Re^2: Regexp substitution using variables

It's working, and crucially supports back-references

That's interesting as I cannot get this to support back-references...this is very similar to my initial attempt. So I have attempted to replicate it:

use strict;

my $pattern     = '(testing)';
my $replacement = 'New \1';
my $flags    = 'i';

my $value    = 'My Testing Text';

$replacement =~ s/\\/\\\\/g;
eval "\$value =~ s/$pattern/$replacement/$flags";

print "$value\n";
[download]

This prints

  My New \1 Text
[download]

It doesn't substitute the capture.

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Re^3: Regexp substitution using variables (updated)
by AnomalousMonk (Archbishop) on Nov 26, 2020 at 03:19 UTC

The OPed sort of problem is tricky, but for this specific iteration:

Win8 Strawberry 5.8.9.5 (32)  Wed 11/25/2020 22:12:13
C:\@Work\Perl\monks
>perl
use strict;
use warnings;

my $pattern     = '(testing)';
my $replacement = 'New \U$1';
my $flags       = 'i';

my $value = 'My Testing Text';

### $replacement =~ s/\\/\\\\/g;
print "replacement '$replacement' \n";

eval "\$value =~ s/$pattern/$replacement/$flags";

print "$value\n";
^Z
replacement 'New \U$1'
My New TESTING Text
[download]

\1

$1

Update: Here's a version of the example code that better illustrates the process of building the evaluation string:

Win8 Strawberry 5.8.9.5 (32)  Wed 11/25/2020 22:45:44
C:\@Work\Perl\monks
>perl
use strict;
use warnings;

my $pattern     = '(testing)';
my $replacement = 'New \U$1';
my $flags       = 'i';

my $value = 'My Testing Text';

print "replacement '$replacement' \n";

my $eval_string = "\$value =~ s/$pattern/$replacement/$flags";
print "eval_string '$eval_string' \n";

eval $eval_string;

print "$value\n";
^Z
replacement 'New \U$1'
eval_string '$value =~ s/(testing)/New \U$1/i'
My New TESTING Text
[download]

Give a man a fish: <%-{-{-{-<

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