How to insert a newline in front of a pattern?

user_72321 0
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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
sed shell
17 Answers

Some of the other answers didn't work for my version of sed. Switching the position of & and \n did work.

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

Edit: This doesn't seem to work on OS X, unless you install gnu-sed.

user_723210

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.

tgamblin_91220

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 .

user_21123820

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'
user_194790
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