Replacing string using perl containing $ sign with other string containing $ sign and bash variable

Question

I want to replace some strings in files that contain $ signs with an other string that also contains $ signs plus the value from a bash variable. The file contains strings like the following.

$Rev: 12345 $ $Author: 12345 $

Lets say I have a bash variable called i containing the string foo. I now want to replace $Rev: 12345 $ with $Rev: $i $. I tried using sed but since sed doesn't support non-greedy regex I switched to perl. Perl works fine when I don't use any bash variables.

# cat file
$Rev: 12345 $ $Author: 12345 $
# perl -p -i -e 's;\$Rev:.*?\$;\$Rev: test \$;g' file
# cat file
$Rev: test $ $Author: 12345 $

But no matter how I escape the $ signs in the command, I cannot get it to work with a bash variable.

# cat file
$Rev: 12345 $ $Author: 12345 $
# i="foo"
# echo $i
foo
# perl -p -i -e "s;\$Rev:.*?\$;\$Rev: $i \$;g" file
Final $ should be \$ or $name at -e line 1, within string
syntax error at -e line 1, near "s;$Rev:.*?$;$Rev: foo $;g"
Execution of -e aborted due to compilation errors.
# perl -p -i -e "s;\\$Rev:.*?\\$;\$Rev: $i \\$;g" file
# cat file
$Rev: foo $ $Author: foo $

Thanks for your help!

These auto-updated version comments were thought to be a good idea back in the RCS and CSV days, but experience taught us that they were not. — tripleee
– tripleee, Commented Sep 29, 2022 at 13:04
Why do you set a bash variable to use as replacement in Perl code? Do you just not know how to set variables in Perl? Or is this a variable that will be used in other code? — TLP
– TLP, Commented Sep 29, 2022 at 15:11

choroba · Accepted Answer · 2022-09-29 15:38:21Z

4

It's better to use a Perl variable inside Perl, not let the shell expand its variable into Perl code.

perl -i~ -pe 's/^\$Rev:.*?\$/\$Rev: $i \$/g' -s -- -i="$i" file

You can then use single quotes around the code which makes it much easier to backslash stuff correctly.

The -s tells Perl to accept -var=val switches and set Perl variable $var to val before running the code.

edited Sep 29, 2022 at 15:38

answered Sep 29, 2022 at 13:05

choroba

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4 Comments

Jim Davis Over a year ago

I like this better than my suggestion.

Chullian Over a year ago

This is in fact a cleaner solution but it will also output the replaced string. I could rederict the output to /dev/null like so "perl -pe 's;\$Rev:.*?\$;\$Rev: $i \$;g' -s -- -i="$i" file> /dev/null 2>&1" but that seems unnecessarily complicated compared to @JimDavis solution. It propably doesn't matter which one I use. What do you think?

TLP Over a year ago

@Chullian What do you mean "it will output the replaced string"? Are you talking about the fact that choroba left out the -i (in-place edit) switch?

choroba Over a year ago

@Chullian: Updated to include the -i.

Jim Davis · Accepted Answer · 2022-09-29 13:05:54Z

0

Try putting $i outside the quotes entirely:

perl -pi -e 's;\$Rev:.*?\$;\$Rev: '$i' \$;g' file
#           ^---------------------^  ^-----^

Or, you could export the variable to sub-processes, and access it via the %ENV hash in Perl:

export i="foo"
perl -pi -e 's;\$Rev:.*?\$;\$Rev: $ENV{i} \$;g' file

answered Sep 29, 2022 at 13:05

Jim Davis

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