Search and replace regex, but only retain a portion of the string

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

Dear monks,

It's been a long time, but I know this is the right place to ask. So here goes. I'm looking for a pointer to the right doc section to read in https://perldoc.perl.org/perlre.html to help me with my test code below:

use Modern::Perl;
use strict;
use warnings;

my $dest = q{UK Mobile - Vodafone [GBRVF] [MSRN]};
$dest =~ s/(\W+)\s?(\[MSRN\])?/_/g;
say $dest;
[download]

which gives me UK_Mobile_Vodafone_GBRVF_MSRN_

I'm trying to turn UK Mobile - Vodafone [GBRVF] [MSRN] into UK_Mobile_Vodafone_GBRVF_, but [MSRN] might not always be there. I want this done all in one substitution too. I think I even want to get to UK_Mobile_Vodafone_GBRVF.

The software I'm doing this in can possibly process the same variable twice, with the second run being a second substitution, so I could make UK Mobile - Vodafone [GBRVF] [MSRN] become UK Mobile - Vodafone [GBRVF], then process it again to become UK_Mobile_Vodafone_GBRVF_

Many thanks,
Gavin.

Comment on Search and replace regex, but only retain a portion of the string Select or Download Code

Replies are listed 'Best First'.
Re: Search and replace regex, but only retain a portion of the string by choroba (Cardinal) on Dec 22, 2018 at 23:11 UTC
Using two substitutions is cleaner: `$dest =~ s/\W+\[MSRN\]\|\W+/_/g; $dest =~ s/_$//;` [download] i.e. replace sequences of non-word characters by underscores, then remove the last underscore. I can't think of a non-hackish solution with one substitution only. `$dest =~ s/(?:\W+(?:MSRN\])?\|\W+)(?(?=$)(?{$%=1})\|(?{$%=0}))/$%?'' +:'_'/ge;` [download] `map{substr$_->[0],$_->[1]\|\|0,1}[\\|\|{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^ARGV,3]`	[reply] [d/l] [select]
Re^2: Search and replace regex, but only retain a portion of the string by ghenry (Vicar) on Dec 23, 2018 at 13:43 UTC
Of course. For goodness sake, an OR. Thanks a lot! I can live with <code>_</code.	[reply]
Re: Search and replace regex, but only retain a portion of the string by tybalt89 (Monsignor) on Dec 23, 2018 at 01:12 UTC
`#!/usr/bin/perl -l # https://perlmonks.org/?node_id=1227618 use strict; use warnings; my $want = q{UK_Mobile_Vodafone_GBRVF}; my $dest = q{UK Mobile - Vodafone [GBRVF] [MSRN]}; print $dest; $dest =~ s/( )\|- \|\[\|\].*/ '_' x !!$1 /ge; print $dest; print $want;` [download]	[reply] [d/l]
Re^2: Search and replace regex, but only retain a portion of the string by ghenry (Vicar) on Dec 23, 2018 at 13:43 UTC
Very nice. Will have a play. Thank you.	[reply]
Re: Search and replace regex, but only retain a portion of the string (updated x4) by AnomalousMonk (Archbishop) on Dec 22, 2018 at 22:51 UTC
I think I even want to get to `UK_Mobile_Vodafone_GBRVF`. Assuming that's really what you want: `c:\@Work\Perl\monks>perl -wMstrict -le "use constant T => q{UK Mobile - Vodafone [GBRVF] [MSRN]}; ;; my $dest = T; $dest =~ s{ \] \s+ \S+ \z }{}xms; print qq{2 steps: a: '$dest'}; ;; $dest =~ s{ \W+ }{_}xmsg; print qq{2 steps: b: '$dest'}; ;; $dest = T; ;; $dest =~ s{ \W+ (\w+ \] \z)? }{ $1 ? '' : '_' }xmsge; print qq{1 step: '$dest'}; " 2 steps: a: 'UK Mobile - Vodafone [GBRVF' 2 steps: b: 'UK_Mobile_Vodafone_GBRVF' 1 step: 'UK_Mobile_Vodafone_GBRVF'` [download] Update 1: Or (with inspiration from tybalt89): `$dest =~ s{ \W+ (\w+ \] \z)? }{ !$1 && '_' }xmsge;` Update 2: Or with no eval: `c:\@Work\Perl\monks>perl -wMstrict -le "my $dest = q{UK Mobile - Vodafone [GBRVF] [MSRN]}; ;; my @xlate = ('', '_'); ;; $dest =~ s{ \W+ (\w+ \] \z)? }{$xlate[! $1]}xmsg; print qq{'$dest'}; " 'UK_Mobile_Vodafone_GBRVF'` [download] With Perl version 5.10+, you could use a persistent state variable: `state $xlate = [ '', '_' ];` (changing the replacement expression, of course). (The original code of this update had a `no warnings 'uninitialized';` statement which turned out to be unneeded. Removed.) Update 3: Or another two-stepper: `c:\@Work\Perl\monks>perl -wMstrict -le "my $dest = q{UK Mobile - Vodafone [GBRVF] [MSRN]}; ;; $dest =~ tr{A-Za-z}{_}cs; $dest =~ s{ _ [^_]+ _ \z }{}xms; ;; print qq{'$dest'}; " 'UK_Mobile_Vodafone_GBRVF'` [download] Update 4: Or: `c:\@Work\Perl\monks>perl -wMstrict -le "my $dest = q{UK Mobile - Vodafone [GBRVF] [MSRN]}; $dest =~ s{ \W+ (\w+ \] \z)? }{ [ '', '_' ]->[! $1] }xmsge; print qq{'$dest'}; " 'UK_Mobile_Vodafone_GBRVF'` [download] I'm going to bed now. (Update: I don't know why I based this solution on an anonymous array; `('', '_')[! $1]` is a bit less obscure and probably slightly faster. But beyond that, this eval-ed solution doesn't really offer anything more than the ternary expression used in the `eval` one-step in my original reply. Well, it was late...) Give a man a fish: `<%-{-{-{-<`	[reply] [d/l] [select]
Re^2: Search and replace regex, but only retain a portion of the string (updated x4) by ghenry (Vicar) on Dec 23, 2018 at 13:42 UTC
Thanks. Amazing effort. Always stunned when someone spends a lot of time on these.	[reply]
Re: Search and replace regex, but only retain a portion of the string by kschwab (Vicar) on Dec 23, 2018 at 20:18 UTC
If the rules are "change any sequence of characters that isn't a-zA-Z into a single underscore", you might like tr// better than a regex: `$ perl my $s="UK Mobile - Vodafone [GBRVF] [MSRN]"; $s=~tr/a-zA-Z/_/cs; print "$s\n"; ^D UK_Mobile_Vodafone_GBRVF_MSRN_` [download] Still needs the second pass to get rid of MSRN_, but clean and straightforward otherwise. Edit: I see someone already updated another comment node to suggest this, apologies for the dupe.	[reply] [d/l]