How can I replace everything after a string using Bash? - language-agnostic

Question

I have a Perl script that uses some local variables as per below:
my $cool_variable="Initial value";
COOLVAR="Initial value for COOLVAR"
I would like to replace the content between the quotes using a bash script.
I got it to work for a non-variable like below:
#!/bin/sh
dummy_var="Replaced value"
sed -i -r "s#^(COOLVAR=).*#\1$dummy_var#" perlscript.pl
But if I replace it with cool_variable or $cool_variable:
sed -i -r "s#^($cool_variable=).*#\1$dummy_var#" perlscript.pl
It does not work..

Answer 1

The are multiple code injection bugs in that snippet. You shouldn't be generating code from the shell or sed.
Say you have
var=COOLVAR
val=coolval
As per How can I process options using Perl in -n or -p mode?, you can use any of
perl -spe's{^$var=\K.*}{"\Q$val\E";};' -- -var="$var" -val="$val" perlscript.pl
var=var val=val perl -pe's{^$ENV{var}=\K.*}{"\Q$ENV{val}\E";};' perlscript.pl
export var
export val
perl -pe's{^$ENV{var}=\K.*}{"\Q$ENV{val}\E";};' perlscript.pl
to transform
COOLVAR="dummy";
HOTVAR="dummy";
into
COOLVAR="coolvar";
HOTVAR="dummy";
The values are passed to the program using arguments to avoid injecting them into the fixer, and the fixer uses Perl's quotemeta (aka \Q..\E) to quote special characters.
Note that $var is assumed to be a valid identifier. No validation checks are performed. This program is absolutely unsafe using untrusted input.
Use -i to modify the file in place.