多看几次这个问题,但似乎无法在我的VBA代码中修复它。
我需要计算给定年份的总周数,符合ISO 8601。
当我使用datediff函数时:<div id="header">
<content id="headerText" select="[header]">111</content>
</div>
它返回52而2015年有53周(ISO 8601)
我怎样才能完成这项工作?
答案 0 :(得分:7)
再次引用https://en.wikipedia.org/wiki/ISO_8601#Week_dates
12月28日总是在同一年的最后一周。
所以它真的像
一样简单(function ($) {
$.fn.extend({
tableExport: function (options) {
var defaults = {
consoleLog: false,
csvEnclosure: '"',
csvSeparator: ',',
csvUseBOM: true,
displayTableName: false,
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答案 1 :(得分:1)
答案 2 :(得分:0)
这里的简单功能将为任何一年提供正确的结果:
Public Function ISO_WeekCount( _
ByVal datYear As Date) _
As Byte
' Calculates number of weeks in year of datYear according to the ISO 8601:1988 standard.
'
' May be freely used and distributed.
' 2001-06-26. Gustav Brock, Cactus Data ApS, CPH
Dim bytISO_Thursday As Byte
' No special error handling.
On Error Resume Next
bytISO_Thursday = Weekday(vbThursday, vbMonday)
datYear = DateSerial(Year(datYear), 12, 31)
' Subtract one week if datYear is in week no. 1 of next year.
datYear = DateAdd("ww", Weekday(datYear, vbMonday) < bytISO_Thursday, datYear)
ISO_WeekCount = DatePart("ww", datYear, vbMonday, vbFirstFourDays)
End Function
答案 3 :(得分:-1)
如果您不介意使用UDF,则此代码将返回它 = ISOWeekNum(“2015年12月31日”)或 = ISOWeekNum(42369)返回53.
http://www.cpearson.com/excel/DateTimeVBA.htm
'http://www.cpearson.com/excel/DateTimeVBA.htm
'John Green, an Excel MVP from Australia.
Public Function ISOWeekNum(AnyDate As Date, Optional WhichFormat As Variant) As Integer
' WhichFormat: missing or <> 2 then returns week number,
' = 2 then YYWW
'
Dim ThisYear As Integer
Dim PreviousYearStart As Date
Dim ThisYearStart As Date
Dim NextYearStart As Date
Dim YearNum As Integer
ThisYear = Year(AnyDate)
ThisYearStart = YearStart(ThisYear)
PreviousYearStart = YearStart(ThisYear - 1)
NextYearStart = YearStart(ThisYear + 1)
Select Case AnyDate
Case Is >= NextYearStart
ISOWeekNum = (AnyDate - NextYearStart) \ 7 + 1
YearNum = Year(AnyDate) + 1
Case Is < ThisYearStart
ISOWeekNum = (AnyDate - PreviousYearStart) \ 7 + 1
YearNum = Year(AnyDate) - 1
Case Else
ISOWeekNum = (AnyDate - ThisYearStart) \ 7 + 1
YearNum = Year(AnyDate)
End Select
If IsMissing(WhichFormat) Then Exit Function
If WhichFormat = 2 Then
ISOWeekNum = CInt(Format(Right(YearNum, 2), "00") & _
Format(ISOWeekNum, "00"))
End If
End Function
Public Function YearStart(WhichYear As Integer) As Date
Dim WeekDay As Integer
Dim NewYear As Date
NewYear = DateSerial(WhichYear, 1, 1)
WeekDay = (NewYear - 2) Mod 7 'Generate weekday index where Monday = 0
If WeekDay < 4 Then
YearStart = NewYear - WeekDay
Else
YearStart = NewYear - WeekDay + 7
End If
End Function
可以在查询中用作: SELECT DISTINCT dDate,ISOWeekNum(dDate)FROM tbl_MyTable