[shell] How to insert a newline in front of a pattern?

How to insert a newline before a pattern within a line?

For example, this will insert a newline behind the regex pattern.

sed 's/regex/&\n/g'

How can I do the same but in front of the pattern?

Given this sample input file, the pattern to match on is the phone number.

some text (012)345-6789

Should become

some text
(012)345-6789

To insert a newline to output stream on Linux, I used:

sed -i "s/def/abc\\\ndef/" file1

Where file1 was:

def

Before the sed in-place replacement, and:

abc
def

After the sed in-place replacement. Please note the use of \\\n. If the patterns have a " inside it, escape using \".

This works in bash and zsh, tested on Linux and OS X:

sed 's/regexp/\'$'\n/g'

In general, for $ followed by a string literal in single quotes bash performs C-style backslash substitution, e.g. $'\t' is translated to a literal tab. Plus, sed wants your newline literal to be escaped with a backslash, hence the \ before $. And finally, the dollar sign itself shouldn't be quoted so that it's interpreted by the shell, therefore we close the quote before the $ and then open it again.

Edit: As suggested in the comments by @mklement0, this works as well:

sed $'s/regexp/\\\n/g'

What happens here is: the entire sed command is now a C-style string, which means the backslash that sed requires to be placed before the new line literal should now be escaped with another backslash. Though more readable, in this case you won't be able to do shell string substitutions (without making it ugly again.)

This works in MAC for me

sed -i.bak -e 's/regex/xregex/g' input.txt sed -i.bak -e 's/qregex/\'$'\nregex/g' input.txt

Dono whether its perfect one...

On my mac, the following inserts a single 'n' instead of newline:

sed 's/regexp/\n&/g'

This replaces with newline:

sed "s/regexp/\\`echo -e '\n\r'`/g"

Hmm, just escaped newlines seem to work in more recent versions of sed (I have GNU sed 4.2.1),

dev:~/pg/services/places> echo 'foobar' | sed -r 's/(bar)/\n\1/;'
foo
bar

in sed you can reference groups in your pattern with "\1", "\2", .... so if the pattern you're looking for is "PATTERN", and you want to insert "BEFORE" in front of it, you can use, sans escaping

sed 's/(PATTERN)/BEFORE\1/g'

i.e.

  sed 's/\(PATTERN\)/BEFORE\1/g'

You can also do this with awk, using -v to provide the pattern:

awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file

This checks if a line contains a given pattern. If so, it appends a new line to the beginning of it.

See a basic example:

$ cat file
hello
this is some pattern and we are going ahead
bye!
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' file
hello
this is some 
pattern and we are going ahead
bye!

Note it will affect to all patterns in a line:

$ cat file
this pattern is some pattern and we are going ahead
$ awk -v patt="pattern" '$0 ~ patt {gsub(patt, "\n"patt)}1' d
this 
pattern is some 
pattern and we are going ahead

sed -e 's/regexp/\0\n/g'

\0 is the null, so your expression is replaced with null (nothing) and then...
\n is the new line

On some flavors of Unix doesn't work, but I think it's the solution to your problem.

echo "Hello" | sed -e 's/Hello/\0\ntmow/g'
Hello
tmow

In my case the below method works.

sed -i 's/playstation/PS4/' input.txt

Can be written as:

sed -i 's/playstation/PS4\nplaystation/' input.txt

PS4
playstation

Consider using \\n while using it in a string literal.

• sed : is stream editor

• -i : Allows to edit the source file

• +: Is delimiter.

I hope the above information works for you .

In this case, I do not use sed. I use tr.

cat Somefile |tr ',' '\012' 

This takes the comma and replaces it with the carriage return.

In vi on Red Hat, I was able to insert carriage returns using just the \r character. I believe this internally executes 'ex' instead of 'sed', but it's similar, and vi can be another way to do bulk edits such as code patches. For example. I am surrounding a search term with an if statement that insists on carriage returns after the braces:

:.,$s/\(my_function(.*)\)/if(!skip_option){\r\t\1\r\t}/

Note that I also had it insert some tabs to make things align better.

In sed, you can't add newlines in the output stream easily. You need to use a continuation line, which is awkward, but it works:

$ sed 's/regexp/\
&/'

Example:

$ echo foo | sed 's/.*/\
&/'

foo

See here for details. If you want something slightly less awkward you could try using perl -pe with match groups instead of sed:

$ echo foo | perl -pe 's/(.*)/\n$1/'

foo

$1 refers to the first matched group in the regular expression, where groups are in parentheses.

echo one,two,three | sed 's/,/\
/g'

After reading all the answers to this question, it still took me many attempts to get the correct syntax to the following example script:

#!/bin/bash
# script: add_domain
# using fixed values instead of command line parameters $1, $2
# to show typical variable values in this example
ipaddr="127.0.0.1"
domain="example.com"
# no need to escape $ipaddr and $domain values if we use separate quotes.
sudo sed -i '$a \\n'"$ipaddr www.$domain $domain" /etc/hosts

The script appends a newline \n followed by another line of text to the end of a file using a single sed command.

You can use perl one-liners much like you do with sed, with the advantage of full perl regular expression support (which is much more powerful than what you get with sed). There is also very little variation across *nix platforms - perl is generally perl. So you can stop worrying about how to make your particular system's version of sed do what you want.

In this case, you can do

perl -pe 's/(regex)/\n$1/'

-pe puts perl into a "execute and print" loop, much like sed's normal mode of operation.

' quotes everything else so the shell won't interfere

() surrounding the regex is a grouping operator. $1 on the right side of the substitution prints out whatever was matched inside these parens.

Finally, \n is a newline.

Regardless of whether you are using parentheses as a grouping operator, you have to escape any parentheses you are trying to match. So a regex to match the pattern you list above would be something like

\(\d\d\d\)\d\d\d-\d\d\d\d

\( or \) matches a literal paren, and \d matches a digit.

Better:

\(\d{3}\)\d{3}-\d{4}

I imagine you can figure out what the numbers in braces are doing.

Additionally, you can use delimiters other than / for your regex. So if you need to match / you won't need to escape it. Either of the below is equivalent to the regex at the beginning of my answer. In theory you can substitute any character for the standard /'s.

perl -pe 's#(regex)#\n$1#'
perl -pe 's{(regex)}{\n$1}'

A couple final thoughts.

using -ne instead of -pe acts similarly, but doesn't automatically print at the end. It can be handy if you want to print on your own. E.g., here's a grep-alike (m/foobar/ is a regex match):

perl -ne 'if (m/foobar/) {print}'

If you are finding dealing with newlines troublesome, and you want it to be magically handled for you, add -l. Not useful for the OP, who was working with newlines, though.

Bonus tip - if you have the pcre package installed, it comes with pcregrep, which uses full perl-compatible regexes.

echo pattern | sed -E -e $'s/^(pattern)/\\\n\\1/'

worked fine on El Captitan with () support