精华区中的“货币大写转换函数”用起来不是很方便，修改如下:

<%
Function ChangeNumToDx(SourceNum)
    Dim String1    ‘’如下定义
    Dim String2    ‘’如下定义
    Dim String3    ‘’从原SourceNum值中取出的值
    Dim I         ‘’循环变量
    Dim J         ‘’SourceNum的值乘以100的字符串长度
    Dim Ch1        ‘’数字的汉语读法
    Dim Ch2        ‘’数字位的汉字读法
    Dim nZero     ‘’用来计算连续的零值是几个

    String1 = "零壹贰叁肆伍陆柒捌玖"
    String2 = "万仟佰拾亿仟佰拾万仟佰拾元角分"
    nZero = 0
    ChangeNumToDx=""

    If Not IsNumeric(SourceNum) Then
        SourceNum="0"
    End If

    If InStr(1, CStr(SourceNum * 100), ".") <> 0 Then
        err.Raise 5000, , "此函数( AtoC() )只能转换小数点后有两位以内的数！"
    End If
    
    J = Len(CStr(SourceNum * 100))
    String2 = Right(String2, J)                             ‘’取出对应位数的STRING2的值
    
    For I = 1 To J
        String3 = Mid(SourceNum * 100, I, 1)                        ‘’取出需转换的某一位的值
        
        If I <> (J - 3) + 1 And I <> (J - 7) + 1 And I <> (J - 11) + 1 And I <>(J - 15) + 1 Then
            If String3 = 0 Then
                Ch1 = ""
                Ch2 = ""
                nZero = nZero + 1
            ElseIf String3 <> 0 And nZero <> 0 Then
                Ch1 = "零" & Mid(String1, clng(String3) + 1, 1)
                Ch2 = Mid(String2, I, 1)
                nZero = 0
            Else
                Ch1 = Mid(String1, clng(String3) + 1, 1)
                Ch2 = Mid(String2, I, 1)
                nZero = 0
            End If
        Else                                                ‘’该位是万亿，亿，万，元位等关键位
            If String3 <> 0 And nZero <> 0 Then
                Ch1 = "零" & Mid(String1, clng(String3) + 1, 1)
                Ch2 = Mid(String2, I, 1)
                nZero = 0
            ElseIf String3 <> 0 And nZero = 0 Then
                Ch1 = Mid(String1, clng(String3) + 1, 1)
                Ch2 = Mid(String2, I, 1)
                nZero = 0
            ElseIf String3 = 0 And nZero >= 3 Then
                Ch1 = ""
                Ch2 = ""
                nZero = nZero + 1
            Else
                Ch1 = ""
                Ch2 = Mid(String2, I, 1)
                nZero = nZero + 1
            End If
            
            If I = (J - 11) + 1 Or I = (J - 3) + 1 Then     ‘’如果该位是亿位或元位，则必须写上
                Ch2 = Mid(String2, I, 1)
            End If
            
        End If
        AtoC = AtoC & Ch1 & Ch2
        
        If I = J And String3 = 0 Then                       ‘’最后一位（分）为0时，加上“整”
            ChangeNumToDx = AtoC & "整"
        End If
        
    Next

    If SourceNum=0 Then
        ChangeNumToDx="零元整"
    End If
    
End Function

‘’以下为例子
Response.Write ChangeNumToDx(100000)
‘’结果为：壹拾万元整
%>

原文转自：http://www.ltesting.net