Convert Base 36 to base 10

I have been given a spreadsheet with transaction numbers converted into base 
36- alpha numeric - I need it in base 10- number format- I have approx 30,000 
of these!- is ther a formula to convert from 1 to another?
-- 
Thanks for your help
0
Utf
2/8/2010 5:10:01 PM
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Are you sure your numbers are Base 36? I ask because I kind of suspect your 
"digits" are these...

0, 1, 2, ...., 9, A, B, ..., X, Y, Z

and if that is the case, then you actually have Base 37 numbers and not Base 
36. For Base36 numbers, the letter Z would not be in your set of digits... Z 
would be the 37th digit because 0 is the first digit. Assuming you really 
have Base36 numbers (no Z), then this function should do what you want...

Function ConvertBase36ToBase10(Base36Number As String) As Long
  Dim X As Long, Total As Long, Digit As String
  For X = Len(Base36Number) To 1 Step -1
    Digit = UCase(Mid(Base36Number, X, 1))
    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
  Next
End Function

If Z is in your set (meaning you have Base37 numbers), then simply change 
all the 36's to 37's.

-- 
Rick (MVP - Excel)


"David" <David@discussions.microsoft.com> wrote in message 
news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
>I have been given a spreadsheet with transaction numbers converted into 
>base
> 36- alpha numeric - I need it in base 10- number format- I have approx 
> 30,000
> of these!- is ther a formula to convert from 1 to another?
> -- 
> Thanks for your help 

0
Rick
2/8/2010 6:36:04 PM
"Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote:
> Are you sure your numbers are Base 36? I ask because I kind of suspect
> your "digits" are these...
> 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
> and if that is the case, then you actually have Base 37

No, that's base 36.  A thru Z represent the 26 values 10 thru 35.  Google 
"base 36" or see http://en.wikipedia.org/wiki/Base_36 .


------ original message -----

"Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
> Are you sure your numbers are Base 36? I ask because I kind of suspect 
> your "digits" are these...
>
> 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
>
> and if that is the case, then you actually have Base 37 numbers and not 
> Base 36. For Base36 numbers, the letter Z would not be in your set of 
> digits... Z would be the 37th digit because 0 is the first digit. Assuming 
> you really have Base36 numbers (no Z), then this function should do what 
> you want...
>
> Function ConvertBase36ToBase10(Base36Number As String) As Long
>  Dim X As Long, Total As Long, Digit As String
>  For X = Len(Base36Number) To 1 Step -1
>    Digit = UCase(Mid(Base36Number, X, 1))
>    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
>                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
>  Next
> End Function
>
> If Z is in your set (meaning you have Base37 numbers), then simply change 
> all the 36's to 37's.
>
> -- 
> Rick (MVP - Excel)
>
>
> "David" <David@discussions.microsoft.com> wrote in message 
> news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
>>I have been given a spreadsheet with transaction numbers converted into 
>>base
>> 36- alpha numeric - I need it in base 10- number format- I have approx 
>> 30,000
>> of these!- is ther a formula to convert from 1 to another?
>> -- 
>> Thanks for your help
> 

0
Joe
2/8/2010 7:49:54 PM
On Mon, 8 Feb 2010 13:36:04 -0500, "Rick Rothstein"
<rick.newsNO.SPAM@NO.SPAMverizon.net> wrote:

>Are you sure your numbers are Base 36? I ask because I kind of suspect your 
>"digits" are these...
>
>0, 1, 2, ...., 9, A, B, ..., X, Y, Z
>
>and if that is the case, then you actually have Base 37 numbers and not Base 
>36. For Base36 numbers, the letter Z would not be in your set of digits... Z 
>would be the 37th digit because 0 is the first digit. Assuming you really 
>have Base36 numbers (no Z), then this function should do what you want...

Are you sure about that Rick?

It seems to me that 10 digits (0-9) + 26 [A-Z] letters --> Base 36

--ron
0
Ron
2/8/2010 7:59:40 PM
Joe, Ron... yes, I screwed that up... thanks for point it out to me.

David... this function will do what you want...

Function ConvertBase36ToBase10(Base36Number As String) As Long
  Dim X As Long, Total As Long, Digit As String
  For X = Len(Base36Number) To 1 Step -1
    Digit = UCase(Mid(Base36Number, X, 1))
    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
              Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
  Next
End Function

-- 
Rick (MVP - Excel)


"Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
> Are you sure your numbers are Base 36? I ask because I kind of suspect 
> your "digits" are these...
>
> 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
>
> and if that is the case, then you actually have Base 37 numbers and not 
> Base 36. For Base36 numbers, the letter Z would not be in your set of 
> digits... Z would be the 37th digit because 0 is the first digit. Assuming 
> you really have Base36 numbers (no Z), then this function should do what 
> you want...
>
> Function ConvertBase36ToBase10(Base36Number As String) As Long
>  Dim X As Long, Total As Long, Digit As String
>  For X = Len(Base36Number) To 1 Step -1
>    Digit = UCase(Mid(Base36Number, X, 1))
>    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
>                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
>  Next
> End Function
>
> If Z is in your set (meaning you have Base37 numbers), then simply change 
> all the 36's to 37's.
>
> -- 
> Rick (MVP - Excel)
>
>
> "David" <David@discussions.microsoft.com> wrote in message 
> news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
>>I have been given a spreadsheet with transaction numbers converted into 
>>base
>> 36- alpha numeric - I need it in base 10- number format- I have approx 
>> 30,000
>> of these!- is ther a formula to convert from 1 to another?
>> -- 
>> Thanks for your help
> 

0
Rick
2/9/2010 4:11:02 AM
Guys thanks for this- 

I have tried the code what I get is for a sample code
O81D8KEURD94I = #value
but
081d8ke = 486026654

Is there any length critera in the function- couldn't spot any

-- 
Thanks for your help


"Rick Rothstein" wrote:

> Joe, Ron... yes, I screwed that up... thanks for point it out to me.
> 
> David... this function will do what you want...
> 
> Function ConvertBase36ToBase10(Base36Number As String) As Long
>   Dim X As Long, Total As Long, Digit As String
>   For X = Len(Base36Number) To 1 Step -1
>     Digit = UCase(Mid(Base36Number, X, 1))
>     ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
>               Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
>   Next
> End Function
> 
> -- 
> Rick (MVP - Excel)
> 
> 
> "Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
> news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
> > Are you sure your numbers are Base 36? I ask because I kind of suspect 
> > your "digits" are these...
> >
> > 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
> >
> > and if that is the case, then you actually have Base 37 numbers and not 
> > Base 36. For Base36 numbers, the letter Z would not be in your set of 
> > digits... Z would be the 37th digit because 0 is the first digit. Assuming 
> > you really have Base36 numbers (no Z), then this function should do what 
> > you want...
> >
> > Function ConvertBase36ToBase10(Base36Number As String) As Long
> >  Dim X As Long, Total As Long, Digit As String
> >  For X = Len(Base36Number) To 1 Step -1
> >    Digit = UCase(Mid(Base36Number, X, 1))
> >    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> >                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
> >  Next
> > End Function
> >
> > If Z is in your set (meaning you have Base37 numbers), then simply change 
> > all the 36's to 37's.
> >
> > -- 
> > Rick (MVP - Excel)
> >
> >
> > "David" <David@discussions.microsoft.com> wrote in message 
> > news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
> >>I have been given a spreadsheet with transaction numbers converted into 
> >>base
> >> 36- alpha numeric - I need it in base 10- number format- I have approx 
> >> 30,000
> >> of these!- is ther a formula to convert from 1 to another?
> >> -- 
> >> Thanks for your help
> > 
> 
> .
> 
0
Utf
2/9/2010 10:18:03 AM
David wrote:
> Guys thanks for this- 
> 
> I have tried the code what I get is for a sample code
> O81D8KEURD94I = #value
> but
> 081d8ke = 486026654
> 
> Is there any length critera in the function- couldn't spot any
> 

Yes. Implicit in the declaration of Long values -2^31 < x <= 2^31-1

You have to explicitly implement some form of long integer arithmetic to 
handle values which go outside this boundary.

Decimal 2146483647 is the largest Long value which in Base36 = ZIK0ZJ
(subject to typos)

You could cut the string into two parts and pray that the leading digit 
is always zero. Unsigned integers can handle 6 base36 digits OK.

The mantissa of Double precision reals would let you do up to 9 digits 
of Base36.

Regards,
Martin Brown
0
Martin
2/9/2010 12:47:34 PM
"David" wrote: 
> I have tried the code what I get is for a sample code
> O81D8KEURD94I = #value

If you had tried my HexTri2Dec function, you wouldn't have gotten that 
problem.


----- original message ------

"David" wrote:
> Guys thanks for this- 
> 
> I have tried the code what I get is for a sample code
> O81D8KEURD94I = #value
> but
> 081d8ke = 486026654
> 
> Is there any length critera in the function- couldn't spot any
> 
> -- 
> Thanks for your help
> 
> 
> "Rick Rothstein" wrote:
> 
> > Joe, Ron... yes, I screwed that up... thanks for point it out to me.
> > 
> > David... this function will do what you want...
> > 
> > Function ConvertBase36ToBase10(Base36Number As String) As Long
> >   Dim X As Long, Total As Long, Digit As String
> >   For X = Len(Base36Number) To 1 Step -1
> >     Digit = UCase(Mid(Base36Number, X, 1))
> >     ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> >               Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
> >   Next
> > End Function
> > 
> > -- 
> > Rick (MVP - Excel)
> > 
> > 
> > "Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
> > news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
> > > Are you sure your numbers are Base 36? I ask because I kind of suspect 
> > > your "digits" are these...
> > >
> > > 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
> > >
> > > and if that is the case, then you actually have Base 37 numbers and not 
> > > Base 36. For Base36 numbers, the letter Z would not be in your set of 
> > > digits... Z would be the 37th digit because 0 is the first digit. Assuming 
> > > you really have Base36 numbers (no Z), then this function should do what 
> > > you want...
> > >
> > > Function ConvertBase36ToBase10(Base36Number As String) As Long
> > >  Dim X As Long, Total As Long, Digit As String
> > >  For X = Len(Base36Number) To 1 Step -1
> > >    Digit = UCase(Mid(Base36Number, X, 1))
> > >    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> > >                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
> > >  Next
> > > End Function
> > >
> > > If Z is in your set (meaning you have Base37 numbers), then simply change 
> > > all the 36's to 37's.
> > >
> > > -- 
> > > Rick (MVP - Excel)
> > >
> > >
> > > "David" <David@discussions.microsoft.com> wrote in message 
> > > news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
> > >>I have been given a spreadsheet with transaction numbers converted into 
> > >>base
> > >> 36- alpha numeric - I need it in base 10- number format- I have approx 
> > >> 30,000
> > >> of these!- is ther a formula to convert from 1 to another?
> > >> -- 
> > >> Thanks for your help
> > > 
> > 
> > .
> > 
0
Utf
2/9/2010 1:06:05 PM
On Tue, 9 Feb 2010 02:18:03 -0800, David <David@discussions.microsoft.com>
wrote:

>Guys thanks for this- 
>
>I have tried the code what I get is for a sample code
>O81D8KEURD94I = #value
>but
>081d8ke = 486026654
>
>Is there any length critera in the function- couldn't spot any

You get a VALUE error because Rick Dim'd is variables as Longs, and your first
entry overflows that.

If you change it to Double, it should work OK:

===================
Function ConvertBase36ToBase10(Base36Number As String) As Double
  Dim X As Long, Total As Double, Digit As String
  For X = Len(Base36Number) To 1 Step -1
    Digit = UCase(Mid(Base36Number, X, 1))
    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
              Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
  Next
End Function
=====================

Of course, Excel is limited to 15 digit precision.  You can get increased
precision in VBA by using the Decimal data type, but the only way to get that
into a worksheet cell would be with a string output.
--ron
0
Ron
2/9/2010 1:12:40 PM
I wrote:
> "David" wrote: 
> > I have tried the code what I get is for a sample code
> > O81D8KEURD94I = #value
> 
> If you had tried my HexTri2Dec function, you wouldn't
> have gotten that problem.

Although my function would do the best we can in converting O81D8KEURD94I, I 
should point that the base10 equivalent is about 1.14778E+20.  Since that is 
more than 15 digits, it cannot be represented exactly as an Excel number.

Since these are transaction ids, not numbers to be used in arithmetic, it 
would be better to use a UDF that returns the exact conversion as text.

Caveat:  Someone might suggest using VBA type Decimal instead of Double.  
That would indeed work for this example.  However, it is not a general 
solution, being limited to 28-digit integers (and some 29-digit integers).

Nevertheless, below is my UDF with that modification.   For your example, 
the result is the string 114779126356831142514.

Note:  This implementation allows only integer base36 numbers.

UDF....


Option Explicit

Function HexTri2Dec(s As String)
Dim c As String * 1, bNeg As Boolean
Dim i As Long, x As Long, d
s = Trim(s)
If Mid(s, 1, 1) = "-" Then
   If Len(s) = 1 Then GoTo badForm
   bNeg = True: i = 2
Else
   bNeg = False: i = 1
End If
c = ""
d = CDec(0)
On Error Resume Next
For i = i To Len(s)
   c = LCase(Mid(s, i, 1))
   If "0" <= c And c <= "9" Then x = Asc(c) - 48 _
   Else If "a" <= c And c <= "z" Then x = Asc(c) - 87 _
   Else: GoTo badForm
   d = d * 36 + x
   If Err.Number <> 0 Then GoTo badNum
Next i

done:
If bNeg Then d = -d
HexTri2Dec = Format(d, "0")    'allow only integers
Exit Function

badNum:
HexTri2Dec = CVErr(xlErrNum)
Exit Function

badForm:
HexTri2Dec = CVErr(xlErrValue)
End Function


----- original message ------

"Joe User" wrote:

> "David" wrote: 
> > I have tried the code what I get is for a sample code
> > O81D8KEURD94I = #value
> 
> If you had tried my HexTri2Dec function, you wouldn't have gotten that 
> problem.
> 
> 
> ----- original message ------
> 
> "David" wrote:
> > Guys thanks for this- 
> > 
> > I have tried the code what I get is for a sample code
> > O81D8KEURD94I = #value
> > but
> > 081d8ke = 486026654
> > 
> > Is there any length critera in the function- couldn't spot any
> > 
> > -- 
> > Thanks for your help
> > 
> > 
> > "Rick Rothstein" wrote:
> > 
> > > Joe, Ron... yes, I screwed that up... thanks for point it out to me.
> > > 
> > > David... this function will do what you want...
> > > 
> > > Function ConvertBase36ToBase10(Base36Number As String) As Long
> > >   Dim X As Long, Total As Long, Digit As String
> > >   For X = Len(Base36Number) To 1 Step -1
> > >     Digit = UCase(Mid(Base36Number, X, 1))
> > >     ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> > >               Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
> > >   Next
> > > End Function
> > > 
> > > -- 
> > > Rick (MVP - Excel)
> > > 
> > > 
> > > "Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
> > > news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
> > > > Are you sure your numbers are Base 36? I ask because I kind of suspect 
> > > > your "digits" are these...
> > > >
> > > > 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
> > > >
> > > > and if that is the case, then you actually have Base 37 numbers and not 
> > > > Base 36. For Base36 numbers, the letter Z would not be in your set of 
> > > > digits... Z would be the 37th digit because 0 is the first digit. Assuming 
> > > > you really have Base36 numbers (no Z), then this function should do what 
> > > > you want...
> > > >
> > > > Function ConvertBase36ToBase10(Base36Number As String) As Long
> > > >  Dim X As Long, Total As Long, Digit As String
> > > >  For X = Len(Base36Number) To 1 Step -1
> > > >    Digit = UCase(Mid(Base36Number, X, 1))
> > > >    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> > > >                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
> > > >  Next
> > > > End Function
> > > >
> > > > If Z is in your set (meaning you have Base37 numbers), then simply change 
> > > > all the 36's to 37's.
> > > >
> > > > -- 
> > > > Rick (MVP - Excel)
> > > >
> > > >
> > > > "David" <David@discussions.microsoft.com> wrote in message 
> > > > news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
> > > >>I have been given a spreadsheet with transaction numbers converted into 
> > > >>base
> > > >> 36- alpha numeric - I need it in base 10- number format- I have approx 
> > > >> 30,000
> > > >> of these!- is ther a formula to convert from 1 to another?
> > > >> -- 
> > > >> Thanks for your help
> > > > 
> > > 
> > > .
> > > 
0
Utf
2/9/2010 1:32:01 PM
Here is the Decimal Data Type version of my function which will handle up to 
a 28-digit Base36 number (max "number" is ZZZZZZZZZZZZZZZZZZ)...

Function ConvertBase36ToBase10(Base36Number As String) As Variant
  Dim x As Long, Digit As String, Power As Variant
  If Len(Base36Number) > 18 Or Base36Number Like "*[!0-9A-Za-z]*" Then
    ConvertBase36ToBase10 = CVErr(xlErrNum)
    Exit Function
  End If
  For x = Len(Base36Number) To 1 Step -1
    Digit = UCase(Mid(Base36Number, x, 1))
    If Len(Base36Number) > 9 Then
      Power = CDec("101559956668416") * (36 ^ (Len(Base36Number) - 9 - x))
    Else
      Power = 36 ^ (Len(Base36Number) - x)
    End If
    ConvertBase36ToBase10 = ConvertBase36ToBase10 + CDec(IIf(IsNumeric( _
                            Digit), Digit, (Asc(Digit) - 55)) * Power)
  Next
End Function

Note that the If..Then handling of the exponent for the 36 base number is 
necessary because raising any number to a power using the caret (^(^(^) 
operator collapses Decimal Data Type values back to Long Data Type values... 
the 101559956668416 value is 36 raised to the 9th power. I also through in 
some error checking as well.

-- 
Rick (MVP - Excel)


"Ron Rosenfeld" <ronrosenfeld@nospam.org> wrote in message 
news:24m2n5t9fnigourdm426m3ue953grv8e2t@4ax.com...
> On Tue, 9 Feb 2010 02:18:03 -0800, David <David@discussions.microsoft.com>
> wrote:
>
>>Guys thanks for this-
>>
>>I have tried the code what I get is for a sample code
>>O81D8KEURD94I = #value
>>but
>>081d8ke = 486026654
>>
>>Is there any length critera in the function- couldn't spot any
>
> You get a VALUE error because Rick Dim'd is variables as Longs, and your 
> first
> entry overflows that.
>
> If you change it to Double, it should work OK:
>
> ===================
> Function ConvertBase36ToBase10(Base36Number As String) As Double
>  Dim X As Long, Total As Double, Digit As String
>  For X = Len(Base36Number) To 1 Step -1
>    Digit = UCase(Mid(Base36Number, X, 1))
>    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
>              Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
>  Next
> End Function
> =====================
>
> Of course, Excel is limited to 15 digit precision.  You can get increased
> precision in VBA by using the Decimal data type, but the only way to get 
> that
> into a worksheet cell would be with a string output.
> --ron 

0
Rick
2/9/2010 3:48:59 PM
Rick Rothstein wrote:
> Here is the Decimal Data Type version of my function which will handle 
> up to a 28-digit Base36 number (max "number" is ZZZZZZZZZZZZZZZZZZ)...
> 
> Function ConvertBase36ToBase10(Base36Number As String) As Variant
>  Dim x As Long, Digit As String, Power As Variant
>  If Len(Base36Number) > 18 Or Base36Number Like "*[!0-9A-Za-z]*" Then
>    ConvertBase36ToBase10 = CVErr(xlErrNum)
>    Exit Function
>  End If
>  For x = Len(Base36Number) To 1 Step -1
>    Digit = UCase(Mid(Base36Number, x, 1))
>    If Len(Base36Number) > 9 Then
>      Power = CDec("101559956668416") * (36 ^ (Len(Base36Number) - 9 - x))
>    Else
>      Power = 36 ^ (Len(Base36Number) - x)
>    End If
>    ConvertBase36ToBase10 = ConvertBase36ToBase10 + CDec(IIf(IsNumeric( _
>                            Digit), Digit, (Asc(Digit) - 55)) * Power)
>  Next
> End Function
> 
> Note that the If..Then handling of the exponent for the 36 base number 
> is necessary because raising any number to a power using the caret 
> (^(^(^) operator collapses Decimal Data Type values back to Long Data 
> Type values... the 101559956668416 value is 36 raised to the 9th power. 
> I also through in some error checking as well.

It may be cleaner to avoid ^ entirely and to do the loop incrementally - 
something along the lines of

ConvertBase36ToBase10 = 0
For x = Len(Base36Number) To 1 Step -1
     Digit = UCase(Mid(Base36Number, x, 1))
     ConvertBase36ToBase10 = ConvertBase36ToBase10*36 + 
CDec(IIf(IsNumeric(Digit), Digit, (Asc(Digit) - 55))
   Next

Regards,
Martin Brown
0
Martin
2/9/2010 3:59:00 PM
Here is a version of my function which will handle up to a 28-digit Base36 
number (max "number" is ZZZZZZZZZZZZZZZZZZ)...

Function ConvertBase36ToBase10(Base36Number As String) As Variant
  Dim x As Long, Digit As String, Power As Variant
  If Len(Base36Number) > 18 Or Base36Number Like "*[!0-9A-Za-z]*" Then
    ConvertBase36ToBase10 = CVErr(xlErrNum)
    Exit Function
  End If
  For x = Len(Base36Number) To 1 Step -1
    Digit = UCase(Mid(Base36Number, x, 1))
    If Len(Base36Number) > 9 Then
      Power = CDec("101559956668416") * (36 ^ (Len(Base36Number) - 9 - x))
    Else
      Power = 36 ^ (Len(Base36Number) - x)
    End If
    ConvertBase36ToBase10 = ConvertBase36ToBase10 + CDec(IIf(IsNumeric( _
                            Digit), Digit, (Asc(Digit) - 55)) * Power)
  Next
End Function

Note that I also added some error checking as well.

-- 
Rick (MVP - Excel)


"David" <David@discussions.microsoft.com> wrote in message 
news:DAEC194E-4145-4AA9-BA5C-50DBC3522CA7@microsoft.com...
> Guys thanks for this-
>
> I have tried the code what I get is for a sample code
> O81D8KEURD94I = #value
> but
> 081d8ke = 486026654
>
> Is there any length critera in the function- couldn't spot any
>
> -- 
> Thanks for your help
>
>
> "Rick Rothstein" wrote:
>
>> Joe, Ron... yes, I screwed that up... thanks for point it out to me.
>>
>> David... this function will do what you want...
>>
>> Function ConvertBase36ToBase10(Base36Number As String) As Long
>>   Dim X As Long, Total As Long, Digit As String
>>   For X = Len(Base36Number) To 1 Step -1
>>     Digit = UCase(Mid(Base36Number, X, 1))
>>     ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), 
>> _
>>               Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
>>   Next
>> End Function
>>
>> -- 
>> Rick (MVP - Excel)
>>
>>
>> "Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message
>> news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
>> > Are you sure your numbers are Base 36? I ask because I kind of suspect
>> > your "digits" are these...
>> >
>> > 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
>> >
>> > and if that is the case, then you actually have Base 37 numbers and not
>> > Base 36. For Base36 numbers, the letter Z would not be in your set of
>> > digits... Z would be the 37th digit because 0 is the first digit. 
>> > Assuming
>> > you really have Base36 numbers (no Z), then this function should do 
>> > what
>> > you want...
>> >
>> > Function ConvertBase36ToBase10(Base36Number As String) As Long
>> >  Dim X As Long, Total As Long, Digit As String
>> >  For X = Len(Base36Number) To 1 Step -1
>> >    Digit = UCase(Mid(Base36Number, X, 1))
>> >    ConvertBase36ToBase10 = ConvertBase36ToBase10 + 
>> > IIf(IsNumeric(Digit), _
>> >                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
>> >  Next
>> > End Function
>> >
>> > If Z is in your set (meaning you have Base37 numbers), then simply 
>> > change
>> > all the 36's to 37's.
>> >
>> > -- 
>> > Rick (MVP - Excel)
>> >
>> >
>> > "David" <David@discussions.microsoft.com> wrote in message
>> > news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
>> >>I have been given a spreadsheet with transaction numbers converted into
>> >>base
>> >> 36- alpha numeric - I need it in base 10- number format- I have approx
>> >> 30,000
>> >> of these!- is ther a formula to convert from 1 to another?
>> >> -- 
>> >> Thanks for your help
>> >
>>
>> .
>> 

0
Rick
2/9/2010 3:59:00 PM
> I also through in some error checking as well.

"through"??? That should have been "threw" instead.

-- 
Rick (MVP - Excel)


"Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
news:OOhXi9ZqKHA.3848@TK2MSFTNGP06.phx.gbl...
> Here is the Decimal Data Type version of my function which will handle up 
> to a 28-digit Base36 number (max "number" is ZZZZZZZZZZZZZZZZZZ)...
>
> Function ConvertBase36ToBase10(Base36Number As String) As Variant
>  Dim x As Long, Digit As String, Power As Variant
>  If Len(Base36Number) > 18 Or Base36Number Like "*[!0-9A-Za-z]*" Then
>    ConvertBase36ToBase10 = CVErr(xlErrNum)
>    Exit Function
>  End If
>  For x = Len(Base36Number) To 1 Step -1
>    Digit = UCase(Mid(Base36Number, x, 1))
>    If Len(Base36Number) > 9 Then
>      Power = CDec("101559956668416") * (36 ^ (Len(Base36Number) - 9 - x))
>    Else
>      Power = 36 ^ (Len(Base36Number) - x)
>    End If
>    ConvertBase36ToBase10 = ConvertBase36ToBase10 + CDec(IIf(IsNumeric( _
>                            Digit), Digit, (Asc(Digit) - 55)) * Power)
>  Next
> End Function
>
> Note that the If..Then handling of the exponent for the 36 base number is 
> necessary because raising any number to a power using the caret (^(^(^) 
> operator collapses Decimal Data Type values back to Long Data Type 
> values... the 101559956668416 value is 36 raised to the 9th power. I also 
> through in some error checking as well.
>
> -- 
> Rick (MVP - Excel)
>
>
> "Ron Rosenfeld" <ronrosenfeld@nospam.org> wrote in message 
> news:24m2n5t9fnigourdm426m3ue953grv8e2t@4ax.com...
>> On Tue, 9 Feb 2010 02:18:03 -0800, David 
>> <David@discussions.microsoft.com>
>> wrote:
>>
>>>Guys thanks for this-
>>>
>>>I have tried the code what I get is for a sample code
>>>O81D8KEURD94I = #value
>>>but
>>>081d8ke = 486026654
>>>
>>>Is there any length critera in the function- couldn't spot any
>>
>> You get a VALUE error because Rick Dim'd is variables as Longs, and your 
>> first
>> entry overflows that.
>>
>> If you change it to Double, it should work OK:
>>
>> ===================
>> Function ConvertBase36ToBase10(Base36Number As String) As Double
>>  Dim X As Long, Total As Double, Digit As String
>>  For X = Len(Base36Number) To 1 Step -1
>>    Digit = UCase(Mid(Base36Number, X, 1))
>>    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), 
>> _
>>              Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
>>  Next
>> End Function
>> =====================
>>
>> Of course, Excel is limited to 15 digit precision.  You can get increased
>> precision in VBA by using the Decimal data type, but the only way to get 
>> that
>> into a worksheet cell would be with a string output.
>> --ron
> 

0
Rick
2/9/2010 3:59:55 PM
Joe / Ron 
Thanks very much for all that- works perfectly
-- 



"Joe User" wrote:

> I wrote:
> > "David" wrote: 
> > > I have tried the code what I get is for a sample code
> > > O81D8KEURD94I = #value
> > 
> > If you had tried my HexTri2Dec function, you wouldn't
> > have gotten that problem.
> 
> Although my function would do the best we can in converting O81D8KEURD94I, I 
> should point that the base10 equivalent is about 1.14778E+20.  Since that is 
> more than 15 digits, it cannot be represented exactly as an Excel number.
> 
> Since these are transaction ids, not numbers to be used in arithmetic, it 
> would be better to use a UDF that returns the exact conversion as text.
> 
> Caveat:  Someone might suggest using VBA type Decimal instead of Double.  
> That would indeed work for this example.  However, it is not a general 
> solution, being limited to 28-digit integers (and some 29-digit integers).
> 
> Nevertheless, below is my UDF with that modification.   For your example, 
> the result is the string 114779126356831142514.
> 
> Note:  This implementation allows only integer base36 numbers.
> 
> UDF....
> 
> 
> Option Explicit
> 
> Function HexTri2Dec(s As String)
> Dim c As String * 1, bNeg As Boolean
> Dim i As Long, x As Long, d
> s = Trim(s)
> If Mid(s, 1, 1) = "-" Then
>    If Len(s) = 1 Then GoTo badForm
>    bNeg = True: i = 2
> Else
>    bNeg = False: i = 1
> End If
> c = ""
> d = CDec(0)
> On Error Resume Next
> For i = i To Len(s)
>    c = LCase(Mid(s, i, 1))
>    If "0" <= c And c <= "9" Then x = Asc(c) - 48 _
>    Else If "a" <= c And c <= "z" Then x = Asc(c) - 87 _
>    Else: GoTo badForm
>    d = d * 36 + x
>    If Err.Number <> 0 Then GoTo badNum
> Next i
> 
> done:
> If bNeg Then d = -d
> HexTri2Dec = Format(d, "0")    'allow only integers
> Exit Function
> 
> badNum:
> HexTri2Dec = CVErr(xlErrNum)
> Exit Function
> 
> badForm:
> HexTri2Dec = CVErr(xlErrValue)
> End Function
> 
> 
> ----- original message ------
> 
> "Joe User" wrote:
> 
> > "David" wrote: 
> > > I have tried the code what I get is for a sample code
> > > O81D8KEURD94I = #value
> > 
> > If you had tried my HexTri2Dec function, you wouldn't have gotten that 
> > problem.
> > 
> > 
> > ----- original message ------
> > 
> > "David" wrote:
> > > Guys thanks for this- 
> > > 
> > > I have tried the code what I get is for a sample code
> > > O81D8KEURD94I = #value
> > > but
> > > 081d8ke = 486026654
> > > 
> > > Is there any length critera in the function- couldn't spot any
> > > 
> > > -- 
> > > Thanks for your help
> > > 
> > > 
> > > "Rick Rothstein" wrote:
> > > 
> > > > Joe, Ron... yes, I screwed that up... thanks for point it out to me.
> > > > 
> > > > David... this function will do what you want...
> > > > 
> > > > Function ConvertBase36ToBase10(Base36Number As String) As Long
> > > >   Dim X As Long, Total As Long, Digit As String
> > > >   For X = Len(Base36Number) To 1 Step -1
> > > >     Digit = UCase(Mid(Base36Number, X, 1))
> > > >     ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> > > >               Digit, (Asc(Digit) - 55)) * (36 ^ (Len(Base36Number) - X))
> > > >   Next
> > > > End Function
> > > > 
> > > > -- 
> > > > Rick (MVP - Excel)
> > > > 
> > > > 
> > > > "Rick Rothstein" <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote in message 
> > > > news:uWTGQ2OqKHA.3848@TK2MSFTNGP06.phx.gbl...
> > > > > Are you sure your numbers are Base 36? I ask because I kind of suspect 
> > > > > your "digits" are these...
> > > > >
> > > > > 0, 1, 2, ...., 9, A, B, ..., X, Y, Z
> > > > >
> > > > > and if that is the case, then you actually have Base 37 numbers and not 
> > > > > Base 36. For Base36 numbers, the letter Z would not be in your set of 
> > > > > digits... Z would be the 37th digit because 0 is the first digit. Assuming 
> > > > > you really have Base36 numbers (no Z), then this function should do what 
> > > > > you want...
> > > > >
> > > > > Function ConvertBase36ToBase10(Base36Number As String) As Long
> > > > >  Dim X As Long, Total As Long, Digit As String
> > > > >  For X = Len(Base36Number) To 1 Step -1
> > > > >    Digit = UCase(Mid(Base36Number, X, 1))
> > > > >    ConvertBase36ToBase10 = ConvertBase36ToBase10 + IIf(IsNumeric(Digit), _
> > > > >                  Digit, Asc(Digit) - 54) * 36 ^ (Len(Base36Number) - X)
> > > > >  Next
> > > > > End Function
> > > > >
> > > > > If Z is in your set (meaning you have Base37 numbers), then simply change 
> > > > > all the 36's to 37's.
> > > > >
> > > > > -- 
> > > > > Rick (MVP - Excel)
> > > > >
> > > > >
> > > > > "David" <David@discussions.microsoft.com> wrote in message 
> > > > > news:23328038-FEB3-461D-A3C6-F0E57A848C33@microsoft.com...
> > > > >>I have been given a spreadsheet with transaction numbers converted into 
> > > > >>base
> > > > >> 36- alpha numeric - I need it in base 10- number format- I have approx 
> > > > >> 30,000
> > > > >> of these!- is ther a formula to convert from 1 to another?
> > > > >> -- 
> > > > >> Thanks for your help
> > > > > 
> > > > 
> > > > .
> > > > 
0
Utf
2/9/2010 4:20:15 PM
Reply:

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