Lapis Lazuli | Level 10

## multiple sustr statements

Hi,

I have multiple subst statements and I want to run them all at once. Can some one help on the best way to combine all these statements in once (I have 30+ statements)

data a (drop=i); set b; array {x} dx:; ar=1=0;

do i=1 to dim(x); if substrn(x{i}, 1, 5) in ('12345', '23456', '30987')

then do; ar=1;

leave;

end;

end;

run;

data a (drop=i); set b; array {x} dx:; arf=1=0;

do i=1 to dim(x); if substrn(x{i}, 1, 4) in ('1234', '2432', '3987')

then do; arf=1;

leave;

end;

end;

run;

data a (drop=i); set b; array {x} dx:; ad=1=0;

do i=1 to dim(x); if substrn(x{i}, 1, 2) in ('12', '24', '39')

leave;

end;

end;

run;

1 ACCEPTED SOLUTION

Accepted Solutions
Diamond | Level 26

## Re: multiple sustr statements

Your code is very hard to read.  What are you trying to do, if its just finding those strings, and setting a flag why not:

```data a (drop=i);
set b;
array a dx:;
ar=0;
arf=0;
do i=1 to dim(x);
ar=ifn(substrn(x{i}, 1, 5) in ('12345', '23456', '30987'),1,ar);
arf=ifn(substrn(x{i},1,4) in ('1234','2432','3987'),1,arf);
end;
run; ```
5 REPLIES 5
Super User

## Re: multiple sustr statements

Since the purpose seems to be to just create the flag variables, you can do all the if's in one do loop and omit the leave statement.

Lapis Lazuli | Level 10

## Re: multiple sustr statements

Thank you for the reply, I am getting an error (one unclosed loop block), would you mind to show me how would you combine them? Thanks

Diamond | Level 26

## Re: multiple sustr statements

Your code is very hard to read.  What are you trying to do, if its just finding those strings, and setting a flag why not:

```data a (drop=i);
set b;
array a dx:;
ar=0;
arf=0;
do i=1 to dim(x);
ar=ifn(substrn(x{i}, 1, 5) in ('12345', '23456', '30987'),1,ar);
arf=ifn(substrn(x{i},1,4) in ('1234','2432','3987'),1,arf);
end;
run; ```
Lapis Lazuli | Level 10

Thank you!

Super User

## Re: multiple sustr statements

If you're just checking for the start of the string to match consider using the : operator.

``if x(i)  in: ('12345', '23456');``

If the length of your X variable can be smaller than the string this likely won't work though.

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