Exclusive pattern matching..

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

How would I go about making a regex that would only match the data provided in a scalar ,nothing more, nothing less. Would I use a specific quantifier??

For example, say that I am matching a scalar provided from STDIN that looks like this:
user1. How would I get it to match ONLY the user1 not the user11 or user12 and so on.....

Any help would be appreciated..

Comment on Exclusive pattern matching..

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Re: Exclusive pattern matching.. by Abigail-II (Bishop) on Jun 21, 2002 at 18:17 UTC
By not using a regex of course! If you want exact matching, you'd use the `eq` operator. Abigail	[reply] [d/l]
Re: Re: Exclusive pattern matching.. by Anonymous Monk on Jun 22, 2002 at 03:09 UTC
that wouldn't work too well for "i am user1" eq "user1", he probably meant something like `m{\b?\Q$word\E\b?}`	[reply] [d/l]
Re: Exclusive pattern matching.. by Dog and Pony (Priest) on Jun 21, 2002 at 18:17 UTC
Use `eq` instead. :) `my $input = shift; if($input eq $scalar) ....` [download] If you really want a regexp for this though, you can use `^` and `$` like this: `$string =~ /^user1$/;` [download] ^ matches only at the beginning of the string, and $ only at the end, so this way you force it to match only if it matches the whole string. You have moved into a dark place. It is pitch black. You are likely to be eaten by a grue.	[reply] [d/l] [select]
Re: Exclusive pattern matching.. by robobunny (Friar) on Jun 21, 2002 at 18:13 UTC
you can indicate the beginning of a string with \A and the end of a string with \Z. ^ matches the beginning of lines, and $ matches the end of lines. so you could use something like /^user1$/ or /\Auser1\Z/. of course, if you just want an exact match, you can always use $var eq 'user1'.	[reply]