topic Re: How do I fix: ERROR 22-322: Syntax error with "else if... then do..." statement in SAS Programming

How do I fix: ERROR 22-322: Syntax error with "else if... then do..." statement

btenney — Tue, 30 Mar 2021 21:40:22 GMT

I keep receiving this error message:

ERROR 22-322: Syntax error, expecting one of the following: !, !!, &, *, **, +, ',', -, /, <, <=, <>, =, >, ><, >=, AND, BY, EQ, GE, GT, IN, LE, LT, MAX, MIN, NE, NG, NL, NOTIN, OR, TO, ^=, |, ||, ~=.

The code is:

data ex7;

INFILE "12q19.character.txt"

dlm='09'x firstobs=2;

input HOUSE Y X1 X2 X3 X4 location $;

if location="intown" then

DO Zlocation1=0;

Zlocation2=0;

Zlocation3=0;

end;

else if location="inner suburbs" then

DO Zlocation1=1;

Zlocation2=0;

Zlocation3=0;

end;

else if location="outer suburbs" then

DO Zlocation1=0;

Zlocation2=1;

Zlocation3=0;

end;

else

Zlocation1=0;

Zlocation2=0;

Zlocation3=1;

end;

x1location1=x1*Zlocation1;

x1location2=x1*Zlocation2;

x1location3=x1*Zlocation3;

run;

This is where the error message is:

Re: How do I fix: ERROR 22-322: Syntax error with "else if... then do..." statement

Tom — Wed, 31 Mar 2021 00:36:07 GMT

Looks like there are gibberish characters before the semi-colon on some of the lines:

502   else if location="inner suburbs" then
503   DO Zlocation1=1?;
                     -
                     388
                     200
                     76
504   Zlocation2=0?;
                  -
                  388
                  200
                  76
505   Zlocation3=0?;
                  -
                  388
                  200
                  76
ERROR 388-185: Expecting an arithmetic operator.

ERROR 200-322: The symbol is not recognized and will be ignored.

ERROR 76-322: Syntax error, statement will be ignored.

506   end;

Also that is a very strange way to code your DO loop. You have essentially coded an iterative DO loop with only one iteration. Instead of using DO ZLOCATION1=0 to 2 or DO ZLOCATION1=1,3,4 you just have used DO ZLOCATION1=0. Normally when you only want to run the DO loop once you do NOT include a iterative variable in the DO statement. Instead just use DO without anything after it and set ZLOCATION1 the same way you are setting ZLOCATION2 and ZLOCATION3.

if location="intown" then DO;
  Zlocation1=0;
  Zlocation2=0;
  Zlocation3=0;
end;
else if location="inner suburbs" then DO;
  Zlocation1=1;
  Zlocation2=0;
  Zlocation3=0;
end;

From your logic it looks like you can eliminate the DO loops and the IF/THEN/ELSE statements and just calculate the 0/1 values as the result of boolean expressions. Your current logic reduces to these three assignment statements.

Zlocation1=location="inner suburbs";
Zlocation2=location="outer suburbs";
Zlocation3=location not in ("intown" "inner suburbs" "outer suburbs");

Re: How do I fix: ERROR 22-322: Syntax error with "else if... then do..." statement

ChrisNZ — Wed, 31 Mar 2021 07:55:12 GMT

You need a semicolon after do.

Re: How do I fix: ERROR 22-322: Syntax error with "else if... then do..." statement

Kurt_Bremser — Wed, 31 Mar 2021 08:19:28 GMT

Maxim 12: Make It Look Nice.

As posted, your code is hard to debug and maintain. This may have been caused by pasting the code into the main posting window. Use the "little running man" button right next to the one indicated for posting code:

(</> is for posting logs and other fixed-width text, e.g. content of text infiles)

Next, replace the IF/THEN/ELSE IF chain with a SELECT block that better shows the purpose.

data ex7;
infile
  "12q19.character.txt"
  dlm='09'x
  firstobs=2
;
input HOUSE Y X1 X2 X3 X4 location :$13.;
select (location);
  when ("intown") do;
    Zlocation1 = 0;
    Zlocation2 = 0;
    Zlocation3 = 0;
  end;
  when ("inner suburbs") do;
    Zlocation1 = 1;
    Zlocation2 = 0;
    Zlocation3 = 0;
  end;
  when ("outer suburbs") do;
    Zlocation1 = 0;
    Zlocation2 = 1;
    Zlocation3 = 0;
  end;
  otherwise do;
    Zlocation1 = 0;
    Zlocation2 = 0;
    Zlocation3 = 1;
  end;
end;
x1location1 = x1 * Zlocation1;
x1location2 = x1 * Zlocation2;
x1location3 = x1 * Zlocation3;
run;

There were some undisplayable characters where the ERROR was indicated.

I also added an informat for the character variable. The $ sign on its own would lead to a defined length of 8, so the values "inner suburbs" and "outer suburbs" would be truncated during the read.