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 & StrDotUnitEnd 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 IfEnd FunctionPrivate 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 = StrRetEnd 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 = StrRetEnd Function
Private Function getDotUpper(ByVal StrVal As String) As String '得到转换后的大写（小数部分）End FunctionPrivate 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 = StrRetEnd FunctionPrivate 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 = StrRetEnd FunctionPrivate 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
