I want to replace the string using a shell variable.
today =`date '+%Y%m%d'`
perl -p -i -e 's/file"20221212"/file="${today}"/g'
The expectation is file="20221215".
But it failed, result is file="".
How to escape this case?
I want to replace the string using a shell variable.
today =`date '+%Y%m%d'`
perl -p -i -e 's/file"20221212"/file="${today}"/g'
The expectation is file="20221215".
But it failed, result is file="".
How to escape this case?
Your issue is shell quoting. Using single quotes, ', to delimit the Perl code disables all variable expansion. You need to use double quotes, ", to get variable expansion.
Using shell without any Perl to illustrate the issue
today=`date '+%Y%m%d'`
echo 'today is $today'
will output this -- with no expansion of $today
today is $today
now with double quotes
today=`date '+%Y%m%d'`
echo "today is $today"
outputs this -- $today has been expanded.
today is 20221215
Applying that to your code (I've removed the -i option to make it easier to see the results) and escaped all double quotes in the perl code with \"
echo 'file"20221212"' >input.txt
perl -p -e "s/file\"20221212\"/file=\"${today}\"/g" input.txt
gives
file="20221215"
Here I am giving you the hint how it should be:
I cannot reproduce the issue from your code.
#!/bin/sh
today=`date '+%Y%m%d'`
echo "today:$today"
file="input.txt";
perl -pi -e 's/file20221212/file='${today}'/g' $file
where input.txt contains:
file20221212
The two earlier answers are horrible. Don't generate Perl code from the shell.
This question was already asked and answered here: How can I process options using Perl in -n or -p mode?
It provides three solution, including
perl -spe's/file\K"20221212"/="$today"/g' -- -today="$today"
TODAY="$today" perl -pe's/file\K"20221212"/="$ENV{today}"/g'
Alternatively, you could also avoid date and passing a value entirely.
perl -MPOSIX -pe'
BEGIN { $today = strftime "%Y%m%d", localtime }
s/file\K"20221212"/="$today"/g
'