在Excel中validationIBAN

有没有准备好的Excel表格来validationIBAN？ 我将进入IBAN，它将显示有效或无效。

我search了一些Adds In，并find了这个

但我不知道如何打开它。 谁能帮忙？

Answer 1

这很简单，只需使用下面的function。

'' Validate IBAN Public Function VALIDATEIBAN(ByVal IBAN As string) As Boolean On Error GoTo Catch Dim objRegExp As Object Dim blnIsValidIBAN As Boolean Set objRegExp = CreateObject("vbscript.regexp") objRegExp.IgnoreCase = True objRegExp.Global = True objRegExp.Pattern = "^[a-zA-Z]{2}\d{2}[ ]\d{4}[ ]\d{4}[ ]\d{4}[ ]\d{4}[ ]\d{4}|CZ\d{22}$" blnIsValidIBAN = objRegExp.Test(IBAN) VALIDATEIBAN = blnIsValidIBAN Exit Function Catch: VALIDATEIBAN = False MsgBox "Module: " & MODULE_NAME & " - VALIDATEIBAN function" & vbCrLf & vbCrLf _ & "Error#: " & Err.Number & vbCrLf & vbCrLf & Err.Description End Function 

如何使用：

 Copy the code. In Excel press Alt + F11 to enter the VBE. Press Ctrl + R to show the Project Explorer. Insert -> Module. Paste code. Save and Exit VBE. 

运行该function：

现在在Excel中有一个用户定义的函数，就像内置的SUM，AVG函数一样。 假设您想在单元格A1中validationIBAN，只需在任何单元格中写入=VALIDATEIBAN(A1). 它将返回TRUE或FALSE。

此外，它将同时适用于：

 ES65 0800 0000 1920 0014 5399 

和

 ES6508000000192000145399 

但不是：

 ES65-0800-0000-1920-0014-5399 

Answer 2

那么这个格式化的问题就解决了，执行了“97检查”：

 Public Function VALIDATEIBAN(ByVal IBAN As String) As String ' Created by : Koen Rijnsent (www.castoro.nl) ' Inspired by : Chris Fannin (AbbydonKrafts) ' Inspired by : bonsvr (http://stackoverflow.com/users/872583/bonsvr) On Error GoTo CatchError Dim objRegExp As Object Dim IBANformat As Boolean Dim IBANNR As String Dim ReplaceChr As String Dim ReplaceBy As String 'Check format IBAN = UCase(IBAN) Set objRegExp = CreateObject("vbscript.regexp") objRegExp.IgnoreCase = True objRegExp.Global = True objRegExp.Pattern = "[a-zA-Z]{2}[0-9]{2}[a-zA-Z0-9]{4}[0-9]{7}([a-zA-Z0-9]?){0,16}" IBANformat = objRegExp.Test(IBAN) 'Validity of country code will not be checked! If IBANformat = False Then VALIDATEIBAN = "FORMAT NOT RECOGNIZED" Else 'Flip first 4 characters to the back IBANNR = Right(IBAN, Len(IBAN) - 4) & Left(IBAN, 4) 'Replace letters by the right numbers For Nr = 10 To 35 ReplaceChr = Chr(Nr + 55) ReplaceBy = Trim(Str(Nr)) IBANNR = Replace(IBANNR, ReplaceChr, ReplaceBy) Next Nr 'Loop through the IBAN, as it is too long to calculate at one go CurrPart = "" Answer = "" For CurrDigit = 1 To Len(IBANNR) CurrPart = CurrPart & Mid$(IBANNR, CurrDigit, 1) CurrNumber = CLng(CurrPart) 'If the number can be divided If 97 <= CurrNumber Then LeftOver = CurrNumber Mod 97 WorkValue = (CurrNumber - LeftOver) / 97 Answer = Answer & CStr(WorkValue) CurrPart = CStr(LeftOver) Else 'If no division occurred, add a trailing zero If Len(Answer) > 0 Then Answer = Answer & "0" 'Exception for the last number If CurrDigit = Len(IBANNR) Then LeftOver = CurrNumber Mod 97 Else End If Else End If End If Next If LeftOver = 1 Then VALIDATEIBAN = "IBAN OK" Else VALIDATEIBAN = "97 CHECK FAILED" End If End If Exit Function CatchError: VALIDATEIBAN = "ERROR: " & Err.Description MsgBox "Module: " & MODULE_NAME & " - VALIDATEIBAN function" & vbCrLf & vbCrLf _ & "Error#: " & Err.Number & vbCrLf & vbCrLf & Err.Description End Function 

Answer 3

我发现从bonsvr有用的答案，谢谢。 从我的代码阅读来看，这似乎是特定于CZ帐户的范围。

由于我主要处理爱尔兰，英国和德国的IBAN代码，所以我开发了这个正则expression式，用objRegExp.Pattern = …代替行。

 objRegExp.Pattern = "^[GB|IE]{2}\d{2}[a-zA-Z]{4}\d{14}|[DE]\d{20}$" 

我希望这可以帮助别人，因为最初的代码帮助了我。 如果你想添加自己的国家，扩展以上。

注：我删除提供的空格（ [ ] ），因为我正在testing的文本没有。 如果您希望每4个字符添加一次，这很容易实现，或者更简单一些，将上面的代码的第一行replace为：

 IBAN = Trim(Ucase(Replace(IBAN, " ", ""))) 

这将消除空间，削减在前面和后面的任何额外的空间，并转换为大写。 （修剪可能是多余的，但皮带和括号…）

正则expression式由[GB | IE]（GB或IE）组成，允许使用ISO国家代码，后跟2位数校验和，4位数字库代码和14位数字，格式相同这两个国家的IBAN格式。 德国允许使用另一个“或”，然后是22位数字。 要添加另一个国家，只需将您的文本放在$符号之前，以|开头。

在这里find其他国家的格式 （维基百科）

Answer 4

 Option Compare Database Option Explicit ' http://en.wikipedia.org/wiki/International_Bank_Account_Number Private Const IbanCountryLengths As String = "AL28AD24AT20AZ28BH22BE16BA20BR29BG22CR21HR21CY28CZ24DK18DO28EE20FO18" & _ "FI18FR27GE22DE22GI23GR27GL18GT28HU28IS26IE22IL23IT27KZ20KW30LV21LB28" & _ "LI21LT20LU20MK19MT31MR27MU30MC27MD24ME22NL18NO15PK24PS29PL28PT25RO24" & _ "SM27SA24RS22SK24SI19ES24SE24CH21TN24TR26AE23GB22VG24QA29" Private Function ValidateIbanCountryLength(CountryCode As String, IbanLength As Integer) As Boolean Dim i As Integer For i = 0 To Len(IbanCountryLengths) / 4 - 1 If Mid(IbanCountryLengths, i * 4 + 1, 2) = CountryCode And _ CInt(Mid(IbanCountryLengths, i * 4 + 3, 2)) = IbanLength Then ValidateIbanCountryLength = True Exit Function End If Next i ValidateIbanCountryLength = False End Function Private Function Mod97(Num As String) As Integer Dim lngTemp As Long Dim strTemp As String Do While Val(Num) >= 97 If Len(Num) > 5 Then strTemp = Left(Num, 5) Num = Right(Num, Len(Num) - 5) Else strTemp = Num Num = "" End If lngTemp = CLng(strTemp) lngTemp = lngTemp Mod 97 strTemp = CStr(lngTemp) Num = strTemp & Num Loop Mod97 = CInt(Num) End Function Public Function ValidateIban(IBAN As String) As Boolean Dim strIban As String Dim i As Integer strIban = UCase(IBAN) ' Remove spaces strIban = Replace(strIban, " ", "") ' Check if IBAN contains only uppercase characters and numbers For i = 1 To Len(strIban) If Not ((Asc(Mid(strIban, i, 1)) <= Asc("9") And Asc(Mid(strIban, i, 1)) >= Asc("0")) Or _ (Asc(Mid(strIban, i, 1)) <= Asc("Z") And Asc(Mid(strIban, i, 1)) >= Asc("A"))) Then ValidateIban = False Exit Function End If Next i ' Check if length of IBAN equals expected length for country If Not ValidateIbanCountryLength(Left(strIban, 2), Len(strIban)) Then ValidateIban = False Exit Function End If ' Rearrange strIban = Right(strIban, Len(strIban) - 4) & Left(strIban, 4) ' Replace characters For i = 0 To 25 strIban = Replace(strIban, Chr(i + Asc("A")), i + 10) Next i ' Check remainder ValidateIban = Mod97(strIban) = 1 End Function 

来源： http ： //www.aswinvanwoudenberg.com/2013/07/18/vba-iban-validator/