pattern matching: why does the following code evaluate true? (and how do i fix it?)

Buckaroo Buddha has asked for the wisdom of the Perl Monks concerning the following question:

        
$key  = "INITIAL.LASTNAME,INITIAL2.LASTNAME2,INITIAL3.LASTNAME3";
$key2 = "INITIAL.LASTNAME,INITIAL2.LASTNAME2,.";

if ($key =~ m/$key2/){ 
                print " [just another perl hacker] \n";
}
[download]


  i'm cycling through two hash tables
  and trying to get the exact match but 
  it seems that i have a less than perfect 
  understanding of pattern matching

  i've got "the perl cookbook" here with me,
  so if i find the answer before this is taken 
  care of i'll post a reply to myself

  but i expected that the above statement would 
  evaluate to FALSE

  (note, i'm not actually making an attempt at a JAPH
   but i thought it was a funny replacement for 
  Execute Block ;)

Comment on pattern matching: why does the following code evaluate true? (and how do i fix it?) Download Code

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Re: pattern matching: why does the following code evaluate true? (and how do i fix it?) by btrott (Parson) on May 31, 2000 at 02:53 UTC
That's not greedy pattern matching. That's just an example of the '.' matching any character (except newlines). To the OP: the regex matches because the '.' matches an "I" character. Because '.' matches anything. Read perlre. And try this, if you wanted a literal period ('.'): `if ($key =~ m/\Q$key2\E/) { print " [just another perl hacker] \n"; }` [download]	[reply] [d/l]
Re: pattern matching: why does the following code evaluate true? (and how do i fix it?) by Buckaroo Buddha (Scribe) on May 31, 2000 at 02:57 UTC
so it should be: `($key =~ m/.$key2/)` [download] i was just about to follow up with `($key =~ m/^$key2$/)` [download] as being how i solved this problem ... (well how O'Reilly solved it) i geuss this means that the '.' in the string is interpreted as a wildcard matching to the end of the line? that's interesting ... please correct me if i'm wrong but if that's the way it works that's a really cool thing to have learned :)!	[reply] [d/l] [select]
RE: Re: pattern matching: why does the following code evaluate true? (and how do i fix it?) by reptile (Monk) on May 31, 2000 at 03:17 UTC
The '.' matches a single character, any character, except \n (unless using the 's' modifier, to treat the expression as a single line). There are two ways to solve the problem, one is how you did it, and another is to use `/\Q$key2/` as others have said, so the '.' is treated as a literal '.' (along with every other metacharacter). `$key =~ m/^$key2$/; # works $key =~ m/\Q$key2/; # also works` [download] Personally I think the second option is better here. The '.' normally matches the literal '.' and anything else as well, but if your $key contained something like `"INITIAL!LASTNAME"` it would match if `"INITIAL.LASTNAME"` where the pattern. Quoting metacharacters with `\Q$pattern\E` is, at least, more correct, even if the first will work ok with the given input. It'll help you avoid bugs to use the second option. `72656B636148206C72655020726568746F6E41207473754A`	[reply] [d/l]
RE: pattern matching: why does the following code evaluate true? (and how do i fix it?) by Adam (Vicar) on May 31, 2000 at 02:42 UTC
Greedy Pattern Matching. The . matches anything (and more importantly - everything.) `{Except \n}` UPDATE: It has been pointed out to me that I am only partly correct. The dot does in fact match anything `{Except \n}` but it does not behave in a greedy manner. Rather, Perl is saying that the substring ($key2) is found in the other string ($key)... which is ok. I apologize if I confused anyone.	[reply]
RE: RE: pattern matching: why does the following code evaluate true? (and how do i fix it?) by takshaka (Friar) on May 31, 2000 at 02:56 UTC
Well, /^$key2$/, wouldn't match--there's nothing greedy in that regex. But since the pattern isn't anchored it looks for the $key2 pattern anywhere in $key. If you're looking for literal periods, do one of the following: `$key2 = "INITIAL\.LASTNAME,INITIAL2\.LASTNAME2,\."; $key2 = quotemeta "INITIAL.LASTNAME,INITIAL2.LASTNAME2,."; $key =~ /\Q$key2/` [download]	[reply] [d/l]