sed help: matching and replacing a literal "\n" (not the newline)

Question

i have a file which contains several instances of \n.

i would like to replace them with actual newlines, but sed doesn't recognize the \n.

i tried

sed -r -e 's/\n/\n/'
sed -r -e 's/\\n/\n/'
sed -r -e 's/[\n]/\n/'

and many other ways of escaping it.

is sed able to recognize a literal \n? if so, how?

is there another program that can read the file interpreting the \n's as real newlines?

Answer 1

Can you please try this

sed -i 's/\\n/\n/g' input_filename

Answer 2

What exactly works depends on your sed implementation. This is poorly specified in POSIX so you see all kinds of behaviors.

The -r option is also not part of the POSIX standard; but your script doesn't use any of the -r features, so let's just take it out. (For what it's worth, it changes the regex dialect supported in the match expression from POSIX "basic" to "extended" regular expressions; some sed variants have an -E option which does the same thing. In brief, things like capturing parentheses and repeating braces are "extended" features.)

On BSD platforms (including MacOS), you will generally want to backslash the literal newline, like this:

sed 's/\\n/\
/g' file

On some other systems, like Linux (also depending on the precise sed version installed -- some distros use GNU sed, others favor something more traditional, still others let you choose) you might be able to use a literal \n in the replacement string to represent an actual newline character; but again, this is nonstandard and thus not portable.

If you need a properly portable solution, probably go with Awk or (gasp) Perl.

perl -pe 's/\\n/\n/g' file

In case you don't have access to the manuals, the /g flag says to replace every occurrence on a line; the default behavior of the s/// command is to only replace the first match on every line.