Question

我在我公司的一项服务中构建了一个由不同人使用的文件。

每个工作表都受密码保护，所有用户条目都使用VBA用户表单处理。当用户修改数据时，所有工作表都受相同密码和我的代码保护/取消保护表的保护。

问题是我在VBA项目中以明文形式存储密码，以便调用ActiveSheet.Protect password方法。 VBA项目也受此密码保护。

是否有安全的方法将该密码存储在VBA项目中？

任何知道如何搜索的人都会找到破解该VBA项目密码并能够阅读它的代码。

编辑：

我想过每次打开文件时通过添加一些随机性来计算新密码。这样就可以在不知道密码的情况下读取代码。添加msgbox可以显示它，但只有在文件重新打开之后才会显示。问题是我无法使用该方法手动取消保护/保护工作表，因为我不知道密码。

Answer 1

这应该可以解决问题。密码为smp2smp2，运行GetPassword时将获得该密码，但该实际值未存储在项目中。它使用代码30555112012321187051111661144119存储，代码CreatePasswordFromCode将使用Option Explicit Function GetPassword() As String 'the password is stored as codes, so the real password is not stored in this project GetPassword = CreatePasswordFromCode("30555112012321187051111661144119") End Function Function CreatePasswordFromCode(ByVal pstrPasswordCode As String) As String Dim intChar As Integer Dim intCode As Integer Dim arrintShifts(0 To 7) As Integer Dim arrlngCharCode(0 To 7) As Long Dim strMessage As String intChar = 0 intCode = 0 For intCode = 0 To 7 'store -8 to -1 into 0-7 arrintShifts(intCode) = intCode - 8 Next intCode 'the code is stored by using the number of the letter of the password in the 4th character. 'the real code of the character is directly behind that. 'so the code 30555112012321187051111661144119 'has on position 3, 055, 5, 112, 0, 123, 2, 118, 7, 051, 1, 116, 6, 114 and 4, 119 'so sorted this is 0, 123, 1, 116, 2, 118, 3, 055, 4, 119, 5, 112, 6, 114, 7, 051 'then there is also the part where those charcode are shifted by adding -8 to -1 to them. 'leading to the real charactercodes: '0, 123-8, 1, 116-7, 2, 118-6, 3, 055-5, 4, 119-4, 5, 112-3, 6, 114-2, 7, 051-1 '0, 115, 1, 109, 2, 112, 3, 050, 4, 115, 5, 109, 6, 112, 7, 050 For intChar = 0 To 7 If Mid(pstrPasswordCode, 1, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 2, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 5, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 6, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 9, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 10, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 13, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 14, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 17, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 18, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 21, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 22, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 25, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 26, 3) + arrintShifts(intChar)) ElseIf Mid(pstrPasswordCode, 29, 1) = intChar Then arrlngCharCode(intChar) = (Mid(pstrPasswordCode, 30, 3) + arrintShifts(intChar)) End If Next intChar 'by getting the charcodes of these values, you create the password CreatePasswordFromCode = Chr(arrlngCharCode(0)) & Chr(arrlngCharCode(1)) & Chr(arrlngCharCode(2)) & Chr(arrlngCharCode(3)) & Chr(arrlngCharCode(4)) & Chr(arrlngCharCode(5)) & Chr(arrlngCharCode(6)) & Chr(arrlngCharCode(7)) End Function转换为实际密码（人类可读）。顺便说一句，我不知道如何轻松获取属于某个密码的代码。这样，它总是8个字符长，除非你调整代码，否则没有改变的余地。我在其他人的旧项目中找到了这个地方，遗憾的是没有提到任何消息来源。

Tabelle3.Range("AV9:AV" & lastrow)

Answer 2

修改了最多99个字符的代码。添加了密码生成器。

但仍然：这只是对真实密码的混淆。

Function CreatePasswordFromCode(ByVal pstrPasswordCode As String) As String
' Original Code https://stackoverflow.com/questions/47990187/securely-store-password-in-a-vba-project?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa
' Modified to extend password length
' Modifications free to use
Dim codeLen As Integer

Dim intChar As Integer
Dim intCode As Integer
Dim arrintShifts() As Integer
Dim arrlngCharCode() As Long
Dim icp As Integer

    ' Initialise Arrays
    icp = IIf(Right(pstrPasswordCode, 1) Mod 2 = 0, 5, 4)
    pstrPasswordCode = Left(pstrPasswordCode, Len(pstrPasswordCode) - IIf(Right(pstrPasswordCode, 1) Mod 2 = 0, 1, 1))
    codeLen = Len(pstrPasswordCode) / icp - 1 ' Array Index starts with 0
    ReDim arrintShifts(codeLen)
    ReDim arrlngCharCode(codeLen)

    intChar = 0
    intCode = 0

    For intCode = 0 To codeLen
        'store -8 to -1 into 0-7
        arrintShifts(intCode) = intCode - (codeLen + 1)
    Next intCode

    'the code is stored by using the number of the letter of the password in the 4th character.
    'the real code of the character is directly behind that.
    'so the code 30555112012321187051111661144119
    'has on position 3, 055, 5, 112, 0, 123, 2, 118, 7, 051, 1, 116, 6, 114 and 4, 119
    'so sorted this is 0, 123, 1, 116, 2, 118, 3, 055, 4, 119, 5, 112, 6, 114, 7, 051
    'then there is also the part where those charcode are shifted by adding -8 to -1 to them.
    'leading to the real charactercodes:
    '0, 123-8, 1, 116-7, 2, 118-6, 3, 055-5, 4, 119-4, 5, 112-3, 6, 114-2, 7, 051-1
    '0, 115, 1, 109, 2, 112, 3, 050, 4, 115, 5, 109, 6, 112, 7, 050

    For intChar = 0 To codeLen
        For intCode = 0 To codeLen
            If CInt(Mid(pstrPasswordCode, intCode * icp + 1, icp - 3)) = intChar Then
                arrlngCharCode(intChar) = (Mid(pstrPasswordCode, (intCode + 1) * icp - 2, 3) + arrintShifts(intChar))
                Exit For
            End If
        Next intCode
    Next intChar

    'by getting the charcodes of these values, you create the password
    CreatePasswordFromCode = ""
    For intChar = 0 To codeLen
        CreatePasswordFromCode = CreatePasswordFromCode & Chr(arrlngCharCode(intChar))
    Next intChar

End Function

Function CreateCodeFromPassword(ByVal pstrPasswordCode As String) As String
' Generator free to use
Dim pwLen As Integer
Dim scp As String   ' String Code Position, for formatting "0" or "00"
Dim icp As Integer  ' marker if pwLen < 10 or > 10
Dim intCode As Integer
Dim arrintShifts() As Integer
Dim arrlngCharCode() As Long
Dim pw() As String

Dim Temp As Variant
Dim arnd() As Variant
Dim irnd As Variant

    Randomize

    ' Initialise Arrays
    pwLen = Len(pstrPasswordCode) - 1 ' Array Index starts with 0
    scp = IIf(pwLen < 10, "0", "00")
    ' Create odd/even marker if we have 1 (odd) or 2 (even) byte index digits (scp), values between 0 and 9
    icp = IIf(pwLen < 10, Int(Rnd() * 5 + 1) * 2 - 1, Int(Rnd() * 5 + 1) * 2)

    ReDim arrintShifts(pwLen)
    ReDim arrlngCharCode(pwLen)
    ReDim pw(pwLen)
    ReDim arnd(pwLen)

    For intCode = 0 To pwLen
        arnd(intCode) = intCode
    Next intCode

    ' randomize the indizes to bring the code into a random order
    For intCode = LBound(arnd) To UBound(arnd)
        irnd = CLng(((UBound(arnd) - intCode) * Rnd) + intCode)
        If intCode <> irnd Then
            Temp = arnd(intCode)
            arnd(intCode) = arnd(irnd)
            arnd(irnd) = Temp
        End If
    Next intCode

    'by getting the charcodes of these values, you create the password
    For intCode = 0 To pwLen
        'get characters
        pw(intCode) = Mid(pstrPasswordCode, intCode + 1, 1)
        'and store -8 to -1 into 0-7 (for additional obfuscation)
        arrintShifts(intCode) = intCode - (pwLen + 1)
    Next intCode

    ' Search for the random index and throw the shifted code at this position
    For intCode = 0 To pwLen
        arrlngCharCode(Application.Match(intCode, arnd, False) - 1) = AscB(pw(intCode)) - arrintShifts(intCode)
    Next intCode

    ' Chain All Codes, combination of arnd(intcode) and arrlngCharCode(intcode) gives the random order
    CreateCodeFromPassword = ""
    For intCode = 0 To pwLen
        CreateCodeFromPassword = CreateCodeFromPassword & Format(arnd(intCode), scp) & Format(arrlngCharCode(intCode), "000")
    Next intCode
    CreateCodeFromPassword = CreateCodeFromPassword & icp

End Function

混淆版本

'VBA code protection using: www.excel-pratique.com/en/vba_tricks/vba-obfuscator.php
Function CreatePasswordFromCode(ByVal z4891679d877f1da36647b21d6197fbfd As String) As String
Dim b2da54ddb60c93bf346493d7e08bc6d08 As Integer
Dim bf56f94eb6ed9a658e82e88591237324d As Integer
Dim bec732ae8e18b7b2ff2e9ccd058f3e8fc As Integer
Dim m06993036154505accc9ce092bdb57b17() As Integer
Dim b8026f9f8f7fe86372be0799d8c9c6691() As Long
Dim q24471047c7a6e466b78de3c6ae66f20f As String
Dim t5f443e88a552a3f943275f985dde03ca As Integer
t5f443e88a552a3f943275f985dde03ca = IIf(Right(z4891679d877f1da36647b21d6197fbfd, 1) Mod 2 = 0, 5, 4)
z4891679d877f1da36647b21d6197fbfd = Left(z4891679d877f1da36647b21d6197fbfd, Len(z4891679d877f1da36647b21d6197fbfd) - IIf(Right(z4891679d877f1da36647b21d6197fbfd, 1) Mod 2 = 0, 1, 1))
b2da54ddb60c93bf346493d7e08bc6d08 = Len(z4891679d877f1da36647b21d6197fbfd) / t5f443e88a552a3f943275f985dde03ca - 1
ReDim m06993036154505accc9ce092bdb57b17(b2da54ddb60c93bf346493d7e08bc6d08)
ReDim b8026f9f8f7fe86372be0799d8c9c6691(b2da54ddb60c93bf346493d7e08bc6d08)
bf56f94eb6ed9a658e82e88591237324d = 0
bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0 To b2da54ddb60c93bf346493d7e08bc6d08
m06993036154505accc9ce092bdb57b17(bec732ae8e18b7b2ff2e9ccd058f3e8fc) = bec732ae8e18b7b2ff2e9ccd058f3e8fc - (b2da54ddb60c93bf346493d7e08bc6d08 + 1)
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
For bf56f94eb6ed9a658e82e88591237324d = 0 To b2da54ddb60c93bf346493d7e08bc6d08
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0 To b2da54ddb60c93bf346493d7e08bc6d08
If CInt(Mid(z4891679d877f1da36647b21d6197fbfd, bec732ae8e18b7b2ff2e9ccd058f3e8fc * t5f443e88a552a3f943275f985dde03ca + 1, t5f443e88a552a3f943275f985dde03ca - 3)) = bf56f94eb6ed9a658e82e88591237324d Then
b8026f9f8f7fe86372be0799d8c9c6691(bf56f94eb6ed9a658e82e88591237324d) = (Mid(z4891679d877f1da36647b21d6197fbfd, (bec732ae8e18b7b2ff2e9ccd058f3e8fc + 1) * t5f443e88a552a3f943275f985dde03ca - 2, 3) + m06993036154505accc9ce092bdb57b17(bf56f94eb6ed9a658e82e88591237324d))
Exit For
End If
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
Next bf56f94eb6ed9a658e82e88591237324d
CreatePasswordFromCode = ""
For bf56f94eb6ed9a658e82e88591237324d = 0 To b2da54ddb60c93bf346493d7e08bc6d08
CreatePasswordFromCode = CreatePasswordFromCode & Chr(b8026f9f8f7fe86372be0799d8c9c6691(bf56f94eb6ed9a658e82e88591237324d))
Next bf56f94eb6ed9a658e82e88591237324d
End Function
Function CreateCodeFromPassword(ByVal z4891679d877f1da36647b21d6197fbfd As String) As String
Dim qe564274d6cab7b91a3393ef092dac78f As Integer
Dim b330c8da5472f3c36b801671ef5a54797 As String
Dim t5f443e88a552a3f943275f985dde03ca As Integer
Dim bec732ae8e18b7b2ff2e9ccd058f3e8fc As Integer
Dim m06993036154505accc9ce092bdb57b17() As Integer
Dim b8026f9f8f7fe86372be0799d8c9c6691() As Long
Dim b343223dcae485b35af2792c7dd91f92b() As String
Dim e0d4cf763c9da42470a729a29b30d7d50 As Variant
Dim b41d8f2e79c0e09113beb7629aa0d8c48() As Variant
Dim b42a57d0c121b9fe34a74143aa279157c As Variant
Randomize
qe564274d6cab7b91a3393ef092dac78f = Len(z4891679d877f1da36647b21d6197fbfd) - 1
b330c8da5472f3c36b801671ef5a54797 = IIf(qe564274d6cab7b91a3393ef092dac78f < 10, "0", "00")
t5f443e88a552a3f943275f985dde03ca = IIf(qe564274d6cab7b91a3393ef092dac78f < 10, Int(Rnd() * 5 + 1) * 2 - 1, Int(Rnd() * 5 + 1) * 2)
ReDim m06993036154505accc9ce092bdb57b17(qe564274d6cab7b91a3393ef092dac78f)
ReDim b8026f9f8f7fe86372be0799d8c9c6691(qe564274d6cab7b91a3393ef092dac78f)
ReDim b343223dcae485b35af2792c7dd91f92b(qe564274d6cab7b91a3393ef092dac78f)
ReDim b41d8f2e79c0e09113beb7629aa0d8c48(qe564274d6cab7b91a3393ef092dac78f)
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0 To qe564274d6cab7b91a3393ef092dac78f
b41d8f2e79c0e09113beb7629aa0d8c48(bec732ae8e18b7b2ff2e9ccd058f3e8fc) = bec732ae8e18b7b2ff2e9ccd058f3e8fc
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = LBound(b41d8f2e79c0e09113beb7629aa0d8c48) To UBound(b41d8f2e79c0e09113beb7629aa0d8c48)
b42a57d0c121b9fe34a74143aa279157c = CLng(((UBound(b41d8f2e79c0e09113beb7629aa0d8c48) - bec732ae8e18b7b2ff2e9ccd058f3e8fc) * Rnd) + bec732ae8e18b7b2ff2e9ccd058f3e8fc)
If bec732ae8e18b7b2ff2e9ccd058f3e8fc <> b42a57d0c121b9fe34a74143aa279157c Then
e0d4cf763c9da42470a729a29b30d7d50 = b41d8f2e79c0e09113beb7629aa0d8c48(bec732ae8e18b7b2ff2e9ccd058f3e8fc)
b41d8f2e79c0e09113beb7629aa0d8c48(bec732ae8e18b7b2ff2e9ccd058f3e8fc) = b41d8f2e79c0e09113beb7629aa0d8c48(b42a57d0c121b9fe34a74143aa279157c)
b41d8f2e79c0e09113beb7629aa0d8c48(b42a57d0c121b9fe34a74143aa279157c) = e0d4cf763c9da42470a729a29b30d7d50
End If
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0 To qe564274d6cab7b91a3393ef092dac78f
b343223dcae485b35af2792c7dd91f92b(bec732ae8e18b7b2ff2e9ccd058f3e8fc) = Mid(z4891679d877f1da36647b21d6197fbfd, bec732ae8e18b7b2ff2e9ccd058f3e8fc + 1, 1)
m06993036154505accc9ce092bdb57b17(bec732ae8e18b7b2ff2e9ccd058f3e8fc) = bec732ae8e18b7b2ff2e9ccd058f3e8fc - (qe564274d6cab7b91a3393ef092dac78f + 1)
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0 To qe564274d6cab7b91a3393ef092dac78f
b8026f9f8f7fe86372be0799d8c9c6691(Application.Match(bec732ae8e18b7b2ff2e9ccd058f3e8fc, b41d8f2e79c0e09113beb7629aa0d8c48, False) - 1) = AscB(b343223dcae485b35af2792c7dd91f92b(bec732ae8e18b7b2ff2e9ccd058f3e8fc)) - m06993036154505accc9ce092bdb57b17(bec732ae8e18b7b2ff2e9ccd058f3e8fc)
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
CreateCodeFromPassword = ""
For bec732ae8e18b7b2ff2e9ccd058f3e8fc = 0 To qe564274d6cab7b91a3393ef092dac78f
CreateCodeFromPassword = CreateCodeFromPassword & Format(b41d8f2e79c0e09113beb7629aa0d8c48(bec732ae8e18b7b2ff2e9ccd058f3e8fc), b330c8da5472f3c36b801671ef5a54797) & Format(b8026f9f8f7fe86372be0799d8c9c6691(bec732ae8e18b7b2ff2e9ccd058f3e8fc), "000")
Next bec732ae8e18b7b2ff2e9ccd058f3e8fc
CreateCodeFromPassword = CreateCodeFromPassword & t5f443e88a552a3f943275f985dde03ca
End Function

Answer 3

从评论中总结出有用的信息：

如果您的代码可以访问密码（甚至直接或通过混淆），那么任何有权访问该代码的人也都可以访问该密码
Excel VBA的密码保护非常薄弱，破解它很简单

结论：无法在Excel VBA中安全存储密码

在VBA项目中安全地存储密码

3 个答案: