Is there a better way to double escape (or escape twice) user input in BASH than calling printf twic


This script will read an old and a new value from the user and then use sed to find and replace them in a file. For example if I entered T<em>T</em>z and B<em>B</em>z it would look for T\<em>T\</em>z in the file and replace with B\<em>B\</em>z. It works but I've been trying to make this more concise.

I don't have any need for the intermediate variables $ESC_OLD_PW and $ESC_NEW_PW. Is there a more <strong><em>sensible</em></strong> way to do this?

#!/bin/bash read -sp "Old:" OLD_PW && echo read -sp "New:" NEW_PW && echo # Add escape characters to what user entered printf -v ESC_OLD_PW "%q" "${OLD_PW}" printf -v ESC_NEW_PW "%q" "${NEW_PW}" # Escape again for the sed evaluation. printf -v ESC_ESC_OLD_PW "%q" "${ESC_OLD_PW}" printf -v ESC_ESC_NEW_PW "%q" "${ESC_NEW_PW}" sed -i -e s/"${ESC_ESC_OLD_PW}"/"${ESC_ESC_NEW_PW}"/g $1

I've tried the following:

~$ OLD_PW="T*T*z" ~$ printf "%q" $OLD_PW | xargs printf "%q" printf: %q: invalid conversion specification ~$

And I've tried many variations on piping things into printf... Any suggestions?


Intermediate variable-free:

sed -i -e "s/$(printf '%q' $(printf '%q' $OLD_PW))/$(printf '%q' $(printf '%q' $NEW_PW))/g" $1


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