解决方法
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
