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; /* Newsgroups: bit.listserv.sas-l Date: Mon, 12 Jun 1995 16:29:22 MET Sender: "SAS(r) Discussion";From: "Wim A.J.G. Lemmens" Subject: Brief summary/answer on : Number of chars in a given string To: Multiple recipients of list SAS-L Dear SAS-LŽers / dear responders, Thank you all for your time spending on my problem. All solutions where based on using a combination of : the sas-functions : Compress - Index - Length - Substr - Tranwrd - Trim . Below is given the macro I'll use in my specific problem. ***********especially the following persons :(sorted on location) adamhndrx @ aol.com Raymond K. Lee gibes @ nutra.monsanto.com Jugdish @ pspltd.demon.co.uk (Jugdish K. Mistry) whitloi1 @ westatpo.westat.com "Zack, Matthew M." dstanle @ IBM.NET Ron Meisenheimer CVRGING @ TECHNION "Richard A. DeVenezia" ************ Greetings from : Wim (W.A.J.G.) Lemmens, Statistical Analyst/Programmer * University of Nijmegen, Faculty of Medical Sciences ------------------* * Department of Medical Informatics, Epidemiology and Statistics * * SAS-version : 6.08 VM/CMS/ESA * * 151-MSA- Phone : (31) -80- 617678 * * P.O.Box 9101 Fax : (31) -80- 613505 * * NL-6500 HB Nijmegen E-mail -1- U439002@vm.uci.kun.nl * * The Netherlands -2- W.Lemmens@mie.kun.nl * *-----------------------------------------------------------------------* * >>> visiting address : Kapittelweg 54 6525 EP Nijmegen <<< * *-----------------------------------------------------------------------* */ %macro strcnt(str=,sub=,nsub=); ** STR : the complete string in which the characters are; ** SUB : the character-substring to be found; **NSUB : the number of times the specified SUB is found; &nsub= ( length(&str) - length( compress( tranwrd( &str, trim( &sub )," " )," " ) ) ) / length(&sub); %mend; *