Hello
Is their a macro around that well do roto13 code?
thanks
Basil

There is a nice one here
http://www.erlandsendata.no/english/...ndownloadtools
The data form is in Norwigian so you will needed to get a few words
translated.
The macro is in English
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
www.stfx.ca/people/bliengme
remove caps from email

"basil" wrote in message
...
Hello
Is their a macro around that well do roto13 code?
thanks
Basil

PS: if you Google for this use ROT13 not ROTO13
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
www.stfx.ca/people/bliengme
remove caps from email

"basil" wrote in message
...
Hello
Is their a macro around that well do roto13 code?
thanks
Basil

It's pretty simple


Public Function encrypt(pInput As String)
Dim n As Integer
Dim i As Integer

n = 13
For i = 1 To Len(pInput)
Mid(pInput, i, 1) = Chr(Asc(Mid(pInput, i, 1)) + n)
Next i
encrypt = pInput

End Function

Public Function decrypt(pInput As String)
Dim n As Integer
Dim i As Integer

n = 13
For i = 1 To Len(pInput)
Mid(pInput, i, 1) = Chr(Asc(Mid(pInput, i, 1)) - n)
Next i
decrypt = pInput

End Function



--
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my addy)

"basil" wrote in message
...
Hello
Is their a macro around that well do roto13 code?
thanks
Basil

Thanks Bernard
I'll check this out.
"Bernard Liengme" wrote in message
...
PS: if you Google for this use ROT13 not ROTO13
best wishes
--
Bernard V Liengme
Microsoft Excel MVP
www.stfx.ca/people/bliengme
remove caps from email

"basil" wrote in message
...
Hello
Is their a macro around that well do roto13 code?
thanks
Basil

Thank you Bob
I will try this also
Basil
"Bob Phillips" wrote in message
...
It's pretty simple


Public Function encrypt(pInput As String)
Dim n As Integer
Dim i As Integer

n = 13
For i = 1 To Len(pInput)
Mid(pInput, i, 1) = Chr(Asc(Mid(pInput, i, 1)) + n)
Next i
encrypt = pInput

End Function

Public Function decrypt(pInput As String)
Dim n As Integer
Dim i As Integer

n = 13
For i = 1 To Len(pInput)
Mid(pInput, i, 1) = Chr(Asc(Mid(pInput, i, 1)) - n)
Next i
decrypt = pInput

End Function



--
HTH

Bob

(there's no email, no snail mail, but somewhere should be gmail in my
addy)

"basil" wrote in message
...
Hello
Is their a macro around that well do roto13 code?
thanks
Basil

It may also be ROT135. In other words, letters ("A".."Z" and "a".."z") are
rotated 13 character positions, and digits are rotated 5 positions.

Digit Becomes
1 6
2 7

etc.

I have a template that I use to do this that has the following function:

'----------------------------------------------------------------------
'ROT135 encodes/decodes text that has been encoded using the ROT135
algorithm.
'Letters "A" thru "M" rotate down the alphabet to become "N" thru "Z" (13
places).
'Letters "N" thru "Z" rotate back around (wrap around) to "A" thru "M".
'Digits rotate 5 places, so that "0" thru "4" become "5" thru "9",
'and digits "5" thru "9" become "0" thru "4".
'If the input argument Text is a number, the encoded/decoded version
'is returned as a string. Use the VALUE worksheet function on the return
'value of ROT135 if a numerical result is needed.

Public Function ROT135(Text As String) As String
Dim strBuffer As String
Dim lngTextLength As Long
Dim lngIndex As Long
Dim strCharacter As String

lngTextLength = Len(Text)
strBuffer = ""

If lngTextLength 0 _
Then
'Check and convert each character.
For lngIndex = 1 To lngTextLength
strCharacter = Mid$(Text, lngIndex, 1)

Select Case strCharacter
'Rotate letters 13 places.
Case "A" To "M", "a" To "m"
strBuffer = strBuffer & Chr$(Asc(strCharacter) + 13)
Case "N" To "Z", "n" To "z"
strBuffer = strBuffer & Chr$(Asc(strCharacter) - 13)

'Rotate digits 5 places.
Case "0" To "4"
strBuffer = strBuffer & Chr$(Asc(strCharacter) + 5)
Case "5" To "9"
strBuffer = strBuffer & Chr$(Asc(strCharacter) - 5)

'Character is a punctuation symbol; skip the conversion.
Case Else
strBuffer = strBuffer & strCharacter
End Select
Next lngIndex
End If

ROT135 = strBuffer
End Function

--
Regards,
Bill Renaud

Thanks Bill
I will also try this out
Basil
"Bill Renaud" wrote in message
. ..
It may also be ROT135. In other words, letters ("A".."Z" and "a".."z") are
rotated 13 character positions, and digits are rotated 5 positions.

Digit Becomes
1 6
2 7

etc.

I have a template that I use to do this that has the following function:

'----------------------------------------------------------------------
'ROT135 encodes/decodes text that has been encoded using the ROT135
algorithm.
'Letters "A" thru "M" rotate down the alphabet to become "N" thru "Z" (13
places).
'Letters "N" thru "Z" rotate back around (wrap around) to "A" thru "M".
'Digits rotate 5 places, so that "0" thru "4" become "5" thru "9",
'and digits "5" thru "9" become "0" thru "4".
'If the input argument Text is a number, the encoded/decoded version
'is returned as a string. Use the VALUE worksheet function on the return
'value of ROT135 if a numerical result is needed.

Public Function ROT135(Text As String) As String
Dim strBuffer As String
Dim lngTextLength As Long
Dim lngIndex As Long
Dim strCharacter As String

lngTextLength = Len(Text)
strBuffer = ""

If lngTextLength 0 _
Then
'Check and convert each character.
For lngIndex = 1 To lngTextLength
strCharacter = Mid$(Text, lngIndex, 1)

Select Case strCharacter
'Rotate letters 13 places.
Case "A" To "M", "a" To "m"
strBuffer = strBuffer & Chr$(Asc(strCharacter) + 13)
Case "N" To "Z", "n" To "z"
strBuffer = strBuffer & Chr$(Asc(strCharacter) - 13)

'Rotate digits 5 places.
Case "0" To "4"
strBuffer = strBuffer & Chr$(Asc(strCharacter) + 5)
Case "5" To "9"
strBuffer = strBuffer & Chr$(Asc(strCharacter) - 5)

'Character is a punctuation symbol; skip the conversion.
Case Else
strBuffer = strBuffer & strCharacter
End Select
Next lngIndex
End If

ROT135 = strBuffer
End Function

--
Regards,
Bill Renaud