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How can I print a newline as \n in bash?

Tags:

bash

shell

Basically, I want to achieve something like the inverse of echo -e.

I have a variable which stores a command output, but I want to print newlines as \n.

like image 439
BubuIIC Avatar asked May 24 '14 18:05

BubuIIC


4 Answers

Here's my solution:

sed 's/$/\\n/' | tr -d '\n'
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Vytenis Bivainis Avatar answered Oct 20 '22 21:10

Vytenis Bivainis


If your input is already in a (Bash) shell variable, say $varWithNewlines:

echo "${varWithNewlines//$'\n'/\\n}"

It simply uses Bash parameter expansion to replace all newline ($'\n') instances with literal '\n' each.


If your input comes from a file, use AWK:

awk -v ORS='\\n' 1

In action, with sample input:

# Sample input with actual newlines created with ANSI C quoting ($'...'),
# which turns `\n` literals into actual newlines.
varWithNewlines=$'line 1\nline 2\nline 3'

# Translate newlines to '\n' literals.
# Note the use of `printf %s` to avoid adding an additional newline.
# By contrast, a here-string - <<<"$varWithNewlines" _always appends a newline_.
printf %s "$varWithNewlines" | awk -v ORS='\\n' 1
  • awk reads input line by line
  • by setting ORS- the output record separator to literal '\n' (escaped with an additional \ so that awk doesn't interpret it as an escape sequence), the input lines are output with that separator
  • 1 is just shorthand for {print}, i.e., all input lines are printed, terminated by ORS.

Note: The output will always end in literal '\n', even if your input does not end in a newline.

This is because AWK terminates every output line with ORS, whether the input line ended with a newline (separator specified in FS) or not.


Here's how to unconditionally strip the terminating literal '\n' from your output.

# Translate newlines to '\n' literals and capture in variable.
varEncoded=$(printf %s "$varWithNewlines" | awk -v ORS='\\n' 1)

# Strip terminating '\n' literal from the variable value
# using Bash parameter expansion.
echo "${varEncoded%\\n}"

By contrast, more work is needed if you want to make the presence of a terminating literal '\n' dependent on whether the input ends with a newline or not.

# Translate newlines to '\n' literals and capture in variable.
varEncoded=$(printf %s "$varWithNewlines" | awk -v ORS='\\n' 1)

# If the input does not end with a newline, strip the terminating '\n' literal.
if [[ $varWithNewlines != *$'\n' ]]; then
  # Strip terminating '\n' literal from the variable value
  # using Bash parameter expansion.
  echo "${varEncoded%\\n}"
else
  echo "$varEncoded"
fi
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mklement0 Avatar answered Oct 20 '22 21:10

mklement0


You can use printf "%q":

eol=$'\n'
printf "%q\n" "$eol"
$'\n'
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anubhava Avatar answered Oct 20 '22 21:10

anubhava


A Bash solution

x=$'abcd\ne fg\nghi'
printf "%s\n" "$x"
abcd
e fg
ghi
y=$(IFS=$'\n'; set -f; printf '%s\\n' $x)
y=${y%??}
printf "%s\n" "$y"
abcd\ne fg\nghi
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iruvar Avatar answered Oct 20 '22 22:10

iruvar