看到前面的金额转换，一时兴起也动手写了一个，写的匆忙支持的位数不多，有错误的地方还请多多指教。入口：getChangedVal
Option Explicit
'总体思路：
'对数字进行分级处理，级长为4
'对分级后的每级分别处理，处理后得到字符串相连
'如：123456=12|3456
'第二级：12=壹拾贰 + “万”
'第一级：3456 =叁千肆百伍拾陆 + “”
Private Const PrvStrNum = "壹贰叁肆伍陆柒捌玖零"
Private Const PrvStrUnit = "万千百拾个"
Private Const PrvStrGradeUnit = "千万亿兆" '"兆亿万千"
Private Const PrvGrade = 4
Public Function getChangedVal(ByVal StrVal As String) As String
Dim StrDotUnit As String
Dim StrIntUnit As String
StrDotUnit = getDotUnit(StrVal) '取小数位
StrIntUnit = getIntUnit(StrVal) '取整数位
StrIntUnit = getIntUpper(StrIntUnit) '整数位转换大写
StrDotUnit = getDotUpper(StrIntUnit) '小数位转换大写
getChangedVal = StrIntUnit & StrDotUnit
End Function
Private Function getDotUnit(ByVal StrVal As String) As String
'得到小数点后的数字
Dim StrRet As String
Dim IntBegin As Integer
Dim IntLen As Integer
IntBegin = InStr(1, StrVal, ".") + 1
IntLen = Len(StrVal) + 1
StrRet = Mid(StrVal, IntBegin, IntLen - IntBegin)
If IntBegin > 1 Then
getDotUnit = StrRet
End If
End Function
Private Function getIntUnit(ByVal StrVal As String) As String
'得到整数数字
Dim StrRet As String
Dim IntBegin As Integer
Dim IntLen As Integer
'取得小数数位的长度
IntBegin = Len(getDotUnit(StrVal))
IntLen = Len(StrVal)
StrRet = Mid(StrVal, 1, IntLen - IntBegin) '总字串长度-小数数位长度=整数数位长度
If Mid(StrRet, Len(StrRet), 1) = "." Then '去除末位小数点
StrRet = Mid(StrRet, 1, Len(StrRet) - 1)
End If
getIntUnit = StrRet
End Function
Private Function getIntUpper(ByVal StrVal As String) As String
'得到转换后的大写（整数部分）
Dim IntGrade As Integer '级次
Dim StrRet As String
Dim StrTmp As String
'得到当前级次，
IntGrade = Fix(Len(StrVal) / PrvGrade)
'调整级次长度
If (Len(StrVal) Mod PrvGrade) <> 0 Then
IntGrade = IntGrade + 1
End If
'MsgBox Mid(PrvStrGradeUnit, IntGrade, 1)
Dim i As Integer
'对每级数字处理
For i = IntGrade To 1 Step -1
StrTmp = getNowGradeVal(StrVal, i) '取得当前级次数字
StrRet = StrRet & getSubUnit(StrTmp) '转换大写
StrRet = dropZero(StrRet) '除零
'加级次单位
If i > 1 Then '末位不加单位
'单位不能相连续
'??????????????????????????????????
'
StrRet = StrRet & Mid(PrvStrGradeUnit, i, 1)
End If
Next
getIntUpper = StrRet
End Function
Private Function getDotUpper(ByVal StrVal As String) As String
'得到转换后的大写（小数部分）
End Function
Private Function dropZero(ByVal StrVal As String) As String
'去除连继的“零”
Dim StrRet As String
Dim StrBefore As String '前一位置字符
Dim StrNow As String '现在位置字符
Dim i As Integer
StrBefore = Mid(StrVal, 1, 1)
StrRet = StrBefore
For i = 2 To Len(StrVal)
StrNow = Mid(StrVal, i, 1)
If StrNow = "零" And StrBefore = "零" Then
'同时为零
Else
StrRet = StrRet & StrNow
End If
StrBefore = StrNow
Next
'末位去零
Dim IntLocate As Integer
IntLocate = Len(StrRet)
'IntLocate = IIf(IntLocate = 0, 1, IntLocate)
If Mid(StrRet, IntLocate, 1) = "零" Then
StrRet = Left(StrRet, Len(StrRet) - 1)
End If
dropZero = StrRet
End Function
Private Function getSubUnit(ByVal StrVal As String) As String
'数值转换
Debug.Print StrVal
Dim IntLen As Integer
Dim i As Integer
Dim StrKey As String
Dim StrRet As String
Dim IntKey As Integer
IntLen = Len(StrVal)
For i = 1 To IntLen
StrKey = Mid(StrVal, i, 1)
IntKey = Val(StrKey)
If IntKey = 0 Then
'“零”作特殊处理
If i <> IntLen Then '转换后数末位不能为零
StrRet = StrRet & "零"
End If
Else
'If IntKey = 1 And i = 2 Then
'“壹拾”作特殊处理
'“壹拾”合理
'Else
StrRet = StrRet & Mid(PrvStrNum, Val(StrKey), 1)
'End If
'追加单位
If i <> IntLen Then '个位不加单位
StrRet = StrRet & Mid(PrvStrUnit, Len(PrvStrUnit) - IntLen + i, 1)
End If
End If
Next
getSubUnit = StrRet
End Function
Private Function getNowGradeVal(ByVal StrVal As String, ByVal IntGrade As Integer) As String
'得到当前级次的串
Dim IntGradeLen As Integer
Dim IntLen As Integer
Dim StrRet As String
IntGradeLen = IntGrade * PrvGrade
IntLen = Len(StrVal)
If IntLen >= IntGradeLen Then
StrRet = Mid(StrVal, IntLen - IntGradeLen + 1, PrvGrade)
Else
StrRet = Mid(StrVal, 1, IntLen - (IntGrade - 1) * PrvGrade)
End If
'MsgBox StrRet
getNowGradeVal = StrRet
End Function
|
本类阅读TOP10