How to replace "\n" string with a new line in Unix Bash script

Question

Cannot seem to find an answer to this one online...

I have a string variable (externally sourced) with new lines "\n" encoded as strings.

I want to replace those strings with actual new line carriage returns. The code below can achieve this...

echo $EXT_DESCR | sed 's/\\n/\n/g'

But when I try to store the result of this into it's own variable, it converts them back to strings

NEW_DESCR=`echo $EXT_DESCR | sed 's/\\n/\n/g'`

How can this be achieved, or what I'm I doing wrong?

Here's my code I've been testing to try get the right results

EXT_DESCR="This is a text\nWith a new line"
echo $EXT_DESCR | sed 's/\\n/\n/g'

NEW_DESCR=`echo $EXT_DESCR | sed 's/\\n/\n/g'`
echo ""
echo "$NEW_DESCR"

Answer 1

No need for sed, using parameter expansion:

$ foo='1\n2\n3'; echo "${foo//'\n'/$'\n'}"  
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With bash 4.4 or newer, you can use the E operator in ${parameter@operator}:

$ foo='1\n2\n3'; echo "${foo@E}"
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Answer 2

Other answers contain alternative solutions. (I especially like the parameter expansion one.)

Here's what's wrong with your attempt:

In

echo $EXT_DESCR | sed 's/\\n/\n/g'

the sed command is in single quotes, so sed gets s/\\n/\n/g as is.

In

NEW_DESCR=`echo $EXT_DESCR | sed 's/\\n/\n/g'`

the whole command is in backticks, so a round of backslash processing is applied. That leads to sed getting the code s/\n/\n/g, which does nothing.

A possible fix for this code:

NEW_DESCR=`echo $EXT_DESCR | sed 's/\\\\n/\\n/g'`

By doubling up the backslashes, we end up with the right command in sed.

Or (easier):

NEW_DESCR=$(echo $EXT_DESCR | sed 's/\\n/\n/g')

Instead of backticks use $( ), which has less esoteric escaping rules.

Note: Don't use ALL_UPPERCASE for your shell variables. UPPERCASE is (informally) reserved for system variables such as HOME and special built-in variables such as IFS or RANDOM.