Regex match and replace

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

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Re: Regex match and replace by kennethk (Abbot) on Mar 04, 2015 at 17:39 UTC
Is there a reason you need to do it as a compound regular expression? It'll probably be easier to do it as a compound operation with index, substr and a substitution: `substr($s, 1+index $s, 'bar') =~ s/bar/foo/g;` [download] You could also run the regular expression repeatedly, using the \K escape. `while ($var =~ s/bar.?\Kbar/foo/) {1}` [download] If you are running an outdated version of Perl, you can do something similar with back references: `while ($var =~ s/(bar.?)bar/$1foo/) {1}` [download] Update: Corrected code, as I had a weird brain fail on reading the spec. #11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.	[reply] [d/l] [select]
Re^2: Regex match and replace by Anonymous Monk on Mar 04, 2015 at 22:25 UTC
Thank you for your detailed answer, that seems like an interesting way of doing it. Thanks!	[reply]
Re: Regex match and replace by AnomalousMonk (Archbishop) on Mar 04, 2015 at 18:04 UTC
`$s = s/bar(?=bar)/foo/;` This statement does a substitution against the string held in the `$_` default scalar, and assigns to `$s` the number of times a substitution occurred. As I do not know what, if anything, was in `$_` when you ran your code, it's hard to tell why the substitution failed (zero substitutions occurred). You may want a statement like `$s =~ s/bar(?=bar)/foo/;` which can actually do a substitution on `$s`. Input foofoo Output foobar I'm confused by this because you write that you want to replace some 'bar' in the input string with 'foo', but there ain't no 'bar' thar! Can you please clarify with exact input and desired output strings? Update: Making some assumptions about what the input string might be, it's possible to do what you want using a positive look-behind: `c:\@Work\Perl>perl -wMstrict -le "my $s = 'barbar'; $s =~ s/(?<=bar)bar/foo/; print qq{'$s'}; " 'barfoo'` [download] (substitute a `'bar'` with `'foo'` if it is preceded by a `'bar'`). Give a man a fish: `<%-(-(-(-<`	[reply] [d/l] [select]
Re^2: Regex match and replace by Anonymous Monk on Mar 04, 2015 at 22:26 UTC
I see the problem that I had now, thanks for your answer.	[reply]
Re: Regex match and replace by hdb (Monsignor) on Mar 04, 2015 at 21:10 UTC
You could also literally count the number of matches and only start replacing from the second match onwards: `use strict; use warnings; my $s="foofoo"; my $ctr = 0; $s =~ s/(foo)/$ctr++?"bar":$1/ge; print "$s\n";` [download]	[reply] [d/l]
Re^2: Regex match and replace by Anonymous Monk on Mar 04, 2015 at 22:30 UTC
That's pretty neat, thanks!	[reply]
Re: Regex match and replace by choroba (Cardinal) on Mar 05, 2015 at 09:20 UTC
TIMTOWTDI: `my $s = 'foo\|foo\|foo'; my $p = do { $s =~ /foo/g; pos $s }; $s =~ s/foo/pos $s >= $p ? 'bar' : 'foo'/ge; print "$s\n";` [download] لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply] [d/l]