Date:         Fri, 14 Mar 2003 08:30:26 -0800
Reply-To:     Max <mjohnson@HSAG.COM>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Max <mjohnson@HSAG.COM>
Organization: http://groups.google.com/
Subject:      Re: length(compress('')) is 1 ?
Content-Type: text/plain; charset=ISO-8859-1

Sig,
This should do what you want.

data _null_;
string='9';
if compress(string,'0987654321')=' ' then do;
number_of_digits = 1 + length(string) - length(compress(string,'0987654321'));
end;
if compress(string,'0987654321') ne ' ' then do;
number_of_digits = length(string) - length(compress(string,'0987654321'));
end;
put number_of_digits;
run;

HTH,
MAX

stigeide@yahoo.com (Stig Eide) wrote in message news:<faa59d77.0303140054.4ccbfbb8@posting.google.com>...
> I don't get this.
> Why is length(compress('')) 1?
> I am trying to count number of digits in a text string,
> what I came up with was:
> number_of_digits = length(string) - length(compress(string,'0987654321'));
>
> But this don't work because the length of compress('9','9') is 1...
>
> Anyone have a better idea on how to count number of digits?
>
> Thanks!
> Stig