернёт разницу в датовом формате 02.03.1980 12:23:02

т.е. разница составила

0 лет
3 месяца
2 дня
12 часов
23 минут
2 секунды