How to insert a newline in front of a pattern?

Question

Not how to insert a newline before a line. This is asking how to insert a newline before a pattern within a line.

For example,

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

will insert a newline behind the regexp pattern.

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

Here is an example input file

somevariable (012)345-6789

Should become

somevariable

(012)345-6789

Answer 1

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.

Answer 2

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'

Answer 3

I tried some of the examples but they didn't seem to work. However, I found a solution. I simply had to switch the position of & and \n.

Thus

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

I'd like to thank you guys anyways though.