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SAS_inquisitive
Lapis Lazuli | Level 10

This works.

 

%put %substr(CHAR_DATA,1,4);

But when used in DATA step it does not work.

data check;
  var=%substr(CHAR_DATA,1,4);
run;
******* or *********;
%macro test;
data check;
  var=%substr(CHAR_DATA,1,4);
run;
%mend test;
%test

 

Thanks for explanation. 

1 ACCEPTED SOLUTION

Accepted Solutions
Astounding
PROC Star

The statement works, but it doesn't do what you are expecting.  Macro language is not processing your data.  It has an entirely different function.  It generates the SAS language statements that will then process your data.  So this statement is legal:

 

%let newvar = %substr(CHAR_VAR, 1, 4);

 

It gives &NEWVAR the value CHAR (the first four characters in "CHAR_VAR").

 

That means that %SUBSTR works, and gives you a program that looks like this:

 

data check;

var = CHAR;

run;

 

Probably, this is not the program that you intended to generate.  So work backwards.  Start with what you would like to see as a SAS program, and then see where macro language might be able to help you get there.

 

Good luck.

View solution in original post

1 REPLY 1
Astounding
PROC Star

The statement works, but it doesn't do what you are expecting.  Macro language is not processing your data.  It has an entirely different function.  It generates the SAS language statements that will then process your data.  So this statement is legal:

 

%let newvar = %substr(CHAR_VAR, 1, 4);

 

It gives &NEWVAR the value CHAR (the first four characters in "CHAR_VAR").

 

That means that %SUBSTR works, and gives you a program that looks like this:

 

data check;

var = CHAR;

run;

 

Probably, this is not the program that you intended to generate.  So work backwards.  Start with what you would like to see as a SAS program, and then see where macro language might be able to help you get there.

 

Good luck.

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