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 = sAnsEnd 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

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 ...

MsgBox DateDiff("d", CDate("18 Oct 2014"), CDate("26 Oct 2014"))

Returns 8

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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 IfEnd Function

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.

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 ...

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 ' 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 = sAnsEnd Function

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

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.

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 IfEnd Function

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