Question

我每周都会获得一个大数据集，包括开始日期和结束日期，开始日期始终是星期一，结束日期始终是星期日。我希望能够按周对不同产品的销售进行YTD滚动。我试图给每周的数据提供一周的数据。但是我的第一周是12/28 / 2015-1 / 3/2016，它一直给我第53周而不是第1周。我在同一个问题上看到了另一个人的帖子，dbDesigner的答案实际上有效：

Get the week number from a given date

但这给了我2016-01一栏。我想要两个列，其中一个用于正确的周数，另一个用于正确的年份，以便能够对我的滚动YTD销售进行子查询。例如我的第一周12/28 / 2015-1 / 3/2016，它将是一周：1年和2016年

感谢。

Answer 1

首先，VBA中的周编号方法的 none 遵循ISO 8601标准。

其次，2015-12-28至2016-01-03周的周数不是2106的第一周，而是2015W53。

您可以使用以下功能检索任何日期的正确ISO 8601周编号：

Public Function ISO_WeekYearNumber( _
  ByVal datDate As Date, _
  Optional ByRef intYear As Integer, _
  Optional ByRef bytWeek As Byte) _
  As String

' Calculates and returns year and week number for date datDate according to the ISO 8601:1988 standard.
' Optionally returns numeric year and week.
' 1998-2007, Gustav Brock, Cactus Data ApS, CPH.    Public Function ISO_WeekNumber( _
  ByVal datDate As Date) _
  As Byte

' Calculates and returns week number for date datDate according to the ISO 8601:1988 standard.
' 1998-2000, Gustav Brock, Cactus Data ApS, CPH.
' May be freely used and distributed.

  Const cbytFirstWeekOfAnyYear  As Byte = 1
  Const cbytLastWeekOfLeapYear  As Byte = 53

  Dim bytWeek                   As Byte
  Dim bytISOThursday            As Byte
  Dim datLastDayOfYear          As Date

  bytWeek = DatePart("ww", datDate, vbMonday, vbFirstFourDays)

  If bytWeek = cbytLastWeekOfLeapYear Then
    bytISOThursday = Weekday(vbThursday, vbMonday)
    datLastDayOfYear = DateSerial(Year(datDate), 12, 31)
    If Weekday(datLastDayOfYear, vbMonday) >= bytISOThursday Then
      ' OK, week count of 53 is caused by leap year.
    Else
      ' Correct for Access97/2000 bug.
      bytWeek = cbytFirstWeekOfAnyYear
    End If
  End If

  ISO_WeekNumber = bytWeek

End Function
' May be freely used and distributed.

  Const cbytFirstWeekOfAnyYear  As Byte = 1
  Const cbytLastWeekOfLeapYear  As Byte = 53
  Const cbytMonthJanuary        As Byte = 1
  Const cbytMonthDecember       As Byte = 12
  Const cstrSeparatorYearWeek   As String = "W"

  Dim bytMonth                  As Byte
  Dim bytISOThursday            As Byte
  Dim datLastDayOfYear          As Date

  intYear = Year(datDate)
  bytMonth = Month(datDate)
  bytWeek = DatePart("ww", datDate, vbMonday, vbFirstFourDays)

  If bytWeek = cbytLastWeekOfLeapYear Then
    bytISOThursday = Weekday(vbThursday, vbMonday)
    datLastDayOfYear = DateSerial(intYear, cbytMonthDecember, 31)
    If Weekday(datLastDayOfYear, vbMonday) >= bytISOThursday Then
      ' OK, week count of 53 is caused by leap year.
    Else
      ' Correct for Access97/2000+ bug.
      bytWeek = cbytFirstWeekOfAnyYear
    End If
  End If

  ' Adjust year where week number belongs to next or previous year.
  If bytMonth = cbytMonthJanuary Then
    If bytWeek >= cbytLastWeekOfLeapYear - 1 Then
      ' This is an early date of January belonging to the last week of the previous year.
      intYear = intYear - 1
    End If
  ElseIf bytMonth = cbytMonthDecember Then
    If bytWeek = cbytFirstWeekOfAnyYear Then
      ' This is a late date of December belonging to the first week of the next year.
      intYear = intYear + 1
    End If
  End If

  ISO_WeekYearNumber = CStr(intYear) & cstrSeparatorYearWeek & Format(bytWeek, "00")

End Function

如果您只需要周数，可以使用这样的函数：

Answer 2

您了解您所指的解决方案吗？因为如果你这样做，你应该注意到年份和星期都是连贯的。所以你应该撤消连接。

如何在Access中获取正确的周数

2 个答案: