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Hi Guys

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

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

but dont know how to resolve the issue ...

MTIA

DWE

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
' build our answer
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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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 ...

```
'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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Start your 7-day free trial49 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.

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 ...

YearsMonthsDays(CDate("06 Aug 1965"), Now, True)Is it still Oct 17th where you are, because here in the USA where it is the 17th I get 49 years, 2 months,

returns 49 years, 2 months, 12 days

49 years, 2 months, 5 days

or

49 years, 2 months, 0 weeks, 5 days

```
'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
```

```
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
' build our answer
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
```

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 ?

so it would seem that your function has some problems,

```
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
```

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 ...

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