Is there a better way to double access (or avoid twice) user input in BASH than calling printf twice?

Question

This script will read the old and new value from the user, and then use sed to find and replace them in the file. For example, if I entered TTz and BBz, it would look for T \ T \ z in the file and replace it with B \ B \ z. This works, but I tried to make it more concise.

I do not need the intermediate variables $ ESC_OLD_PW and $ ESC_NEW_PW. Is there a reasonable 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 tried the following:

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

And I tried a lot of variations on things in printf ... Any suggestions?

Answer 1

Intermediate variable:

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