虽然DateDiff()函数似乎是计算年龄的合理选择,但遗憾的是它并不计算两个日期之间经过的整年或月数.例如,假设一个婴儿出生于2014年12月31日,并在2015年1月2日48小时后进行了检查.也就是说,

DateOfBirth = DateSerial(2014,12,31)
DateOfExam = DateSerial(2015,1,2)

如果我们只是使用DateDiff()来计算考试时的年和月的“年龄”,我们就会得到

?DateDiff("yyyy",DateOfBirth,DateOfExam)
 1 
?DateDiff("m",DateOfExam)
 1

因此,我们会报告婴儿是1岁零1个月,而实际上她只有2天大.

适当的年龄计算需要比这更复杂.以下VBA函数将计算年和月的“年龄”,返回“2年1个月”之类的字符串：

Public Function AgeInYearsAndMonths(StartDate As Variant,EndDate As Variant) As Variant
    Dim Date1 As Date,Date2 As Date
    Dim mm1 As Integer,dd1 As Integer,mm2 As Integer,dd2 As Integer
    Dim ageYears As Integer,ageMonths As Integer,rtn As Variant
    rtn = Null
    If Not (IsNull(StartDate) Or IsNull(EndDate)) Then
        If StartDate <= EndDate Then
            Date1 = StartDate
            Date2 = EndDate
        Else
            Date1 = EndDate
            Date2 = StartDate
        End If
        mm1 = Month(Date1)
        dd1 = Day(Date1)
        mm2 = Month(Date2)
        dd2 = Day(Date2)
        ageYears = DateDiff("yyyy",Date1,Date2)
        If (mm1 > mm2) Or (mm1 = mm2 And dd1 > dd2) Then
            ageYears = ageYears - 1
        End If
        ageMonths = DateDiff("m",Date2) Mod 12
        If dd1 > dd2 Then
            If ageMonths = 0 Then
                ageMonths = 12
            End If
            ageMonths = ageMonths - 1
        End If
        If ageYears = 0 And ageMonths = 0 Then
            rtn = "less than 1 month"
        Else
            rtn = ageYears & " year" & IIf(ageYears = 1,"","s") & " and " & ageMonths & " month" & IIf(ageMonths = 1,"s")
        End If
    End If
    AgeInYearsAndMonths = rtn
End Function