Question

我试图让这个函数返回一个蒙版数字。假设我的号码是123456789.我试图让它返回* 6789如果我给出一个未屏蔽的值4，或者* 789如果我给它一个未屏蔽的值3.目前它显示的是数字中的位数，这就是我试图隐藏的东西。我一直在玩弄这个，但我不能让它做我想做的事。

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

Answer 1

如果您想要将数字掩盖为整数（比如iNumber）而不是字符串，则可以使用

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

（请注意，在vb.net中^是指数。）

如果您不需要使用sNumber，请使用

"*" & Right(sNumber, iUnmaskedLength)

Right()允许iUnmaskedLength大于字符串的长度;在这种情况下，它返回输入字符串。

Answer 2

进行一些错误检查。

测试代码

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

功能

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

通过显示前面带有1个掩码字符的最后“x”数字来屏蔽数字

2 个答案: