Re: Likely trivial regex question

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Re^2: Likely trivial regex question by moodywoody (Novice) on Nov 09, 2011 at 07:15 UTC
For example I thought this would be it `$re = qr/beer=(\d{2}).(vodka=(\d{2})){0,1}.chips=(\d{3})/;` [download] but it doesn't capture $2,$3 in str1 (why?).	[reply] [d/l]
Re^3: Likely trivial regex question by graff (Chancellor) on Nov 09, 2011 at 09:54 UTC
Well, first off, by putting parens around "vodka=(\d{2})", you've created another capture, but the approach that might have been "correct" also doesn't work: `my $re = qr/beer=(\d{2}).(?:vodka=(\d{2}))?.chips=(\d{3})/;` [download] That "matches" both strings, but doesn't capture the number that follows "vodka" in the first string. Sorry, but I can't explain why. So I'd be inclined to take a different approach: `#!/usr/bin/perl use strict; use warnings; my @strs = ( "beer=10&otherstuff&vodka=20&otherstuff&chips=100", "beer=10&otherstuff&juice=20&otherstuff&chips=100" ); my @targets = qw/beer vodka chips/; for ( @strs ) { my @matched; for my $target ( @targets ) { push @matched, $1 if ( /$target=(\d+)/ ); } if ( @matched ) { print "matched: @matched\n"; } }` [download] UPDATE: Here's a variant on that approach, which I think is closer to what your own snippet would actually do if it worked (in case that's really what you want): `my @targets = qw/beer=(\d{2}) (vodka=(\d{2})) chips=(\d{3})/; for ( @strs ) { my @matched; for my $target ( @targets ) { push @matched, ( /$target/ ); } if ( @matched ) { print "matched: @matched\n"; } }` [download]	[reply] [d/l] [select]
Re^4: Likely trivial regex question by moodywoody (Novice) on Nov 09, 2011 at 10:35 UTC
Hi graff, thanks for your efforts. As it happens I'm particularly interested in the part about which you wrote that you cannot explain it either (optional grouping not captured). As a workaround I've written to regexes coupled with if: `my $re1 = qr/beer=(\d{2}).chips=(\d{3})/; my $re2 = qr/vodka=(\d{2})/; if ( $str =~ /$re1 ) { my $m1 = $1; my $m2 = $2; if ( $str =~ /$re2/ ) { #use that capture and go on... } }` [download] This does exactly what I want it to do; it's just that I'd like to learn if it's possible to achieve with a single regex. And if not, then why? While reading up and googling this I realized that I basically do not understand much about greedy/non-greedy quantifier and/or optional groups. For example I am also puzzled why `"cat:dog" =~ /(cat)/;` [download] captures "cat", but `"dog:cat" =~ /(cat)*/;` [download] doesn't. Cheers	[reply] [d/l] [select]
Re^5: Likely trivial regex question by ww (Archbishop) on Nov 09, 2011 at 12:38 UTC
Re^5: Likely trivial regex question by choroba (Cardinal) on Nov 09, 2011 at 11:00 UTC
Re^4: Likely trivial regex question by choroba (Cardinal) on Nov 09, 2011 at 10:56 UTC
Sorry, but I can't explain why. Because the `.` before "vodka" eats "vodka" as well, as you can easily see if you add () around `.`: `my $re = qr/beer=(\d{2})(.)(?:vodka=(\d{2}))?(.)chips=(\d{3})/; my $str1 = "beer=10&otherstuff&vodka=20&otherstuff&chips=100"; my $str2 = "beer=10&otherstuff&juice=20&otherstuff&chips=100"; say for $str1 =~ /$re/g; say "=="; say for $str2 =~ /$re/g;` [download] Changing the first `.` to `.?` does not help, either, because then the second `.*` eats "vodka".	[reply] [d/l] [select]
Re^3: Likely trivial regex question by vinian (Beadle) on Nov 11, 2011 at 14:48 UTC
hi, you can add `use re 'debug';` and see the output what happen when the string match.	[reply] [d/l]
Re^2: Likely trivial regex question by moodywoody (Novice) on Nov 09, 2011 at 07:09 UTC
Sorry, the example might have been suboptimal. The "juice" part is never to be captured. The string will either contains a vodka=xx part or not. If it is there I the regex like to capture it in $2 and $3. If it isn't there I would like the regex to match (say "matches) and give "unitialized" for $2,$3 but still capture $1 and $4.	[reply]