CODEWITHSUNDEEP
vb.netmasking
Derek Meyer
Derek Meyer

Reputation: 507

Masking a number by showing last "x" digits with 1 mask character in front

Im attempting to get this function to return a masked number. Say i have number 123456789. Im trying to get it to return *6789 if i give in an unmasked value of 4, or *789 if i give it an unmasked value of 3. Currently it is showing the number of digits in the number, which is what im trying to hide. I have been toying around with this but i cant quite get it to do what I want.

Public Function GetMaskedNumber(ByVal sNumber As String, ByVal iUnmaskedLength As Integer, ByVal sMaskChar As String) As String
    sMaskChar = Trim(sMaskChar)
    If iUnmaskedLength > 0 AndAlso Len(sMaskChar) > 0 Then
        GetMaskedNumber = New String(sMaskChar(0), iUnmaskedLength)
        If iUnmaskedLength < Len(sNumber) Then
            Mid(GetMaskedNumber, (Len(sNumber) - iUnmaskedLength), iUnmaskedLength + 1) = Right(sNumber, iUnmaskedLength)
        Else
            GetMaskedNumber = sNumber
        End If
    Else
        GetMaskedNumber = sNumber
    End If
End Function

Upvotes: 0

Views: 2509

Answers (2)

dbasnett
dbasnett

Reputation: 11773

With some error checking.

test code

Private Sub Button1_Click(sender As Object, e As EventArgs) Handles Button1.Click
    'test
    Dim i As Long = Long.MaxValue
    Dim s As String = GetMaskedNumber(i, 12)
    s = GetMaskedNumber(i, 20)
End Sub

the function

Public Function GetMaskedNumber(ByVal theNum As Long, _
                                ByVal UnmaskedLength As Integer, _
                                Optional MaskChar As String = "*") As String
    Dim retval As String = theNum.ToString
    If retval.Length > UnmaskedLength Then
        retval = String.Format("{0}{1}", MaskChar, retval.Substring(retval.Length - UnmaskedLength))
    Else
        'ERROR???
        Stop
    End If
    Return retval
End Function

Upvotes: 0

Bathsheba
Bathsheba

Reputation: 234715

If you have the number you want to mask as an Integer (say iNumber) rather than a string, you could use

"*" & CStr(iNumber mod (10 ^ iUnmaskedLength))

(Note that in vb.net ^ is exponentation.)

If you don't and need to work with sNumber then use

"*" & Right(sNumber, iUnmaskedLength)

Right() allows iUnmaskedLength to be larger than the length of the string; in such cases it returns the input string.

Upvotes: 3

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