# vb6 Date Difference Calculations

Hi Guys

The following function returns a string representing the date difference between two dates:

``````Public Function DateDiffEx(StartDate As Variant, FinishDate As Variant) As String

Dim yer As Integer, mon As Integer, d As Integer
Dim dt As Date
Dim fd As Date

Dim sAns  As String

dt = CDate(StartDate)       ' our start date
fd = CDate(FinishDate)      ' our finish date

yer = Year(dt)              ' year of our start date
mon = Month(dt)             ' month of ..
d = Day(dt)                 ' day of ...

yer = Year(fd) - yer        ' year of our finish date
mon = Month(fd) - mon       ' month of ..
d = Day(fd) - d             ' day of ...

If Sgn(d) = -1 Then         ' are we less than 0
d = 30 - Abs(d)
mon = mon - 1
End If

If Sgn(mon) = -1 Then
mon = 12 - Abs(mon)
yer = yer - 1
End If

Dim wek As Integer          ' work our how many weeks
If d > 7 Then              ' if the number of days is larger than 7 - theres a week or more in it
wek = d / 7             ' wek = days / 7
d = d Mod 7             ' d = gets the mod (remiander) after taking out whole weeks
wek = wek - 1           ' take 1 away
End If

If yer > 0 Then
sAns = yer & " year" & IIf(yer > 1, "(s), ", ", ")
End If

If mon > 0 Then
sAns = sAns & mon & " month" & IIf(mon > 1, "(s), ", ", ")
End If

If wek > 0 Then
sAns = sAns & wek & " Week" & IIf(wek > 1, "(s), ", ", ")
End If

If d > 0 Then
sAns = sAns & d & " Day" & IIf(d > 1, "(s)", "")
End If

'sAns = yer & " year(s) " & mon & " month(s) " & wek & " Week(s), " & d & " day(s)"

DateDiffEx = sAns

End Function
``````

It works for dates where there is more than 1 months difference.

For example,

msgbox DateDiffEx(CDate("06 Aug 1965"), CDate(now))

returns 49 years, 2 months, 1 week, 5 days

However, the following returns 1 day which is a week out of kilter

msgbox DateDiffEx(CDate("18 Oct 2014"), CDate("26 Oct 2014"))

returns 1 day

I know the problem exists because I subtract 1 from wek

``````     If d > 7 Then              ' if the number of days is larger than 7 - theres a week or more in it
wek = d / 7             ' wek = days / 7
d = d Mod 7             ' d = gets the mod (remiander) after taking out whole weeks
wek = wek - 1           ' take 1 away
End If
``````

but dont know how to resolve the issue ...

MTIA

DWE
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Older than dirtCommented:
Why not use the built-in DateDiff Function?
Author Commented:
Hi Martin,

I found it inaccurate - I started off using it and ran into problems ... if you could show me how datediff would be used to return the same sort of string I'd be interested ...
Older than dirtCommented:
MsgBox DateDiff("d", CDate("18 Oct 2014"), CDate("26 Oct 2014"))

Returns 8
Older than dirtCommented:
This function is from the web.
``````'Usage:
MsgBox YearsMonthsDays(CDate("06 Aug 1965"), Now, True)
``````

``````Function YearsMonthsDays(Date1 As Date, Date2 As Date, Optional ShowAll As _
Boolean = False, Optional Grammar As Boolean = True)

' This function returns a string "X years, Y months, Z days" showing the time
' between two dates.  This function may be used in any VBA or VB project

' Date1 and Date2 must either be dates, or strings that can be implicitly
' converted to dates.  If these arguments have time portions, the time portions
' are ignored. If Date1 > Date2 (after ignoring time portions), the function
' returns an empty string

' ShowAll indicates whether all portions of the string "X years, Y months, Z days"
' are included in the output.  If ShowAll = True, all portions of the string are
' always included.  If ShowAll = False, then if the year portion is zero the year
' part of the string is omitted, and if the year portion and month portion are both
' zero, than both year and month portions are omitted.  The day portion is always
' included, and if at least one year has passed then the month portion is always
' included

' Grammar indicates whether to test years/months/days for singular or plural

' By definition, a "full month" means that the day number in Date2 is >= the day
' number in Date1, or Date1 and Date2 occur on the last days of their respective
' months. A "full year" means that 12 "full months" have passed.

Dim TestYear As Long, TestMonth As Long, TestDay As Long
Dim TargetDate As Date, Last1 As Date, Last2 As Date

' Strip time portions
Date1 = Int(Date1)
Date2 = Int(Date2)

' Test for invalid dates
If Date1 > Date2 Then
YearsMonthsDays = ""
Exit Function
End If

' Test for whether the calendar year is the same
If Year(Date2) > Year(Date1) Then

' Different calendar year.

' Test to see if calendar month is the same.  If it is, we have to look at the
' day to see if a full year has passed
If Month(Date2) = Month(Date1) Then
If Day(Date2) >= Day(Date1) Then
TestYear = DateDiff("yyyy", Date1, Date2)
Else
TestYear = DateDiff("yyyy", Date1, Date2) - 1
End If

' In this case, a full year has definitely passed
ElseIf Month(Date2) > Month(Date1) Then
TestYear = DateDiff("yyyy", Date1, Date2)

' A full year has not passed
Else
TestYear = DateDiff("yyyy", Date1, Date2) - 1
End If

' Calendar year is the same, so a full year has not passed
Else
TestYear = 0
End If

' Test to see how many full months have passed, in excess of the number of full
' years
TestMonth = (DateDiff("m", DateSerial(Year(Date1), Month(Date1), 1), _
DateSerial(Year(Date2), Month(Date2), 1)) + IIf(Day(Date2) >= _
Day(Date1), 0, -1)) Mod 12

' See how many days have passed, in excess of the number of full months.  If the day
' number for Date2 is >= that for Date1, it's simple
If Day(Date2) >= Day(Date1) Then
TestDay = Day(Date2) - Day(Date1)

' If not, we have to test for end of the month
Else
Last1 = DateSerial(Year(Date2), Month(Date2), 0)
Last2 = DateSerial(Year(Date2), Month(Date2) + 1, 0)
TargetDate = DateSerial(Year(Date2), Month(Date2) - 1, Day(Date1))
If Last2 = Date2 Then
If TestMonth = 11 Then
TestMonth = 0
TestYear = TestYear + 1
Else
TestMonth = TestMonth + 1
End If
Else
TestDay = DateDiff("d", IIf(TargetDate > Last1, Last1, TargetDate), Date2)
End If
End If

If ShowAll Or TestYear >= 1 Then
YearsMonthsDays = TestYear & IIf(TestYear = 1 And Grammar, " year, ", _
" years, ") & TestMonth & IIf(TestMonth = 1 And Grammar, " month, ", _
" months, ") & TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
Else
If TestMonth >= 1 Then
YearsMonthsDays = TestMonth & IIf(TestMonth = 1 And Grammar, " month, ", _
" months, ") & TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
Else
YearsMonthsDays = TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
End If
End If

End Function
``````

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Older than dirtCommented:
BTW the above returns Years, Months, Days, so you get in the above example

49 years, 2 months, 11 days

which is one day different from your result but it seems to be correct because if you add two months to Aug 6 you get Oct 6 and there are 11 days between then and now (Oct 17).

BTW the function could easily be changed so that instead of 11 days it would show 1 week 4 days.
Author Commented:
Hi Martin, you've got the idea - the function I use does the same thing but doesn't have weeks in it ...

YearsMonthsDays(CDate("06 Aug 1965"), Now, True)

returns 49 years, 2 months, 12 days

its the number of weeks I am looking to include ... and I am having difficulty with ... using the example you gave, 12 days should convert to 1 week and 5 days - thats where my difficulty lay ...
Author Commented:
BTW - I am in Australian time, and you're on USA time, so the date difference of a day is explainable :-)
Older than dirtCommented:
YearsMonthsDays(CDate("06 Aug 1965"), Now, True)

returns 49 years, 2 months, 12 days
Is it still Oct 17th where you are, because here in the USA where it is the 17th I get 49 years, 2 months, 11 days
Author Commented:
;-) ... no its 18 October here ...
Older than dirtCommented:
OK, if using different dates "my" function gave for example 49 years, 2 months, 5 days, do you want the result to be

49 years, 2 months, 5 days

or

49 years, 2 months, 0 weeks, 5 days
Author Commented:
if you use your function and use the start date as 1 August 1965 - I would want the result to say 49 years, 2 months, 2 weeks, 3 days
Older than dirtCommented:
Here's a start. Add these lines at the bottom of the function. The "weeks"/"week" and "Days"/"Day" isn't handled but I assume you can fix that.

``````    'new
Dim intPos As Integer
Dim strTemp As String
intPos = InStrRev(YearsMonthsDays, ",")
strTemp = Mid(YearsMonthsDays, intPos + 2)
If Val(strTemp) > 7 Then
strTemp = Val(strTemp) \ 7 & " weeks, " & Val(strTemp) Mod 7 & " days"
End If
YearsMonthsDays = Left\$(YearsMonthsDays, intPos) & strTemp
``````
Author Commented:
Martin, using the following function - do you get the correct result ?

``````Public Function DateDiffEx(StartDate As Variant, FinishDate As Variant) As String

Dim yer As Integer, mon As Integer, d As Integer
Dim dt As Date
Dim fd As Date

Dim sAns  As String

dt = CDate(StartDate)       ' our start date
fd = CDate(FinishDate)      ' our finish date

yer = Year(dt)              ' year of our start date
mon = Month(dt)             ' month of ..
d = Day(dt)                 ' day of ...

yer = Year(fd) - yer        ' year of our finish date
mon = Month(fd) - mon       ' month of ..
d = Day(fd) - d             ' day of ...

If Sgn(d) = -1 Then         ' are we less than 0
d = 30 - Abs(d)
mon = mon - 1
End If

If Sgn(mon) = -1 Then
mon = 12 - Abs(mon)
yer = yer - 1
End If

Dim wek As Integer          ' work our how many weeks
If d > 7 Then              ' if the number of days is larger than 7 - theres a week or more in it
Debug.Print "days = " & d
wek = d / 7             ' wek = days / 7
d = (d + 1) Mod 7           ' d = gets the mod (remiander) after taking out whole weeks
wek = IIf(mon > 0, wek - 1, wek)         ' take 1 away
End If

If yer > 0 Then
sAns = yer & " year" & IIf(yer > 1, "(s), ", ", ")
End If

If mon > 0 Then
sAns = sAns & mon & " month" & IIf(mon > 1, "(s), ", ", ")
End If

If wek > 0 Then
sAns = sAns & wek & " Week" & IIf(wek > 1, "(s), ", ", ")
End If

If d > 0 Then
sAns = sAns & d & " Day" & IIf(d > 1, "(s)", "")
End If

'sAns = yer & " year(s) " & mon & " month(s) " & wek & " Week(s), " & d & " day(s)"

DateDiffEx = sAns

End Function
``````
Older than dirtCommented:
I get

49 year(s), 2 month(s), 1 Week, 5 Day(s)
Author Commented:
where StartDate = 06 Aug 1965 and FinishDate = 17 Oct 2014 ?
if so its returning the correct result ...

but the value of wek is not consistantly correct? When using smaller date differences ... such as about 3 weeks ...

Can you see why ?
Older than dirtCommented:
I haven't looked at it in detail but I just did a test and while

YearsMonthsDays(CDate("31-Jan-2006"), CDate("1-Mar-2006"), True) -> 0 years, 1 month, 1 day
DateDiffEx(CDate("31-Jan-2006"), CDate("1-Mar-2006")) -> 1 month,

so it would seem that your function has some problems,
Author Commented:
yes I recognised that at the outset ... thats why I posted a query and sought some assistance ..
Older than dirtCommented:
Why not use YearsMonthsDays (modified to suit your output requirements) instead of DateDiffEx?
Author Commented:
Can you tell me - why is wek = 2 when it should be equal to 1. ie 13/7 is returning 2

``````where TestDay = 13 ...

Dim wek As Integer             ' work our how many weeks
If TestDay >= 7 Then             ' if the number of days is larger than or equal to 7 - theres a week or more in it
Debug.Print "TestDay = " & TestDay
wek = TestDay / 7
Debug.Print "Wek = " & wek
TestDay = TestDay Mod 7             ' d = gets the mod (remiander) after taking out whole weeks
Debug.Print "TestDay = " & TestDay
End If
``````
Older than dirtCommented:
Because TestDay / 7 yields 1.85714285714286 which is automatically rounded by VB to 2. If you do TestDay \ 7 (which is integer division) it will yield 1.
Author Commented:
thanks - I picked up that a moment after I posted it ... must be tired ...

heres the function you posted modded to include weeks ...

``````Function YearsMonthsDays(Date1 As Date, Date2 As Date, Optional ShowAll As _
Boolean = False, Optional Grammar As Boolean = True)

' This function returns a string "X years, Y months, Z days" showing the time
' between two dates.  This function may be used in any VBA or VB project

' Date1 and Date2 must either be dates, or strings that can be implicitly
' converted to dates.  If these arguments have time portions, the time portions
' are ignored. If Date1 > Date2 (after ignoring time portions), the function
' returns an empty string

' ShowAll indicates whether all portions of the string "X years, Y months, Z days"
' are included in the output.  If ShowAll = True, all portions of the string are
' always included.  If ShowAll = False, then if the year portion is zero the year
' part of the string is omitted, and if the year portion and month portion are both
' zero, than both year and month portions are omitted.  The day portion is always
' included, and if at least one year has passed then the month portion is always
' included

' Grammar indicates whether to test years/months/days for singular or plural

' By definition, a "full month" means that the day number in Date2 is >= the day
' number in Date1, or Date1 and Date2 occur on the last days of their respective
' months. A "full year" means that 12 "full months" have passed.

Dim TestYear As Long, TestMonth As Long, TestDay As Long
Dim TargetDate As Date, Last1 As Date, Last2 As Date

' Strip time portions
Date1 = Int(Date1)
Date2 = Int(Date2)

' Test for invalid dates
If Date1 > Date2 Then
YearsMonthsDays = ""
Exit Function
End If

' Test for whether the calendar year is the same
If Year(Date2) > Year(Date1) Then

' Different calendar year.

' Test to see if calendar month is the same.  If it is, we have to look at the
' day to see if a full year has passed
If Month(Date2) = Month(Date1) Then
If Day(Date2) >= Day(Date1) Then
TestYear = DateDiff("yyyy", Date1, Date2)
Else
TestYear = DateDiff("yyyy", Date1, Date2) - 1
End If

' In this case, a full year has definitely passed
ElseIf Month(Date2) > Month(Date1) Then
TestYear = DateDiff("yyyy", Date1, Date2)

' A full year has not passed
Else
TestYear = DateDiff("yyyy", Date1, Date2) - 1
End If

' Calendar year is the same, so a full year has not passed
Else
TestYear = 0
End If

' Test to see how many full months have passed, in excess of the number of full
' years
TestMonth = (DateDiff("m", DateSerial(Year(Date1), Month(Date1), 1), _
DateSerial(Year(Date2), Month(Date2), 1)) + IIf(Day(Date2) >= _
Day(Date1), 0, -1)) Mod 12

' See how many days have passed, in excess of the number of full months.  If the day
' number for Date2 is >= that for Date1, it's simple
If Day(Date2) >= Day(Date1) Then
TestDay = Day(Date2) - Day(Date1)

' If not, we have to test for end of the month
Else
Last1 = DateSerial(Year(Date2), Month(Date2), 0)
Last2 = DateSerial(Year(Date2), Month(Date2) + 1, 0)
TargetDate = DateSerial(Year(Date2), Month(Date2) - 1, Day(Date1))
If Last2 = Date2 Then
If TestMonth = 11 Then
TestMonth = 0
TestYear = TestYear + 1
Else
TestMonth = TestMonth + 1
End If
Else
TestDay = DateDiff("d", IIf(TargetDate > Last1, Last1, TargetDate), Date2)
End If
End If

' ok work out if there is a weeks calculation to be slotted in
Dim wek As Integer             ' work our how many weeks
If TestDay >= 7 Then             ' if the number of days is larger than or equal to 7 - theres a week or more in it
wek = TestDay \ 7
TestDay = TestDay Mod 7             ' d = gets the mod (remiander) after taking out whole weeks
End If

If ShowAll Or TestYear >= 1 Then
YearsMonthsDays = TestYear & IIf(TestYear = 1 And Grammar, " year, ", " years, ") & _
TestMonth & IIf(TestMonth = 1 And Grammar, " month, ", " months, ") & _
wek & IIf(wek = 1 And Grammar, " week, ", " weeks, ") & _
TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
Else
If TestMonth >= 1 Then
YearsMonthsDays = TestMonth & IIf(TestMonth = 1 And Grammar, " month, ", " months, ") & _
TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
Else
YearsMonthsDays = TestDay & IIf(TestDay = 1 And Grammar, " day", " days")
End If
End If

End Function
``````

Thanks for the patient assistance ...
Older than dirtCommented:
You're welcome and I'm glad I was able to help.

In my profile you'll find links to some articles I've written that may interest you.
Marty - MVP 2009 to 2014
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