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Re: Regex backreference

by AnomalousMonk (Archbishop)
on Jun 12, 2013 at 18:26 UTC ( [id://1038537]=note: print w/replies, xml ) Need Help??


in reply to Regex backreference

This is regarding regex backreference.

I'm not sure you're using the right terminology, but here's an example in which a regex with a backreference is entered, matches and captures:

>perl -wMstrict -le "$_ = '1:A silly sentence (495,a), silly but useful.(3)'; ;; print 'Enter a regular expression:'; my $expression = <STDIN>; chomp($expression); print qq{expression is '$expression'}; ;; if (/$expression/) { print 'The expression matches the string'; print qq{\$1 is '$1'} if defined $1; print qq{\$2 is '$2'} if defined $2; print qq{\$3 is '$3'} if defined $3; } else { print 'The expression does not match'; } " Enter a regular expression: (\w+).*(\1) expression is '(\w+).*(\1)' The expression matches the string $1 is 'silly' $2 is 'silly'

See discussion of backreferences in Capture groups in perlre.

Replies are listed 'Best First'.
Re^2: Regex backreference
by vyeddula (Acolyte) on Jun 12, 2013 at 20:49 UTC

    I copied and executed the same script but it is not prompting me what stored in $1 or $2 in spite of pattern match.

      BTW: Here's a version that handles any number of capture groups.
      Question: Why does capture group 2 ($2) in the  (\w+).*(\d{2,}).*(\1) example only capture '95'? Shouldn't  (\d{2,}) match and capture "the maximum of 2 or more decimal digits", i.e., '495', as it did in the third example?

      >perl -wMstrict -le "$_ = '1:A silly sentence (495,a), silly but useful.(3)'; ;; EXPRESSION: { print qq{\n}; print 'Enter a regular expression:'; my $expression = <STDIN>; last EXPRESSION unless $expression =~ m{ \S }xms; chomp($expression); print qq{Expression is '$expression'}; ;; if (! defined($expression = eval qq{qr/$expression/})) { print qq{Regex error: $@}; redo EXPRESSION; } ;; if ($_ !~ $expression) { print 'The expression does not match the string'; redo EXPRESSION; } print 'The expression matches the string'; ;; if ($#- < 1) { print qq{No capture groups}; redo EXPRESSION; } ;; for my $cg (1 .. $#-) { printf qq{capture group \$$cg is '%s' starting at offset %d \n}, substr($_, $-[$cg], $+[$cg]-$-[$cg]), $-[$cg]; } redo EXPRESSION; } ;; print 'done'; " Enter a regular expression: foo Expression is 'foo' The expression does not match the string Enter a regular expression: \d{2,} Expression is '\d{2,}' The expression matches the string No capture groups Enter a regular expression: (\d{2,}) Expression is '(\d{2,})' The expression matches the string capture group $1 is '495' starting at offset 20 Enter a regular expression: (\w+).*(\1) Expression is '(\w+).*(\1)' The expression matches the string capture group $1 is 'silly' starting at offset 4 capture group $2 is 'silly' starting at offset 28 Enter a regular expression: (\w+).*(\d{2,}).*(\1) Expression is '(\w+).*(\d{2,}).*(\1)' The expression matches the string capture group $1 is 'silly' starting at offset 4 capture group $2 is '95' starting at offset 21 capture group $3 is 'silly' starting at offset 28 Enter a regular expression: \d*** Expression is '\d***' Regex error: Nested quantifiers in regex; marked by <-- HERE in m/\d** <-- HERE */ at (eval 6) line 1, <STDIN> line 6. Enter a regular expression: silly Expression is 'silly' The expression matches the string No capture groups Enter a regular expression: done

      what pattern did you enter?

        Made a mistake..Rectified..Thanks

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