Can someone explain how this Excel 2010 or earlier password protector works
Can someone please explain how this works, why do i need counts of 65 to 66, where are there 12 counts and what do all the Chr(variable) give?
I'm not trying to code this, but i did find this on the web when someone asked how to unlock their sheet after they forgot their password. I would like to understand what is actually happening
Sub PasswordBreaker()
Dim i As Integer, j As Integer, k As Integer
Dim l As Integer, m As Integer, n As Integer
Dim i1 As Integer, i2 As Integer, i3 As Integer
Dim i4 As Integer, i5 As Integer, i6 As Integer
On Error Resume Next
For i = 65 To 66: For j = 65 To 66: For k = 65 To 66
For l = 65 To 66: For m = 65 To 66: For i1 = 65 To 66
For i2 = 65 To 66: For i3 = 65 To 66: For i4 = 65 To 66
For i5 = 65 To 66: For i6 = 65 To 66: For n = 32 To 126
ActiveSheet.Unprotect Chr(i) & Chr(j) & Chr(k) & _
Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & Chr(i3) & _
Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
If ActiveSheet.ProtectContents = False Then
MsgBox "One usable password is " & Chr(i) & Chr(j) & _
Chr(k) & Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & _
Chr(i3) & Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
Exit Sub
End If
Next: Next: Next: Next: Next: Next
Next: Next: Next: Next: Next: Next
End Sub
Thank You
excel vba excel-vba
asked Nov 21 at 11:10
Mike Hill
488
add a comment |
Can someone please explain how this works, why do i need counts of 65 to 66, where are there 12 counts and what do all the Chr(variable) give?
I'm not trying to code this, but i did find this on the web when someone asked how to unlock their sheet after they forgot their password. I would like to understand what is actually happening
Sub PasswordBreaker()
Dim i As Integer, j As Integer, k As Integer
Dim l As Integer, m As Integer, n As Integer
Dim i1 As Integer, i2 As Integer, i3 As Integer
Dim i4 As Integer, i5 As Integer, i6 As Integer
On Error Resume Next
For i = 65 To 66: For j = 65 To 66: For k = 65 To 66
For l = 65 To 66: For m = 65 To 66: For i1 = 65 To 66
For i2 = 65 To 66: For i3 = 65 To 66: For i4 = 65 To 66
For i5 = 65 To 66: For i6 = 65 To 66: For n = 32 To 126
ActiveSheet.Unprotect Chr(i) & Chr(j) & Chr(k) & _
Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & Chr(i3) & _
Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
If ActiveSheet.ProtectContents = False Then
MsgBox "One usable password is " & Chr(i) & Chr(j) & _
Chr(k) & Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & _
Chr(i3) & Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
Exit Sub
End If
Next: Next: Next: Next: Next: Next
Next: Next: Next: Next: Next: Next
End Sub
Thank You
excel vba excel-vba
asked Nov 21 at 11:10
Mike Hill
488
add a comment |
Can someone please explain how this works, why do i need counts of 65 to 66, where are there 12 counts and what do all the Chr(variable) give?
I'm not trying to code this, but i did find this on the web when someone asked how to unlock their sheet after they forgot their password. I would like to understand what is actually happening
Sub PasswordBreaker()
Dim i As Integer, j As Integer, k As Integer
Dim l As Integer, m As Integer, n As Integer
Dim i1 As Integer, i2 As Integer, i3 As Integer
Dim i4 As Integer, i5 As Integer, i6 As Integer
On Error Resume Next
For i = 65 To 66: For j = 65 To 66: For k = 65 To 66
For l = 65 To 66: For m = 65 To 66: For i1 = 65 To 66
For i2 = 65 To 66: For i3 = 65 To 66: For i4 = 65 To 66
For i5 = 65 To 66: For i6 = 65 To 66: For n = 32 To 126
ActiveSheet.Unprotect Chr(i) & Chr(j) & Chr(k) & _
Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & Chr(i3) & _
Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
If ActiveSheet.ProtectContents = False Then
MsgBox "One usable password is " & Chr(i) & Chr(j) & _
Chr(k) & Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & _
Chr(i3) & Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
Exit Sub
End If
Next: Next: Next: Next: Next: Next
Next: Next: Next: Next: Next: Next
End Sub
Thank You
excel vba excel-vba
asked Nov 21 at 11:10
Mike Hill
488
Can someone please explain how this works, why do i need counts of 65 to 66, where are there 12 counts and what do all the Chr(variable) give?
I'm not trying to code this, but i did find this on the web when someone asked how to unlock their sheet after they forgot their password. I would like to understand what is actually happening
Sub PasswordBreaker()
Dim i As Integer, j As Integer, k As Integer
Dim l As Integer, m As Integer, n As Integer
Dim i1 As Integer, i2 As Integer, i3 As Integer
Dim i4 As Integer, i5 As Integer, i6 As Integer
On Error Resume Next
For i = 65 To 66: For j = 65 To 66: For k = 65 To 66
For l = 65 To 66: For m = 65 To 66: For i1 = 65 To 66
For i2 = 65 To 66: For i3 = 65 To 66: For i4 = 65 To 66
For i5 = 65 To 66: For i6 = 65 To 66: For n = 32 To 126
ActiveSheet.Unprotect Chr(i) & Chr(j) & Chr(k) & _
Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & Chr(i3) & _
Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
If ActiveSheet.ProtectContents = False Then
MsgBox "One usable password is " & Chr(i) & Chr(j) & _
Chr(k) & Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & _
Chr(i3) & Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
Exit Sub
End If
Next: Next: Next: Next: Next: Next
Next: Next: Next: Next: Next: Next
End Sub
Thank You
excel vba excel-vba
excel vba excel-vba
asked Nov 21 at 11:10
Mike Hill
488
asked Nov 21 at 11:10
Mike Hill
488
asked Nov 21 at 11:10
Mike Hill
488
asked Nov 21 at 11:10
Mike Hill
488
asked Nov 21 at 11:10
Mike Hill
488
488
add a comment |
add a comment |
1 Answer
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This will generate multiple 12 character long passwords like AAAAAAAAAAA or AAAAAAAAAABA and so on and try if they work to unprotect the worksheet.
The Chr function Chr(65) gives you the character with the ASCII code 65 which is an A. So actually the loops produce all permutations of A and B for the first 11 characters and the last loop has some more characters from ASCII 32 to 126 for the 12ᵗʰ character of the password.
Because the old Excel version uses a not secure hash function to store the password there is more than one password to produce the same hash. So that's just a cheap way to find one of the passwords that work.
For more detailed info see: Understanding Excel's Password Security Methodology
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
add a comment |
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1 Answer
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oldest
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1 Answer
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active
oldest
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oldest
votes
active
oldest
votes
This will generate multiple 12 character long passwords like AAAAAAAAAAA or AAAAAAAAAABA and so on and try if they work to unprotect the worksheet.
The Chr function Chr(65) gives you the character with the ASCII code 65 which is an A. So actually the loops produce all permutations of A and B for the first 11 characters and the last loop has some more characters from ASCII 32 to 126 for the 12ᵗʰ character of the password.
Because the old Excel version uses a not secure hash function to store the password there is more than one password to produce the same hash. So that's just a cheap way to find one of the passwords that work.
For more detailed info see: Understanding Excel's Password Security Methodology
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
add a comment |
This will generate multiple 12 character long passwords like AAAAAAAAAAA or AAAAAAAAAABA and so on and try if they work to unprotect the worksheet.
The Chr function Chr(65) gives you the character with the ASCII code 65 which is an A. So actually the loops produce all permutations of A and B for the first 11 characters and the last loop has some more characters from ASCII 32 to 126 for the 12ᵗʰ character of the password.
Because the old Excel version uses a not secure hash function to store the password there is more than one password to produce the same hash. So that's just a cheap way to find one of the passwords that work.
For more detailed info see: Understanding Excel's Password Security Methodology
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
add a comment |
This will generate multiple 12 character long passwords like AAAAAAAAAAA or AAAAAAAAAABA and so on and try if they work to unprotect the worksheet.
The Chr function Chr(65) gives you the character with the ASCII code 65 which is an A. So actually the loops produce all permutations of A and B for the first 11 characters and the last loop has some more characters from ASCII 32 to 126 for the 12ᵗʰ character of the password.
Because the old Excel version uses a not secure hash function to store the password there is more than one password to produce the same hash. So that's just a cheap way to find one of the passwords that work.
For more detailed info see: Understanding Excel's Password Security Methodology
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
This will generate multiple 12 character long passwords like AAAAAAAAAAA or AAAAAAAAAABA and so on and try if they work to unprotect the worksheet.
The Chr function Chr(65) gives you the character with the ASCII code 65 which is an A. So actually the loops produce all permutations of A and B for the first 11 characters and the last loop has some more characters from ASCII 32 to 126 for the 12ᵗʰ character of the password.
Because the old Excel version uses a not secure hash function to store the password there is more than one password to produce the same hash. So that's just a cheap way to find one of the passwords that work.
For more detailed info see: Understanding Excel's Password Security Methodology
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
edited Nov 21 at 12:34
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
answered Nov 21 at 11:19
Pᴇʜ
20.2k42650
20.2k42650
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
add a comment |
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
Thank you for your explanation, really helped
– Mike Hill
22 hours ago
add a comment |
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