The naming of SAS library has 3 rules:

1. no more than 8 character;
2. may consist with underscore, numbers and English letters;
3. start with underscore or English letters;

Here comes my question: How to validate a string include invalid library name or not using perl regular expression?

The string is consist with words, which are separated by one space, like the following:

sasuser work sashelp
sasuser work 7z sashelp
sasuser work dictionary

7z and dictionary not statisfy the rules, so I want an output, with 0, 1, 1 corresponding with the three input strings.

I have trying this in SAS, but it doesn't work:

data test;
  input string&$42.;
  x=prxmatch('/\b(?=\S )(?![A-Za-z_][A-Za-z0-9_]{0,7})\b/',string);
  put x=;
  cards;
sasuser work sashelp
sasuser work 7z sashelp
sasuser work dictionary
;
run;

Thanks for any hint.

Edit:2022-11-11
I am really looking for a regex way, you may use SAS language or not. I have a thought as following:

1. Judge if the string contains a word or not;
2. The word mismatch a regular expression;
3. The regular expression discribe the rules of SAS library naming;

Is that possible?

CodePudding user response：

You appear to be testing if ANY of the words in the list are valid librefs. Instead test each word in the string separately.

Note that SAS already has a function, NVALID(), to test if a name is valid, but you need to add an additional test to make sure the length is not too long to use as a libref or fileref.

data test;
  input string $80. ;
  do index=1 to countw(string,' ');
    word = scan(string,index,' ');
    nvalid=nvalid(word,'v7') and lengthn(word) in (1:8);
    x=prxmatch('\b[A-Za-z_][A-Za-z0-9_]{0,7}\b/',word);
    output;
  end;
cards;
sasuser work sashelp
sasuser work 7z sashelp
sasuser work dictionary
;

Result

Obs            string             index    word          nvalid    x

  1    sasuser work sashelp         1      sasuser          1      1
  2    sasuser work sashelp         2      work             1      1
  3    sasuser work sashelp         3      sashelp          1      1
  4    sasuser work 7z sashelp      1      sasuser          1      1
  5    sasuser work 7z sashelp      2      work             1      1
  6    sasuser work 7z sashelp      3      7z               0      0
  7    sasuser work 7z sashelp      4      sashelp          1      1
  8    sasuser work dictionary      1      sasuser          1      1
  9    sasuser work dictionary      2      work             1      1
 10    sasuser work dictionary      3      dictionary       0      0

CodePudding user response：

If you want to do this without perl regex then here is a solution:

First let's get some more sample data:

data test;
  input string&$42.;
  cards;
sasuser work  sashelp
sas_user _work 7z sashelp
sasuser work77 dictionary
;
run;

Here, the resulting column "valid" consists of a list of flags (1 for valid, 0 for invalid):

data validation (drop=i txt);
set test;
length valid $12;
do i=1 to countw(string);
  txt=scan(string,i,' ');
  if txt ne '' then do;
    if (length(txt) gt 8 
        or substr(txt,1,1) eq compress(substr(txt,1,1),'_' , 'a')
        or txt ne compress(txt, ,'kan')
        ) 
      then valid=catx(', ',valid,'0');
      else valid=catx(', ',valid,'1');
  end;
end;
run;

Result:

string                    valid
-----------------------------------------
sasuser work  sashelp       1, 1, 0
sas_user _work 7z sashelp   1, 1, 0, 1
sasuser work77 dictionary   1, 1, 0