handling hypens(-) in word boundary matching

[arun]

Hi Perl gurus,

Iam trying to replace the sub string "feat-ha" with 1 in a string but
it's failing because of hypen (-) please
help me how to handle it. I have pasted my code snippet below.


#!c:\Perl\site\bin

use strict;

my $string = qq((feat-bgp-mpls-vpn AND feat-ha-sso AND span));

my $pattern = qq(feat-ha);

$string=~s/\b$pattern\b/1/g;

print $string;

output : (feat-bgp-mpls-vpn AND 1-sso AND span)


In the above programme iam trying to replace the word "feat-ha" with 1
if it matches the exact word in the string "(feat-bgp-mpls-vpn AND
feat-ha-sso AND span)" but it's replacing feat-ha-sso as 1-sso which
shouldn't happen please help me.

Thanks in Advance..

Regards
-Arun

 

[Tad J McClellan]

Hi Perl gurus,

Iam trying to replace the sub string "feat-ha" with 1 in a string but
it's failing because of hypen (-) please
help me how to handle it. I have pasted my code snippet below.


#!c:\Perl\site\bin

use strict;

my $string = qq((feat-bgp-mpls-vpn AND feat-ha-sso AND span));

my $pattern = qq(feat-ha);

$string=~s/\b$pattern\b/1/g;

print $string;

output : (feat-bgp-mpls-vpn AND 1-sso AND span)


In the above programme iam trying to replace the word "feat-ha" with 1
if it matches the exact word in the string "(feat-bgp-mpls-vpn AND
feat-ha-sso AND span)" but it's replacing feat-ha-sso as 1-sso which
shouldn't happen please help me.

Use a negative lookahead assertion with a positive char class:

$string =~ s/\b$pattern(?![\w-])/1/g;

or use a positive lookahead assertion with a negated char class:

$string =~ s/\b$pattern(?=[^\w-])/1/g;

If $pattern might occur at the end of the string, then you'll need
to account for that as well:

$string =~ s/\b$pattern(?!([\w-]|$))/1/g;


You will likely need to do something similar to the beginning of your pattern.

 

[sln]

Hi Perl gurus,

Iam trying to replace the sub string "feat-ha" with 1 in a string but
it's failing because of hypen (-) please
help me how to handle it. I have pasted my code snippet below.


#!c:\Perl\site\bin

use strict;

my $string = qq((feat-bgp-mpls-vpn AND feat-ha-sso AND span));

my $pattern = qq(feat-ha);

$string=~s/\b$pattern\b/1/g;

print $string;

output : (feat-bgp-mpls-vpn AND 1-sso AND span)


In the above programme iam trying to replace the word "feat-ha" with 1
if it matches the exact word in the string "(feat-bgp-mpls-vpn AND
feat-ha-sso AND span)" but it's replacing feat-ha-sso as 1-sso which
shouldn't happen please help me.

Thanks in Advance..

Regards
-Arun

From perlretut:
An anchor useful in basic regexps is the word anchor \b.
This matches a boundary between a word character and a non-word character \w\W or \W\w

Where:
\d is a digit and represents [0-9]

\s is a whitespace character and represents [\ \t\r\n\f]

\w is a word character (alphanumeric or _) and represents [0-9a-zA-Z_]

\D is a negated \d; it represents any character but a digit [^0-9]

\S is a negated \s; it represents any non-whitespace character [^\s]

\W is a negated \w; it represents any non-word character [^\w]

The period '.' matches any character but ``\n''

------------------------------------------------

So feat-ha-sso matches because between 'a' and '-' is a word boundry going from \w\W.

Visually, it appears you want this -> s/(\s)$pattern(\s)/$1\1$2/g
if say you don't wan't to do extended pattern constructs.

Possibly -> s/\b$pattern([^\w-]|$))/1$1/g

But doing it this way negates the variability of $pattern since the last character
in the pattern must be a word character.

Still though, its better to be specific. To just use \b, a word boundry as bookend
delimeters might yeild more than you want (or less than you want). It all depends upon
what data you expect to be fed to it.

Oh I guess you could force a prevailing rule of \b, boundries, test the pattern, create
pre/post delimeter character classes then put it all together in a regular expression.

Generally, \b is not something reliable given the variable nature of possiblillities it could
match with complex source text.

sln