MS Access：设定值为第14周周日的日期

Question

数学和编码都不是我的事情，所以我真的希望有人能帮助我解决这个问题。我深深地被困在这个问题上......

情景：客户达到13周的就业后，我们开始计算保留月份。所以在WeeksTotal达到13之后，接下来的星期日标志着保留的第一个月的开始。

简而言之

所以...

当WeeksTotal是< 13： MonthsCurrentJob应等于0

当WeeksTotal> = 13：

1. RetentionStartdate值应被设置为第14周周日的日期

2. MonthsCurrentJob应该使用RetentionSta rtdate来计算月至今

3. 1应加入MonthsCurrentJob

   价值

这很可能是（也许）是去这样做这个完全错误的方式。但我希望这解释了我想要完成的事情。如果任何人都能指引我走向正确的方向，我会牺牲一只山羊以纪念你。好吧，也许不是......但你会有我永恒的赞赏和钦佩！

Answer 1

我想你可以计算出这样的：使用这样的函数

RetentionMonths = Months(RetentionStartDate, Date) 

：

RetentionStartdate = DateAdd("ww", 14, DateAdd("d", vbSaturday - Weekday(HireDate), HireDate)) 

从此，你可以指望整月

Public Function Months(_ 
    ByVal datDate1 As Date, _ 
    ByVal datDate2 As Date, _ 
    Optional ByVal booLinear As Boolean) _ 
    As Integer 

' Returns the difference in full months between datDate1 and datDate2. 
' 
' Calculates correctly for: 
' negative differences 
' leap years 
' dates of 29. February 
' date/time values with embedded time values 
' negative date/time values (prior to 1899-12-29) 
' 
' Optionally returns negative counts rounded down to provide a 
' linear sequence of month counts. 
' For a given datDate1, if datDate2 is decreased stepwise one month from 
' returning a positive count to returning a negative count, one or two 
' occurrences of count zero will be returned. 
' If booLinear is False, the sequence will be: 
' 3, 2, 1, 0, 0, -1, -2 
' If booLinear is True, the sequence will be: 
' 3, 2, 1, 0, -1, -2, -3 
' 
' If booLinear is False, reversing datDate1 and datDate2 will return 
' results of same absolute Value, only the sign will change. 
' This behaviour mimics that of Fix(). 
' If booLinear is True, reversing datDate1 and datDate2 will return 
' results where the negative count is offset by -1. 
' This behaviour mimics that of Int(). 

' DateAdd() is used for check for month end of February as it correctly 
' returns Feb. 28. when adding a count of months to dates of Feb. 29. 
' when the resulting year is a common year. 
' 
' 2010-03-30. Cactus Data ApS, CPH. 

    Dim intDiff As Integer 
    Dim intSign As Integer 
    Dim intMonths As Integer 

    ' Find difference in calendar months. 
    intMonths = DateDiff("m", datDate1, datDate2) 
    ' For positive resp. negative intervals, check if the second date 
    ' falls before, on, or after the crossing date for a 1 month period 
    ' while at the same time correcting for February 29. of leap years. 
    If DateDiff("d", datDate1, datDate2) > 0 Then 
    intSign = Sgn(DateDiff("d", DateAdd("m", intMonths, datDate1), datDate2)) 
    intDiff = Abs(intSign < 0) 
    Else 
    intSign = Sgn(DateDiff("d", DateAdd("m", -intMonths, datDate2), datDate1)) 
    If intSign <> 0 Then 
     ' Offset negative count of months to continuous sequence if requested. 
     intDiff = Abs(booLinear) 
    End If 
    intDiff = intDiff - Abs(intSign < 0) 
    End If 

    ' Return count of months as count of full 1 month periods. 
    Months = intMonths - intDiff 

End Function 

现在，用于查询中，创建一个帮助函数：

Public Function RetentionMonths(ByVal HireDate As Date) As Integer 

    Dim RetentionStartdate As Date 

    RetentionStartdate = DateAdd("ww", 14, DateAdd("d", vbSaturday - Weekday(HireDate), HireDate)) 

    RetentionMonths = Months(RetentionStartDate, Date) 

End Function 

保存此功能的代码模块中，和您的查询可能看起来像：

Select *, RetentionMonths([HireDate]) As RetentionMonths 
From YourTable