i wanted to capture \n in a variable (assume a) and substitute $a in front of search pattern in a file (abc.txt)

can somebody suggest me the solution ?

> cat abc.txt
hi how are you how is your health
> echo $a

expected output ( search pattern - how)

hi \
how are you \
how is your health
  • a=$(printf "\n") ? – Raman Sailopal Nov 21 '17 at 13:47
  • @RamanSailopal, that wouldn't work as command substitution strips trailing newline characters. – Stéphane Chazelas Nov 21 '17 at 13:51
  • What does "in front of a search pattern in a file" mean? – B Layer Nov 21 '17 at 13:54

In a Bourne-like or rc-like shell

' perl -pe 's/pattern/$ENV{a}$&/g' < abc.txt

would pass a a environment variable to perl containing a newline character and perl would insert the content of that variable before each occurrence of pattern in the content of abc.txt.

To assign any character x other than ' to a variable in Bourne-like shells, you can use:


So for backslash:


For newline:


For control characters like newline, but that would be even worse for backspace or carriage-return for instance, however, that renders the code less legible.

ksh93 introduced a new $'...' form of quotes to specify those characters in an escaped form. Other shells like bash, zsh, mksh or FreeBSD sh have copied and extended (like the $'\uXXXX' form added by zsh) it since.

In those shells, you can use:

a=$'\n' a=$'\b' a=$'\r' a=$'\\' a=$'\''

To assign a newline, backspace, carriage return, backslash or single quote character.

That quoting operator is not standard yet, so you can't use it in a portable sh script. A standard alternative is to use:

a=$(printf '\b')

for backspace for instance, but that doesn't work for newline as command substitution ($(cmd)) strips trailing newline characters from the output of cmd. One has to resort to hacks like:

nl=$(printf '\n.'); nl=${nl%.}

(that is add a . and remove it afterwards) in order to preserve it.

Note that perl also understands those \n notations in its code, so you can do:

perl -pe 's/pattern/\n$&/g' < abc.txt

The GNU implementation of sed (and a few others nowadays) also support it:

sed 's/pattern/\n&/g' < abc.txt

but that's not standard. Standardly, a newline in the replacement part of the s command in sed has to be expressed with a backslash followed by an actually newline, so that would be:

backslash='\' newline='
sed "s/pattern/$backslash$newline&/g" < abc.txt

Or without the variables:

sed 's/pattern/\
&/g' < abc.txt

So, if you want to insert both a backslash and a newline in front of pattern, you need:

sed 's/pattern/\\\
&/g' < abc.txt

3 backslashes in total, two for a literal backslash and one for the newline. Or with perl:

perl -pe 's/pattern/\\\n$&/g' < abc.txt

Or with your a variable:

' perl -pe 's/pattern/$ENV{a}$&/g' < abc.txt
  • ======= > echo $a \ > a=' ' perl -pe 's/how/$ENV{a}$&/g' < abc.txt hi how are you =========== @StéphaneChazelas .. this is not printing \ (backslash) – Vikas Venna Nov 21 '17 at 14:42
  • @Vika, see edit. I see that you've now clarified your question. – Stéphane Chazelas Nov 21 '17 at 14:45
  • it worked .. thank you ============== > sed 's/how/\\\ > &/g' < abc.txt hi \ how are you \ how is your health ============ when I use sed -i (to edit the same file ) it is complaining "sed: no input files". can you help me here ? – Vikas Venna Nov 21 '17 at 15:06

You can also use the own bash commands using:

cat abc.txt | sed -e 's/how/\nhow/g' |  while read line; do echo "$line \\"; done
  • this is printing \ at the end of all lines. – Vikas Venna Nov 22 '17 at 9:20

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