Substr problem 1

[mikeymay]

I am trying to use the following code where Region1 is a string variable


RegLen = length(Region1);

if substr(Region1,RegLen-7,8) = "Division" then
Region2 = cat(Region1," Support");

For some reason this won't work and the error occures at 'RegLen-7' in the substr function. when I replace this with a number, 14 for example' it works, but as Region1 can be any number of characters long I need this calculation in there.


Thanks

 

[Sastronaut]

check the actual length of region1, if it is 7 or less then you will get an error.

try using a condition to check the length before the conditional statement

RegLen = length(Region1);

if reglen > 7 then if substr(Region1,RegLen-7,8) = "Division" then
Region2 = cat(Region1," Support");

 

[kdt82]

Sastronaut,

Your exact example code wont work as it's missing do/end structures. If you want to do it in a one liner you can use the ifc function.

I agree that the most likely problem of the OP is region being too short, that's why example data is really important.

if length(Region1)>7 then Region2=ifc(substr(Region1,RegLen-7,8)="Division",cat(Region1," Support"),'');

 

[ChrisW75]

Can you include the error message and what the data was when the error was generated? There should be a mini-dump in the log showing the values of the variables when the error occured.
An alternative methods, rather than using substr is to use SCAN.
First off, this works:-

data test;

  region1 = 'Wine Division';
  RegLen = length(Region1);

  if substr(Region1,RegLen-7,8) = "Division" then
      Region2 = cat(Region1," Support");
run;

so I'm not sure what your error was.
However, this will also work...

data test;

  region1 = 'Wine Division';

  if scan(Region1,-1,' ') = "Division" then
      Region2 = cat(Region1," Support");
run;

And might get around your length issues...

Chris
Business Analyst, Code Monkey, Data Wrangler.
SAS Guru.

 

[Sastronaut]

kdt82,

FYI - the code i posted works perfectly in SAS. there is no do/end required for daisy chaining if then if then statements like that.

 

[kdt82]

Sastronaut,

My apologies, you are right. I don't believe I have seen this before. It even works for the else statement

Curiously, where did you find out about this?

data test;
region1 ='1Divishion';
RegLen = length(Region1);
if reglen > 7 then if substr(Region1,RegLen-7,8) = "Division" then
Region2 = cat(Region1," Support"); else region2 = 'Other'; 
run;
proc print;run;

output;

                       Reg
                                 Obs     region1      Len    Region2

                                  1     1Divishion     10     Other

 

[Sastronaut]

hi kdt82,

just tried it one day years ago and it worked