So the problem is length(' ')=1;

When not simply goes,

if x=' ' then digits=0; else digits=...;

Just my $0.02.

EdHeaton@WESTAT.COM (Ed Heaton) wrote in message news:<9B501B3774931C469BCCCC021BE5372231E6E0@remailnt2-re01.westat.com>... > Stig, > > A little trick I learned from the ell. This works when the list of > non-wanted characters is longer than the list of wanted characters. > > %let max = 50 ; > %let seed = 574111 ; > Data _null_ ; > /* Create a test string. */ > Length x $&max ; > Do i=1 to &max ; > x = trim(x) || byte( floor( ranUni(&seed) * 256 ) ) ; > End ; > /* Here is your solution. */ > digits = length( "x" || compress( > x > , compress( x , "0123456789" ) > ) ) -1 ; > /* Show the results. */ > Put x= / digits= ; > Run ; > > That solves your task. Others have answered your question about why. > > Ed > > Edward Heaton, Senior Systems Analyst, > Westat (An Employee-Owned Research Corporation), > 1600 Research Boulevard, Room RW-3541, Rockville, MD 20850-3195 > Voice: (301) 610-4818 Fax: (301) 610-5128 > mailto:EdHeaton@westat.com http://www.westat.com > > > -----Original Message----- > From: Stig Eide [mailto:stigeide@YAHOO.COM] > Sent: Friday, March 14, 2003 3:55 AM > To: SAS-L@LISTSERV.UGA.EDU > Subject: length(compress('')) is 1 ? > > > 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