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Obsidian | Level 7 scb
Obsidian | Level 7

In the below code, I can get 2018 for variable A and 2019 for variable B.  May I know how to get 2018/2019 for variable C? Thanks.

 

%LET RUNDATE = TODAY();

 

data  _null_;
call symput("A",put(intnx('YEAR',&RUNDATE,-1),YEAR4.));
call symput("B",put(intnx('YEAR',&RUNDATE,0),YEAR4.));
run;

 

%PUT A = &A;
%PUT B = &B;

2 REPLIES 2
RW9
Diamond | Level 26 RW9
Diamond | Level 26

Yes, bur first off, please avoid shouting code at us, and finish macro variables with a dot, use the code window, and indentations!!

data  _null_;
  a=put(intnx('year',today(),-1),year4.);
  b=put(intnx('year',today(),0),year4.);
  call symput("A",a);
  call symput("B",b);
  call symput("C",catx('/',a,b));
run;

%put A=&a.;
%put B=&b.;
%put C=&c.;

The next question is, why put the today() result in a macro variable at all, or in fact the other two data items?  Each of these elements can simply be found by querying the today() function at the appropriate moment?

 

Astounding
PROC Star

Once you have this much:

 

%LET RUNDATE = TODAY();

 

data  _null_;
call symput("A",put(intnx('YEAR',&RUNDATE,-1),YEAR4.));
run;

 

That lets you continue with:

 

%let B = %eval(&A + 1);

%let C = &A/&B;

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