Passing a regex substitution as a variable in Perl?

Question

I need to pass a regex substitution as a variable:

sub proc {
    my $pattern = shift;
    my $txt = \"foo baz\";

    $txt =~ $pattern;
}

my $pattern = \'s         


        

Answer 1

sub proc {
    my($match, $subst) = @_;
    my $txt = "foo baz";
    $txt =~ s/$match/$subst/;
    print "$txt\n";
}

my $matcher = qr/foo/;
my $sub_str = "bar";

proc($matcher, $sub_str);

This rather directly answers your question. You can do more - but when I used a qr// term instead of the $sub_str as a simple literal, then the expanded regex was substituted.

I recently needed to create a parser (test parser) for statements with some peculiar (dialect of) SQL types, recognizing lines such as this, splitting it into three type names:

input: datetime year to second,decimal(16,6), integer

The script I used to demo this used quoted regexes.

#!/bin/perl -w
use strict;
while (<>)
{
    chomp;
    print "Read: <$_>\n";
    my($r1) = qr%^input\s*:\s*%i;
    if ($_ =~ $r1)
    {
        print "Found input:\n";
        s%$r1%%;
        print "Residue: <$_>\n";
        my($r3) = qr%(?:year|month|day|hour|minute|second|fraction(?:\([1-5]\))?)%;
        my($r2) = qr%
                        (?:\s*,?\s*)?   # Commas and spaces
                        (
                            (?:money|numeric|decimal)(?:\(\d+(?:,\d+)?\))?   |
                            int(?:eger)?  |
                            smallint      |
                            datetime\s+$r3\s+to\s+$r3
                        )
                    %ix;
        while ($_ =~ m/$r2/)
        {
            print "Got type: <$1>\n";
            s/$r2//;
        }
        print "Residue 2: <$_>\n";
    }
    else
    {
        print "No match:\n";
    }
    print "Next?\n";
}

We can argue about the use of names like $r1, etc. But it did the job...it was not, and is not, production code.