Perl one-liner substitution without further expansion of a shell variable?

Question

I have a script to replace a specific email address in various files. The replacement address is the first parameter to the script:

#!/bin/bash
perl -pi -e s/'name\@domain\.org'/$1/ file-list

This doesn't work, as the @ character in $1 is substituted by perl. Is there a straightforward fix for this? Running the script as subst foo\@bar.com, subst foo\\@bar.com, subst "[email protected]", and so on, doesn't work. Is there a sed script that could handle this more easily?

Answer 1

Instead of expanding a shell variable directly in the perl code you can pass it as an argument by setting the s switch:

#!/usr/bin/env bash

perl -i -spe 's/name\@domain\.org/$replacement/' -- -replacement="$1" file1.txt file2.txt

In perl's s///, without the use of e or ee modifiers, variables in the replacement part are treated as literals so you don't need to escape them.

Answer 2

This works, but needs you to pass the new mail address to the script with the @ character preceded by \\:

#!/bin/bash
perl -pi -e "s/name\@domain.org/$1/" file-list

If the script is subst, run as:

subst newname\\@example.com

This is a better alternative, which uses sed to carry out the escaping:

#!/bin/bash
ADR="$(echo "$1" | sed -e 's/@/\\\@/')"
perl -pi -e "s/name\@domain.org/$ADR/" file-list

Of course, in this case it's probably better to use sed to do the whole thing.