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有効なIMEIを確認する方法を知っている人はいますか?

このページでチェックする関数を見つけました: http://www.dotnetfunda.com/articles/article597-imeivalidator-in-vbnet-.aspx

ただしfalse、有効な IMEI (fe 352972024585360) に対しては返されます。このページでオンラインで検証できます: http://www.numberingplans.com/?page=analysis&sub=imeinr

特定の IMEI が有効かどうかを確認する正しい方法 (VB.Net) は何ですか?

PS: 上記のページのこの関数は、何らかの形で正しくないに違いありません:

Public Shared Function isImeiValid(ByVal IMEI As String) As Boolean
    Dim cnt As Integer = 0
    Dim nw As String = String.Empty
    Try
        For Each c As Char In IMEI
            cnt += 1
            If cnt Mod 2 <> 0 Then
                nw += c
            Else
                Dim d As Integer = Integer.Parse(c) * 2 ' Every Second Digit has to be Doubled '
                nw += d.ToString() ' Genegrated a new number with doubled digits '
            End If
        Next
        Dim tot As Integer = 0
        For Each ch As Char In nw.Remove(nw.Length - 1, 1)
            tot += Integer.Parse(ch) ' Adding all digits together '
        Next
        Dim chDigit As Integer = 10 - (tot Mod 10) ' Finding the Check Digit my Finding the Remainder of the sum and subtracting it from 10 '
        If chDigit = Integer.Parse(IMEI(IMEI.Length - 1)) Then ' Checking the Check Digit with the last digit of the Given IMEI code '
            Return True
        Else
            Return False
        End If
    Catch ex As Exception
        Return False
    End Try
End Function

編集:これは私の作業中の「checkIMEI」関数です:

Public Shared Function checkIMEI(ByRef IMEI As String) As Boolean
    Const allowed As String = "0123456789"

    Dim cleanNumber As New System.Text.StringBuilder
    For i As Int32 = 0 To IMEI.Length - 1
        If (allowed.IndexOf(IMEI.Substring(i, 1)) >= 0) Then
            cleanNumber.Append(IMEI.Substring(i, 1))
        End If
    Next

    If cleanNumber.Length <> 15 Then
        Return False
    Else
        IMEI = cleanNumber.ToString
    End If

    For i As Int32 = cleanNumber.Length + 1 To 16
        cleanNumber.Insert(0, "0")
    Next

    Dim multiplier As Int32, digit As Int32, sum As Int32, total As Int32 = 0
    Dim number As String = cleanNumber.ToString()

    For i As Int32 = 1 To 16
        multiplier = 1 + (i Mod 2)
        digit = Int32.Parse(number.Substring(i - 1, 1))
        sum = digit * multiplier
        If (sum > 9) Then
            sum -= 9
        End If
        total += sum
    Next

    Return (total Mod 10 = 0)
End Function
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2 に答える 2

11

IMEI 番号は、Luhnアルゴリズムを使用して検証されます。リンクされたページには、さまざまな言語での実装があります。この投稿には、さらにいくつかの実装と、Luhn アルゴリズムを解決する方法に関する一般的な方法論も含まれています。

于 2010-03-25T15:02:02.933 に答える