If in the previous case it is showing '0-6 months' then length can not be changed in between. and moreover if a variable is created by assignment the length of that variable can not be changed with a length statement. ... View more

If I understand correctly, you just want to get the year of the Last_dep_date variable?  I would assume that you have the variable stored as a numeric variable with a date format.  If that is the case, then this code will capture just the year: data deposit; input client $ Last_dep_date DDMMYY8. ; format last_dep-date ddmmyy8. ; datalines ; A 27/12/07 B 04/08/07 C 02/04/08 ; run ; data deposit; set deposit; year=year(last_dep_date); run ; If your last_dep_date variable happens to be character, then you can use scan(last_dep_date,3,'/').  This means to treat the date variable as though it is 3 "words" separated by a "/" and you want to keep the 3rd "word" (or the year). ... View more

Wouldn't you want to take the number of days divided by 30, not 12, to get an approximation of the number of months? 270 days / 12  = 22.5 270 days / 30 = 9 months ... View more

Here is a simple way which can change all alpha character into UPPERCASE. filename tran 'c:\temp.dat'; proc cport data=sashelp.class file=tran outtype=upcase; run; proc cimport infile=tran data=want; run; Ksharp ... View more