具有最小值和最大值的 Excel VBA yearfrac

Excel VBA yearfrac with mins and maxes

我试图在 Excel VBA 中查找两个日期之间的周数(中间有一些最小/最大功能)，出现以下类型不匹配错误(运行时错误 \\'13\\')行：

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WeeksWorked = Application.WorksheetFunction.RoundDown _
(52 * Application.WorksheetFunction.YearFrac _
(Application.WorksheetFunction.Max(DOH, DateValue("Jan 1, 2012")), _
DateValue("Dec 31, 2012")), 0)

任何人对我做错的事情有任何指导，将不胜感激！

相关讨论

不确定为什么需要在 VBA 中使用它，您可以尝试以下方法。

在 Excel 中：

假设开始日期在 A1，结束日期在 A2，然后是 A3，

1	=(NETWORKINGDAYS(A1,A2))/5

现在是从工作日的angular来看，因此每周有 5 天。如果您需要每周 7 天和正常的日子，

1	=WEEKNUM(A3)-WEEKNUM(A2)

The function WEEKNUM() in the Analysis Toolpack addin calculates the correct week number for a given date, if you are in the U.S. The user defined function below will calculate the correct week number depending on the national language settings on your computer.

如果你仍然需要使用 VBA，试试这个：(正如 Tim 指出的那样，DateDiff 非常方便。)或者你甚至可以使用 Evaluate to trigger WEEKNUM.

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Option Explicit

Function numWeeks(startDate As Date, endDate As Date)
numWeeks = DateDiff("ww", startDate, endDate)
End Function

在 WEEKNUM 上使用评估：

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Function numWeeks(startDate As Range, endDate As Range)
Dim s As Integer, t As Integer
s = Application.Evaluate("=WEEKNUM(" & startDate.Address &",1)")
t = Application.Evaluate("=WEEKNUM(" & endDate.Address &",1)")
numWeeks = t - s
End Function

Excel VBA DataTime 函数参考

相关讨论

enter image description here 试试下面的代码：

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Sub example()

Dim dob As Date
dob = #7/31/1986#

Dim todaydt As Date
todaydt = Date

Dim numWeek As Long
numWeek = DateDiff("ww", dob, todaydt) ' Difference in weeks

End Sub

按照评论中的建议，您可以这样做：

1	debug.print DateDiff("ww", DateValue("Jan 1, 2012"), DateValue("Dec 31, 2012"))

如果出于某种原因您想自己滚动，您可以截断以下商：

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| day1 - day2 |
---------------
7

示例代码：

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Sub test_numWeeks_fn()
Call numWeeks(DateValue("Jan 1, 2012"), DateValue("Dec 31, 2012"))
End Sub

Function numWeeks(d1 As Date, d2 As Date)
Dim numDays As Long
numDays = Abs(d1 - d2)
numWeeks = Int(numDays / 7)
Debug.Print numWeeks
End Function

结果：