On Tue, 23 Oct 2012 09:38:19 -0700, Gord Dibben wrote:

Try a UDF

Function RemAlpha(str As String) As String
'Remove Alphas from a string
Dim re As Object
Set re = CreateObject("vbscript.regexp")
re.Global = True
re.Pattern = "\D"
RemAlpha = re.Replace(str, "")
End Function


Function RemDigits(str As String) As String
'Remove numbers from a string
Dim re As Object
Set re = CreateObject("vbscript.regexp")
re.Global = True
re.Pattern = "\d+"
RemDigits = re.Replace(str, "")
End Function


Gord





Just a few points, depending, of course, on what the OP really means. My guess is that your routines will satisfy his requirements, but I'm in a nit-pickey mood today :-))

I assume by "alphabets" he means [A-Za-z].

Your first expression will remove all non-digits, not just the "alphabets".
Your second expression will remove all digits, leaving not only the "alphabets" but also various special characters.

So, the \D+ will serve to return all the digits.

But \d+ will return all non-digits, which can include punctuation, etc.

To only return the "alphabets" I would suggest [^A-Za-z]+