How to escape special characters when echoing variable in Bash script?

Question

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How should I handle escaping special characters in a string like this:

^VfAro@khm=Y)@5,^AyxPO[[$2jW#+Vg;Paj2ycIj8VUr5z1,}qR~JnK7r_0@$ov

I am using this string in a variable which is piped to cryptsetup luksOpen inside the Bash script:

echo "${string}" | cryptsetup luksOpen UUID={...}

But when executing the script, I got some characters stripped. When echoing the plain string, all the characters are preserved.

I have tried with different variable enclosing as well as with printf "%q" but without any viable result.

Answer 1

If you do want a trailing newline on the input to cryptsetup:

s='^VfAro@khm=Y)@5,^AyxPO[[$2jW#+Vg;Paj2ycIj8VUr5z1,}qR~JnK7r_0@$ov'
printf '%s\n' "$s" | cryptsetup luksOpen

If you don't want a trailing newline:

s='^VfAro@khm=Y)@5,^AyxPO[[$2jW#+Vg;Paj2ycIj8VUr5z1,}qR~JnK7r_0@$ov'
printf '%s' "$s" | cryptsetup luksOpen

In both cases, we don't use echo when we care about byte-for-byte perfect output.