There is an issue with the use of Format or DatePart functions. The last Monday in some calendar years can be returned as week 53 when it should be week 1.
Symptoms
When you use either the Format or DatePart function to determine the week number for dates using the following syntax:
Format(AnyDate, "ww", vbMonday, vbFirstFourDays)
DatePart("ww", AnyDate, vbMonday, vbFirstFourDays)
the last Monday in some calendar years is returned as week 53 when it should be week 1.
Cause
When determining the week number of a date according to the ISO 8601 standard, the underlying function call to the Oleaut32.dll file mistakenly returns week 53 instead of week 1 for the last Monday in certain years.
Resolution
Use a user-defined function to return the Week number based on the rules for the ISO 8601 standard. An example is included in this article.
More Information
The ISO 8601 standard is used extensively in Europe and includes the following:
ISO 8601 "Data elements and interchange formats - Information interchange - Representation of dates and times"
ISO 8601 : 1988 (E) paragraph 3.17:
"week, calendar: A seven day period within a calendar year, starting on a Monday and identified by its ordinal number within the year; the first calendar week of the year is the one that includes the first Thursday of that year. In the Gregorian calendar, this is equivalent to the week which includes 4 January."
This can be implemented by applying these rules for Calendar weeks:
- A year is divided into either 52 or 53 calendar weeks.
- A calendar week has 7 days. Monday is day 1, Sunday is day 7.
- The first calendar week of a year is the one containing at least 4 days.
- If a year is not concluded on a Sunday, either its 1-3 last days belong to next year's first calendar week or the first 1-3 days of next year belong to the present year's last calendar week.
- Only a year starting or concluding on a Thursday has 53 calendar weeks.
In Visual Basic and Visual Basic for Applications, all date functionality, except for the DateSerial function, comes from calls to the Oleaut32.dll file. Because both the Format() and DatePart() functions can return the calendar week number for a given date, both are affected by this bug. To avoid this problem, you must use the alternative code that this article provides.
Steps to Reproduce Behavior
- Open the Visual Basic project within an Office application (Alt + F11).
- From the Project menu, add a new module.
- Paste the following code into the module:
Option Explicit Public Function Test1() ' This code tests a "problem" date and the days around it Dim DateValue As Date Dim i As Integer Debug.Print " Format function:" DateValue = #12/27/2003# For i = 1 To 4 ' examine the last 4 days of the year DateValue = DateAdd("d", 1, DateValue) Debug.Print "Date: " & DateValue & " Day: " & _ Format(DateValue, "ddd") & " Week: " & _ Format(DateValue, "ww", vbMonday, vbFirstFourDays) Next i End Function Public Function Test2() ' This code lists all "Problem" dates within a specified range Dim MyDate As Date Dim Years As Long Dim days As Long Dim woy1 As Long Dim woy2 As Long Dim ToPrint As String For Years = 1850 To 2050 For days = 0 To 3 MyDate = DateSerial(Years, 12, 28 + days) woy1 = Format(MyDate, "ww", vbMonday, vbFirstFourDays) woy2 = Format(MyDate, "ww", vbMonday, vbFirstFourDays) If woy2 > 52 Then If Format(MyDate + 7, "ww", vbMonday, vbFirstFourDays) = 2 Then woy2 = 1 End If If woy1 <> woy2 Then ToPrint = MyDate & String(13 - Len(CStr(MyDate)), " ") ToPrint = ToPrint & Format(MyDate, "dddd") & _ String(10 - Len(Format(MyDate, "dddd")), " ") ToPrint = ToPrint & woy1 & String(5 - Len(CStr(woy1)), " ") ToPrint = ToPrint & woy2 Debug.Print ToPrint End If Next days Next Years End Function
- Use (Ctrl + G) to open the Immediate Window if it's not already open.
- Type
?Test1
in the Immediate window and hit Enter, note the following results in the Immediate window:Format function: Date: 12/28/03 Day: Sun Week: 52 Date: 12/29/03 Day: Mon Week: 53 Date: 12/30/03 Day: Tue Week: 1 Date: 12/31/03 Day: Wed Week: 1
With this format, all weeks start with Monday, so that 12/29/2003 should be considered the start of Week 1 and not part of Week 53.
- Type
?Test2
in the Immediate window and hit Enter to see a list of dates in the specified range that experience this problem. The list includes the date, Week day (always Monday), the Week # returned by Format (53), and the Week number it should return (1.) For example:12/29/1851 Monday 53 1 12/31/1855 Monday 53 1 12/30/1867 Monday 53 1 12/29/1879 Monday 53 1 12/31/1883 Monday 53 1 12/30/1895 Monday 53 1 ...
Workarounds
If you use the Format or DatePart functions, you need to check the return value. When it is 53, run another check and force a return of 1, if necessary. This code sample demonstrates one way to do this:
Function WOY(MyDate As Date) As Integer ' Week Of Year
WOY = Format(MyDate, "ww", vbMonday, vbFirstFourDays)
If WOY > 52 Then
If Format(MyDate + 7, "ww", vbMonday, vbFirstFourDays) = 2 Then WOY = 1
End If
End Function
You can avoid using these functions to determine Week number by writing code that implements the ISO 8601 rules described above. The following example demonstrates a replacement function to return the Week number.
Step by Step Example
- Open the Visual Basic project within an Office application (Alt + F11).
- From the Project menu, add a new module.
- Paste the following code into the module:
Option Explicit Function WeekNumber(InDate As Date) As Integer Dim DayNo As Integer Dim StartDays As Integer Dim StopDays As Integer Dim StartDay As Integer Dim StopDay As Integer Dim VNumber As Integer Dim ThurFlag As Boolean DayNo = Days(InDate) StartDay = Weekday(DateSerial(Year(InDate), 1, 1)) - 1 StopDay = Weekday(DateSerial(Year(InDate), 12, 31)) - 1 ' Number of days belonging to first calendar week StartDays = 7 - (StartDay - 1) ' Number of days belonging to last calendar week StopDays = 7 - (StopDay - 1) ' Test to see if the year will have 53 weeks or not If StartDay = 4 Or StopDay = 4 Then ThurFlag = True Else ThurFlag = False VNumber = (DayNo - StartDays - 4) / 7 ' If first week has 4 or more days, it will be calendar week 1 ' If first week has less than 4 days, it will belong to last year's ' last calendar week If StartDays >= 4 Then WeekNumber = Fix(VNumber) + 2 Else WeekNumber = Fix(VNumber) + 1 End If ' Handle years whose last days will belong to coming year's first ' calendar week If WeekNumber > 52 And ThurFlag = False Then WeekNumber = 1 ' Handle years whose first days will belong to the last year's ' last calendar week If WeekNumber = 0 Then WeekNumber = WeekNumber(DateSerial(Year(InDate) - 1, 12, 31)) End If End Function Function Days(DayNo As Date) As Integer Days = DayNo - DateSerial(Year(DayNo), 1, 0) End Function Public Function Test3() Dim DateValue As Date, i As Integer Debug.Print " WeekNumber function:" DateValue = #12/27/2003# For i = 1 To 4 ' examine the last 4 days of the year DateValue = DateAdd("d", 1, DateValue) Debug.Print "Date: " & DateValue & " Day: " & _ Format(DateValue, "ddd") & " Week: " & WeekNumber(DateValue) Next i End Function
- Use (Ctrl + G) to open the Immediate Window if it's not already open.
- Type
?Test3
in the Immediate window and hit Enter, note the following results in the Immediate window:WeekNumber function: Date: 12/28/03 Day: Sun Week: 52 Date: 12/29/03 Day: Mon Week: 1 Date: 12/30/03 Day: Tue Week: 1 Date: 12/31/03 Day: Wed Week: 1
Note that Monday is considered to be Week 1 as it should be.