Bash – echoing captured variable removes newline

September 23, 2023

I have a private key I want to both print to stdout and transform to a public key which should also be printed.

Based on this answer: Piping echo output to another echo

echo "pkcs12 here" |\ 
  openssl rsa |\
  { var=$(cat); echo "$var\n$(echo $var | openssl rsa -pubout)"; }

This works fine in ZSH, but I’ve found out that in Bash the innermost echo $var removes newline characters, which causes the openssl command to fail

Now that we know what I want to do, let’s simplify the problem a bit:

$ echo "a\nb\nc"
a
b
c

echo "a\nb\nc" | { var=$(cat); echo "$(echo $var)"; }
a b c

How do I preserve new lines in the simplified example?

>Solution :

Like this:

echo "a\nb\nc" | { var=$(cat); echo "$(echo -e "$var")"; }
a
b  
c

Learn how to quote properly in shell, it’s very important :

"Double quote" every literal that contains spaces/metacharacters and every expansion: "$var", "$(command "$var")", "${array[@]}", "a & b". Use 'single quotes' for code or literal $'s: 'Costs $5 US', ssh host 'echo "$HOSTNAME"'. See
http://mywiki.wooledge.org/Quotes
http://mywiki.wooledge.org/Arguments
https://web.archive.org/web/20230224010517/https://wiki.bash-hackers.org/syntax/words
when-is-double-quoting-necessary