Question

我想只在MS访问的文本框中显示5到12之间的年龄。年龄字段不是表格而是计算的。这是代码

= Int（（Now（） - [出生日期]）/ 365.25）

现在验证无效。这是代码：

= 5且＆lt; = 12

当年龄> 12时，验证文字没有出现？我该如何解决？

由于

Answer 1

首先，你需要一个适当的函数来计算年龄 - 像这样：

Public Function AgeSimple( _
  ByVal datDateOfBirth As Date) _
  As Integer

' Returns the difference in full years from datDateOfBirth to current date.
'
' Calculates correctly for:
'   leap years
'   dates of 29. February
'   date/time values with embedded time values
'
' DateAdd() is used for check for month end of February as it correctly
' returns Feb. 28. when adding a count of years to dates of Feb. 29.
' when the resulting year is a common year.
' After an idea of Markus G. Fischer.
'
' 2007-06-26. Cactus Data ApS, CPH.

  Dim datToday  As Date
  Dim intAge    As Integer
  Dim intYears  As Integer

  datToday = Date
  ' Find difference in calendar years.
  intYears = DateDiff("yyyy", datDateOfBirth, datToday)
  If intYears > 0 Then
    ' Decrease by 1 if current date is earlier than birthday of current year
    ' using DateDiff to ignore a time portion of datDateOfBirth.
    intAge = intYears - Abs(DateDiff("d", datToday, DateAdd("yyyy", intYears, datDateOfBirth)) > 0)
  End If

  AgeSimple = intAge

End Function

然后验证应该是：

>=5 And <=12

但验证仅适用于输入，这不会发生，因为控件绑定到表达式。当你考虑它时，可以采取什么行动，并且＆＃34;无效＆＃34;从表达式输出？

Answer 2

拳头，为了在几天之间获得可中继的时间间隔，请使用DateDiff功能。您的年龄函数应如下所示：Int(DateDiff("d",[Date Of Birth],Now()/365.25)
第二，您需要条件显示，使用IIf功能。这应该是你的控制权：

=IIf(Int(DateDiff("d",[Date Of Birth],Now())/365.25)>=5 and Int(DateDiff("d",[Date Of Birth],Now())/365.25)<=12,Int(DateDiff("d",[Date Of Birth],Now())/365.25),Null)

正如古斯塔夫指出的那样，验证文本在这种情况下并不相关。

访问验证无效

2 个答案: