Re^3: Regexp substitution using variables

Well, this gets me much of what I need ... but I can't get back-references, either with $1 or \1. In either case, they appear as literals. Any ideas, other than eval?

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Re^4: Regexp substitution using variables by LanX (Saint) on Nov 25, 2020 at 23:56 UTC
> but I can't get back-references, This works for me `DB<44> $pat='(x)\1' DB<45> $mod='i' DB<46> 'xX' =~ /(?$mod)$pat/; say "$&:$1" xX:x DB<47>` [download] Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l]
Re^5: Regexp substitution using variables by MikeTaylor (Sexton) on Nov 26, 2020 at 10:34 UTC
Sorry, I think I was using the term "back-reference" incorrectly. I didn't mean referring back within the pattern to something earlier in the pattern, but referring in the substitution string to a captured sub-expression of the pattern. Something like `s/(foo\|bar)/He said "$1"/`.	[reply]