Regex to match certain words but not others - regex

Question

I am working with regexes in perl, and I'm trying to make a regex that finds two words where one ends with d and the next word starts with p (but not ph). Here is my regex, which works:
d\s(p[^h])}
However, I'd also like this to exclude the word "and" (but only within this pattern) so I have tried to use a negative lookahead, so my code looks like this:
if ($text =~ m{d\s(p[^h])} && $text =~ m{(?:(?!\sand\s))}) {
print "Yes\n";
} else {
}
However, this doesn't seem to work.
Here are some sample input/output:
sand pet -> yes
sand phone -> no
go and pet -> no
sand pet and -> yes
Any help with this is greatly appreciated!

Answer 1

You can accomplish what you need with a single regex:
/(?<!\ban)d\s(p[^h]\w+)/
Where:
\b is the word boundary anchor, doesn't consume any chars but assure that the excluded words is and and not sand. It matches between \w (word chars: [a-zA-Z0-9_]) and \W (not in word chars) and in the same position of ^ and $.
(?<!\ban)d a d not preceded by an isolated an, technically speaking is almost equivalent to (?<!\Wan).
Online Demo
If you doesn't need to extract the first and second word separately you may remove the capturing groups also and add some tolerance (one or more spaces between words):
if ( $input =~ m/(?<!\ban)d\s+p(?!h)/ )
print "Yes\n";
else
print "No\n";
Note: this regex is actually search for a d (not preceded by a non-substring an) separated by one or more spaces from a p not followed by a h. It tells nothing aboud words. If you want to make sure there are words of more than one char you can add a leading and trailing \w+.
Another Demo

Answer 2

You're getting too complicated. That negative lookahead is applied to the string, and matches against any substring. So it will match any substring which doesn't contain \sand\s which is always going to work, because zero length substrings are 'ok'.
You can see this at work with turning on debugging:
#!/usr/bin/env perl
use strict;
use warnings;
use re 'debug';
while ( <DATA> ) {
print if m{(?:(?!\sand\s))};
}
__DATA__
sand pet
sand phone
go and pet
sand pet and
empty
That lookahead is used with another pattern to say 'match this, but only if this is (or isn't) next'.
So something like:
m{d\s(p[^h])} and not m{\sand\s};
Might do what you want - or alternatively, just break it down into phases:
#!/usr/bin/env perl
use strict;
use warnings;
#use re 'debug';
while (<DATA>) {
my ($capture) = m{d\s(p[^h])};
if ( $capture and not $capture =~ m/\sand\s/ ) {
print $capture, " => ", $_, "\n";
}
}
__DATA__
sand pet
sand phone
go and pet
sand pet and
empty

Answer 3

It's often inappropriate to try to get everything working in a single regex. This program has a subroutine ok_words that checks a pair of words to see if your criteria apply. The calling code takes every pair of words in the string and prints a yes if the test is true for any pair, otherwise no
These are your tests, together with the Perl code that checks for them
The first ends in a d — /\d\z/
...but isn't and — `ne 'and'
The second starts with a p, but not ph — /\Ap(?!h)/
And this is the program that applies them
use strict;
use warnings 'all';
use List::MoreUtils qw/ any /;
while ( <DATA> ) {
chomp;
my #w = split;
if ( any { ok_words( $w[$_], $w[$_+1] ) } 0 .. $#w-1 ) {
print "$_ -> yes\n";
}
else {
print "$_ -> no\n";
}
}
sub ok_words {
my ($this, $next) = map lc, #_;
$this =~ /d\z/ and $this ne 'and' and $next =~ /\Ap(?!h)/;
}
__DATA__
sand pet
sand phone
go and pet
sand pet and
output
sand pet -> yes
sand phone -> no
go and pet -> no
sand pet and -> yes