Perl string replace: Match, but not replace, part of regex

Question

Say I have a string in Perl I am trying to match and replace with stuff:

$string =~ s/[^a-zA-Z]$find[^a-zA-Z]/$replace/g;

So as shown, I want to replace everything that is surrounded on both sides by nonletter characters. However, when I replace the string, I do NOT want to also replace these characters: they are just necessary for correct matching. How can I tell the Perl regex to avoid replacing the things surrounding $find?

Do you want to locate text that is sandwiched between non-alphabetic characters? Or do you want to match a pattern but excluding alphabetic characters? The two tests are not the same at the beginning or end of the string. — 200_success
– 200_success, Commented Nov 10, 2015 at 22:46
My intent was to do the first of the two. Both sln's and hjpotter92's answers accomplish this, not sure about ergonaut's. — Brandon Slaght
– Brandon Slaght, Commented Nov 10, 2015 at 23:41

clemep · Accepted Answer · 2022-02-04 14:57:55Z

4

Use perl lookaround assertions.

s/(?<=[^a-zA-Z])$find(?=[^a-zA-Z])/$replace/g

edited Feb 4, 2022 at 14:57

clemep

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answered Nov 10, 2015 at 22:33

user557597

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hjpotter92 · Accepted Answer · 2015-11-10 22:27:58Z

4

Store them as a matched group, and reference them in the replacement string:

$string =~ s/([^a-z])$find([^A-Z])/\1$replace\2/gi;

answered Nov 10, 2015 at 22:27

hjpotter92

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1 Comment

ThisSuitIsBlackNot Over a year ago

Although using \1 and \2 on the RHS of a substitution works, perldoc perlre calls it "a dirty habit to get into." You should use $1 and $2 instead.

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