How can I replace a newline ("\n") with a space ("") using the sed command?
I unsuccessfully tried:
sed 's#\n# #g' file
sed 's#^$# #g' file
How do I fix it?
sed is intended to be used on line-based input. Although it can do what you need.
A better option here is to use the tr command as follows:
tr '\n' ' ' < input_filename
or remove the newline characters entirely:
tr -d '\n' < input.txt > output.txt
or if you have the GNU version (with its long options)
tr --delete '\n' < input.txt > output.txt
Use this solution with GNU sed:
sed ':a;N;$!ba;s/\n/ /g' file
This will read the whole file in a loop (':a;N;$!ba), then replaces the newline(s) with a space (s/\n/ /g). Additional substitutions can be simply appended if needed.
Explanation:
sed starts by reading the first line excluding the newline into the pattern space.
Create a label via :a.
Append a newline and next line to the pattern space via N.
If we are before the last line, branch to the created label $!ba ($! means not to do it on the last line. This is necessary to avoid executing N again, which would terminate the script if there is no more input!).
Finally the substitution replaces every newline with a space on the pattern space (which is the whole file).
Here is cross-platform compatible syntax which works with BSD and OS X's sed (as per #Benjie comment):
sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/ /g' file
As you can see, using sed for this otherwise simple problem is problematic. For a simpler and adequate solution see this answer.
Fast answer
sed ':a;N;$!ba;s/\n/ /g' file
:a create a label 'a'
N append the next line to the pattern space
$! if not the last line, ba branch (go to) label 'a'
s substitute, /\n/ regex for new line, / / by a space, /g global match (as many times as it can)
sed will loop through step 1 to 3 until it reach the last line, getting all lines fit in the pattern space where sed will substitute all \n characters
Alternatives
All alternatives, unlike sed will not need to reach the last line to begin the process
with bash, slow
while read line; do printf "%s" "$line "; done < file
with perl, sed-like speed
perl -p -e 's/\n/ /' file
with tr, faster than sed, can replace by one character only
tr '\n' ' ' < file
with paste, tr-like speed, can replace by one character only
paste -s -d ' ' file
with awk, tr-like speed
awk 1 ORS=' ' file
Other alternative like "echo $(< file)" is slow, works only on small files and needs to process the whole file to begin the process.
Long answer from the sed FAQ 5.10
5.10. Why can't I match or delete a newline using the \n escape
sequence? Why can't I match 2 or more lines using \n?
The \n will never match the newline at the end-of-line because the
newline is always stripped off before the line is placed into the
pattern space. To get 2 or more lines into the pattern space, use
the 'N' command or something similar (such as 'H;...;g;').
Sed works like this: sed reads one line at a time, chops off the
terminating newline, puts what is left into the pattern space where
the sed script can address or change it, and when the pattern space
is printed, appends a newline to stdout (or to a file). If the
pattern space is entirely or partially deleted with 'd' or 'D', the
newline is not added in such cases. Thus, scripts like
sed 's/\n//' file # to delete newlines from each line
sed 's/\n/foo\n/' file # to add a word to the end of each line
will NEVER work, because the trailing newline is removed before
the line is put into the pattern space. To perform the above tasks,
use one of these scripts instead:
tr -d '\n' < file # use tr to delete newlines
sed ':a;N;$!ba;s/\n//g' file # GNU sed to delete newlines
sed 's/$/ foo/' file # add "foo" to end of each line
Since versions of sed other than GNU sed have limits to the size of
the pattern buffer, the Unix 'tr' utility is to be preferred here.
If the last line of the file contains a newline, GNU sed will add
that newline to the output but delete all others, whereas tr will
delete all newlines.
To match a block of two or more lines, there are 3 basic choices:
(1) use the 'N' command to add the Next line to the pattern space;
(2) use the 'H' command at least twice to append the current line
to the Hold space, and then retrieve the lines from the hold space
with x, g, or G; or (3) use address ranges (see section 3.3, above)
to match lines between two specified addresses.
Choices (1) and (2) will put an \n into the pattern space, where it
can be addressed as desired ('s/ABC\nXYZ/alphabet/g'). One example
of using 'N' to delete a block of lines appears in section 4.13
("How do I delete a block of specific consecutive lines?"). This
example can be modified by changing the delete command to something
else, like 'p' (print), 'i' (insert), 'c' (change), 'a' (append),
or 's' (substitute).
Choice (3) will not put an \n into the pattern space, but it does
match a block of consecutive lines, so it may be that you don't
even need the \n to find what you're looking for. Since GNU sed
version 3.02.80 now supports this syntax:
sed '/start/,+4d' # to delete "start" plus the next 4 lines,
in addition to the traditional '/from here/,/to there/{...}' range
addresses, it may be possible to avoid the use of \n entirely.
A shorter awk alternative:
awk 1 ORS=' '
Explanation
An awk program is built up of rules which consist of conditional code-blocks, i.e.:
condition { code-block }
If the code-block is omitted, the default is used: { print $0 }. Thus, the 1 is interpreted as a true condition and print $0 is executed for each line.
When awk reads the input it splits it into records based on the value of RS (Record Separator), which by default is a newline, thus awk will by default parse the input line-wise. The splitting also involves stripping off RS from the input record.
Now, when printing a record, ORS (Output Record Separator) is appended to it, default is again a newline. So by changing ORS to a space all newlines are changed to spaces.
GNU sed has an option, -z, for null-separated records (lines). You can just call:
sed -z 's/\n/ /g'
The Perl version works the way you expected.
perl -i -p -e 's/\n//' file
As pointed out in the comments, it's worth noting that this edits in place. -i.bak will give you a backup of the original file before the replacement in case your regular expression isn't as smart as you thought.
Who needs sed? Here is the bash way:
cat test.txt | while read line; do echo -n "$line "; done
In order to replace all newlines with spaces using awk, without reading the whole file into memory:
awk '{printf "%s ", $0}' inputfile
If you want a final newline:
awk '{printf "%s ", $0} END {printf "\n"}' inputfile
You can use a character other than space:
awk '{printf "%s|", $0} END {printf "\n"}' inputfile
tr '\n' ' '
is the command.
Simple and easy to use.
Three things.
tr (or cat, etc.) is absolutely not needed. (GNU) sed and (GNU) awk, when combined, can do 99.9% of any text processing you need.
stream != line based. ed is a line-based editor. sed is not. See sed lecture for more information on the difference. Most people confuse sed to be line-based because it is, by default, not very greedy in its pattern matching for SIMPLE matches - for instance, when doing pattern searching and replacing by one or two characters, it by default only replaces on the first match it finds (unless specified otherwise by the global command). There would not even be a global command if it were line-based rather than STREAM-based, because it would evaluate only lines at a time. Try running ed; you'll notice the difference. ed is pretty useful if you want to iterate over specific lines (such as in a for-loop), but most of the times you'll just want sed.
That being said,
sed -e '{:q;N;s/\n/ /g;t q}' file
works just fine in GNU sed version 4.2.1. The above command will replace all newlines with spaces. It's ugly and a bit cumbersome to type in, but it works just fine. The {}'s can be left out, as they're only included for sanity reasons.
Why didn't I find a simple solution with awk?
awk '{printf $0}' file
printf will print the every line without newlines, if you want to separate the original lines with a space or other:
awk '{printf $0 " "}' file
The answer with the :a label ...
How can I replace a newline (\n) using sed?
... does not work in freebsd 7.2 on the command line:
( echo foo ; echo bar ) | sed ':a;N;$!ba;s/\n/ /g'
sed: 1: ":a;N;$!ba;s/\n/ /g": unused label 'a;N;$!ba;s/\n/ /g'
foo
bar
But does if you put the sed script in a file or use -e to "build" the sed script...
> (echo foo; echo bar) | sed -e :a -e N -e '$!ba' -e 's/\n/ /g'
foo bar
or ...
> cat > x.sed << eof
:a
N
$!ba
s/\n/ /g
eof
> (echo foo; echo bar) | sed -f x.sed
foo bar
Maybe the sed in OS X is similar.
Easy-to-understand Solution
I had this problem. The kicker was that I needed the solution to work on BSD's (Mac OS X) and GNU's (Linux and Cygwin) sed and tr:
$ echo 'foo
bar
baz
foo2
bar2
baz2' \
| tr '\n' '\000' \
| sed 's:\x00\x00.*:\n:g' \
| tr '\000' '\n'
Output:
foo
bar
baz
(has trailing newline)
It works on Linux, OS X, and BSD - even without UTF-8 support or with a crappy terminal.
Use tr to swap the newline with another character.
NULL (\000 or \x00) is nice because it doesn't need UTF-8 support and it's not likely to be used.
Use sed to match the NULL
Use tr to swap back extra newlines if you need them
You can use xargs:
seq 10 | xargs
or
seq 10 | xargs echo -n
cat file | xargs
for the sake of completeness
If you are unfortunate enough to have to deal with Windows line endings, you need to remove the \r and the \n:
tr '\r\n' ' ' < $input > $output
I'm not an expert, but I guess in sed you'd first need to append the next line into the pattern space, bij using "N". From the section "Multiline Pattern Space" in "Advanced sed Commands" of the book sed & awk (Dale Dougherty and Arnold Robbins; O'Reilly 1997; page 107 in the preview):
The multiline Next (N) command creates a multiline pattern space by reading a new line of input and appending it to the contents of the pattern space. The original contents of pattern space and the new input line are separated by a newline. The embedded newline character can be matched in patterns by the escape sequence "\n". In a multiline pattern space, the metacharacter "^" matches the very first character of the pattern space, and not the character(s) following any embedded newline(s). Similarly, "$" matches only the final newline in the pattern space, and not any embedded newline(s). After the Next command is executed, control is then passed to subsequent commands in the script.
From man sed:
[2addr]N
Append the next line of input to the pattern space, using an embedded newline character to separate the appended material from the original contents. Note that the current line number changes.
I've used this to search (multiple) badly formatted log files, in which the search string may be found on an "orphaned" next line.
In response to the "tr" solution above, on Windows (probably using the Gnuwin32 version of tr), the proposed solution:
tr '\n' ' ' < input
was not working for me, it'd either error or actually replace the \n w/ '' for some reason.
Using another feature of tr, the "delete" option -d did work though:
tr -d '\n' < input
or '\r\n' instead of '\n'
I used a hybrid approach to get around the newline thing by using tr to replace newlines with tabs, then replacing tabs with whatever I want. In this case, " " since I'm trying to generate HTML breaks.
echo -e "a\nb\nc\n" |tr '\n' '\t' | sed 's/\t/ <br> /g'`
You can also use this method:
sed 'x;G;1!h;s/\n/ /g;$!d'
Explanation
x - which is used to exchange the data from both space (pattern and hold).
G - which is used to append the data from hold space to pattern space.
h - which is used to copy the pattern space to hold space.
1!h - During first line won't copy pattern space to hold space due to \n is
available in pattern space.
$!d - Clear the pattern space every time before getting the next line until the
the last line.
Flow
When the first line get from the input, an exchange is made, so 1 goes to hold space and \n comes to pattern space, appending the hold space to pattern space, and a substitution is performed and deletes the pattern space.
During the second line, an exchange is made, 2 goes to hold space and 1 comes to the pattern space, G append the hold space into the pattern space, h copy the pattern to it, the substitution is made and deleted. This operation is continued until EOF is reached and prints the exact result.
Bullet-proof solution. Binary-data-safe and POSIX-compliant, but slow.
POSIX sed
requires input according to the
POSIX text file
and
POSIX line
definitions, so NULL-bytes and too long lines are not allowed and each line must end with a newline (including the last line). This makes it hard to use sed for processing arbitrary input data.
The following solution avoids sed and instead converts the input bytes to octal codes and then to bytes again, but intercepts octal code 012 (newline) and outputs the replacement string in place of it. As far as I can tell the solution is POSIX-compliant, so it should work on a wide variety of platforms.
od -A n -t o1 -v | tr ' \t' '\n\n' | grep . |
while read x; do [ "0$x" -eq 012 ] && printf '<br>\n' || printf "\\$x"; done
POSIX reference documentation:
sh,
shell command language,
od,
tr,
grep,
read,
[,
printf.
Both read, [, and printf are built-ins in at least bash, but that is probably not guaranteed by POSIX, so on some platforms it could be that each input byte will start one or more new processes, which will slow things down. Even in bash this solution only reaches about 50 kB/s, so it's not suited for large files.
Tested on Ubuntu (bash, dash, and busybox), FreeBSD, and OpenBSD.
In some situations maybe you can change RS to some other string or character. This way, \n is available for sub/gsub:
$ gawk 'BEGIN {RS="dn" } {gsub("\n"," ") ;print $0 }' file
The power of shell scripting is that if you do not know how to do it in one way you can do it in another way. And many times you have more things to take into account than make a complex solution on a simple problem.
Regarding the thing that gawk is slow... and reads the file into memory, I do not know this, but to me gawk seems to work with one line at the time and is very very fast (not that fast as some of the others, but the time to write and test also counts).
I process MB and even GB of data, and the only limit I found is line size.
Finds and replaces using allowing \n
sed -ie -z 's/Marker\n/# Marker Comment\nMarker\n/g' myfile.txt
Marker
Becomes
# Marker Comment
Marker
You could use xargs — it will replace \n with a space by default.
However, it would have problems if your input has any case of an unterminated quote, e.g. if the quote signs on a given line don't match.
On Mac OS X (using FreeBSD sed):
# replace each newline with a space
printf "a\nb\nc\nd\ne\nf" | sed -E -e :a -e '$!N; s/\n/ /g; ta'
printf "a\nb\nc\nd\ne\nf" | sed -E -e :a -e '$!N; s/\n/ /g' -e ta
To remove empty lines:
sed -n "s/^$//;t;p;"
Using Awk:
awk "BEGIN { o=\"\" } { o=o \" \" \$0 } END { print o; }"
A solution I particularly like is to append all the file in the hold space and replace all newlines at the end of file:
$ (echo foo; echo bar) | sed -n 'H;${x;s/\n//g;p;}'
foobar
However, someone said me the hold space can be finite in some sed implementations.
Replace newlines with any string, and replace the last newline too
The pure tr solutions can only replace with a single character, and the pure sed solutions don't replace the last newline of the input. The following solution fixes these problems, and seems to be safe for binary data (even with a UTF-8 locale):
printf '1\n2\n3\n' |
sed 's/%/%p/g;s/#/%a/g' | tr '\n' # | sed 's/#/<br>/g;s/%a/#/g;s/%p/%/g'
Result:
1<br>2<br>3<br>
It is sed that introduces the new-lines after "normal" substitution. First, it trims the new-line char, then it processes according to your instructions, then it introduces a new-line.
Using sed you can replace "the end" of a line (not the new-line char) after being trimmed, with a string of your choice, for each input line; but, sed will output different lines. For example, suppose you wanted to replace the "end of line" with "===" (more general than a replacing with a single space):
PROMPT~$ cat <<EOF |sed 's/$/===/g'
first line
second line
3rd line
EOF
first line===
second line===
3rd line===
PROMPT~$
To replace the new-line char with the string, you can, inefficiently though, use tr , as pointed before, to replace the newline-chars with a "special char" and then use sed to replace that special char with the string you want.
For example:
PROMPT~$ cat <<EOF | tr '\n' $'\x01'|sed -e 's/\x01/===/g'
first line
second line
3rd line
EOF
first line===second line===3rd line===PROMPT~$
Related
I can't figure how to tell sed dot match new line:
echo -e "one\ntwo\nthree" | sed 's/one.*two/one/m'
I expect to get:
one
three
instead I get original:
one
two
three
sed is line-based tool. I don't think these is an option.
You can use h/H(hold), g/G(get).
$ echo -e 'one\ntwo\nthree' | sed -n '1h;1!H;${g;s/one.*two/one/p}'
one
three
Maybe you should try vim
:%s/one\_.*two/one/g
If you use a GNU sed, you may match any character, including line break chars, with a mere ., see :
.
Matches any character, including newline.
All you need to use is a -z option:
echo -e "one\ntwo\nthree" | sed -z 's/one.*two/one/'
# => one
# three
See the online sed demo.
However, one.*two might not be what you need since * is always greedy in POSIX regex patterns. So, one.*two will match the leftmost one, then any 0 or more chars as many as possible, and then the rightmost two. If you need to remove one, then any 0+ chars as few as possible, and then the leftmost two, you will have to use perl:
perl -i -0 -pe 's/one.*?two//sg' file # Non-Unicode version
perl -i -CSD -Mutf8 -0 -pe 's/one.*?two//sg' file # S&R in a UTF8 file
The -0 option enables the slurp mode so that the file could be read as a whole and not line-by-line, -i will enable inline file modification, s will make . match any char including line break chars, and .*? will match any 0 or more chars as few as possible due to a non-greedy *?. The -CSD -Mutf8 part make sure your input is decoded and output re-encoded back correctly.
You can use python this way:
$ echo -e "one\ntwo\nthree" | python -c 'import re, sys; s=sys.stdin.read(); s=re.sub("(?s)one.*two", "one", s); print s,'
one
three
$
This reads the entire python's standard input (sys.stdin.read()), then substitutes "one" for "one.*two" with dot matches all setting enabled (using (?s) at the start of the regular expression) and then prints the modified string (the trailing comma in print is used to prevent print from adding an extra newline).
This might work for you:
<<<$'one\ntwo\nthree' sed '/two/d'
or
<<<$'one\ntwo\nthree' sed '2d'
or
<<<$'one\ntwo\nthree' sed 'n;d'
or
<<<$'one\ntwo\nthree' sed 'N;N;s/two.//'
Sed does match all characters (including the \n) using a dot . but usually it has already stripped the \n off, as part of the cycle, so it no longer present in the pattern space to be matched.
Only certain commands (N,H and G) preserve newlines in the pattern/hold space.
N appends a newline to the pattern space and then appends the next line.
H does exactly the same except it acts on the hold space.
G appends a newline to the pattern space and then appends whatever is in the hold space too.
The hold space is empty until you place something in it so:
sed G file
will insert an empty line after each line.
sed 'G;G' file
will insert 2 empty lines etc etc.
How about two sed calls:
(get rid of the 'two' first, then get rid of the blank line)
$ echo -e 'one\ntwo\nthree' | sed 's/two//' | sed '/^$/d'
one
three
Actually, I prefer Perl for one-liners over Python:
$ echo -e 'one\ntwo\nthree' | perl -pe 's/two\n//'
one
three
Below discussion is based on Gnu sed.
sed operates on a line by line manner. So it's not possible to tell it dot match newline. However, there are some tricks that can implement this. You can use a loop structure (kind of) to put all the text in the pattern space, and then do the operation.
To put everything in the pattern space, use:
:a;N;$!ba;
To make "dot match newline" indirectly, you use:
(\n|.)
So the result is:
root#u1804:~# echo -e "one\ntwo\nthree" | sed -r ':a;N;$!ba;s/one(\n|.)*two/one/'
one
three
root#u1804:~#
Note that in this case, (\n|.) matches newline and all characters. See below example:
root#u1804:~# echo -e "oneXXXXXX\nXXXXXXtwo\nthree" | sed -r ':a;N;$!ba;s/one(\n|.)*two/one/'
one
three
root#u1804:~#
Given a set of replacement strings in a file replacements.txt like
s/string1/replacement1/g;
s/string2/replacement2/g;
s/string3/replacement3/g;
s/string4/replacement4/g;
s/string5/replacement5/g;
I would like to obtain the equivalent of
sed -f replacements.txt infile.txt
my file is so big that sed cant handle it, while I know that perl could do the trick.
Also the replacements are really a lot, and change from time to time. ( I need to run this a dozen of times)
Note that the replacements are fixed strings, so I do not really need those to be regular expressions.
sed has problems only when the regexp has globs and the input file is a single large line.
The perl equivalent of your sed command is:
perl -p replacements.txt infile.txt
It should work with your sample replacements.txt, given that the s statements are properly ;-terminated (note that sed would recognize the end of the line by itself as the statement terminator).
The real problem, however, is that the entire large file is a single line, so the key to avoiding running out of memory is to:
temporarily break that line into many short lines,
send these short lines through the pipeline and perform the string replacements on them,
and then re-join the modified short lines to form a single line again.
If there's a character in the data that delimits records (units of data), in a away that doesn't interfere with the string replacements, breaking the long line into multiple ones with the help of tr is a viable approach; I'll use } as an example, because Kuzeko states that the data is JSON-like:
If you have GNU sed (Linux; verify with sed --version):
tr '}' '\0' < infile.txt | sed -z -f replacements.txt | tr '\0' '}'
Having tr output NUL-separated "lines" (\0) and sed read them accordingly (-z) is the most robust way to handle the chunking.
Unfortunately, the -z / --null-data option is not POSIX-compliant and the BSD/macOS implementation does not support it.
Otherwise (e.g., on macOS):
tr '}' '\n' < infile.txt | perl -p replacements.txt infile.txt | tr '\n' '}'
Caveat: If the single line in infile.txt has a trailing \n, you'll end up with an extra } char. at the end; to prevent that, add an initial tr stage to the pipeline that deletes the \n:
tr -d '\n' < infile.txt | tr '}' '\n' | ...
perl is still needed, because - unlike BSD/macOS sed - it preserves the trailing-\n-or-not status of the input's last line.
if I have the following:
>AB ABABABA
>AC ACACACA
how do I shift everything onto a newline after the space i.e.
>AB
ABABABABA
>AC
ACACACACA
I have tried:
cat file | sed 's/ /\n/g'
cat file | tr ' ' '\n'
however I get the exact same output.
** UPDATE **
Upon inspecting the file using less and nano, the output was different to using cat. The file contains some terminal escape characters that aren't displayed in cat, but are in less. (how does this even happen?)
This was a terrible bug to spot and everyone has actually posted corrected answers based on the output of cat. So thank you for your help. Could the mods close this one?
It seems that you need to replace (any kind of) space with a newline
perl -pe 's/\s+/\n/' data.txt
This produces the required output in my tests. The -p sets up the loop over input (opening files or using STDIN) and sets $_ to the current line. It also prints $_ each time after processing.
If there are multiple spaces, each to be replaced by \n, add /g modifier.
If there may be more to do you can also capture patterns and replace them
perl -pe 's/\s+(.*)/\n$1/' data.txt
Following the observation in the answer by glenn jackman and looking "more closely" it appears that the first word on the line need be copied to the next line. Then the above is modified to
perl -pe 's/^>(\S+)\K\s+/\n$1';
The \K is a particular form of the positive lookbehind, which asserts that the pattern preceeds the current match position but it discards all previous matches (so you don't have to capture and copy them). You can find it in perlre. Without it the >(\S+) would be consumed so it would have to be copied back in the replacement part, as />$1\n$1/.
Are you trying to move the content from before the space onto the next line as well?
As in >A BC becomes:
>A
ABC
Then one can use sed like this:
$ sed 'h;s/^>\([^ ]*\) /\1/;x;s/ .*/ /;G' file
>AB
ABABABABA
>AC
ACACACACA
Breakdown:
h; # Copy pattern space to hold buffer
s/^>\([^ ]*\) /\1/; # Convert >A BC to ABC
x; # eXchange hold buffer and pattern space
s/ .*/ /; # Remove everything after, but including the
# first space: >A BC -> >A
G # Append hold buffer to pattern space
Looking more closely it looks like you want to repeat the first word on the next line: to transform this
>foo bar
>baz qux
into this
>foo
foobar
>baz
bazqux
If that's true, you can do
sed -r 's/^>([^ ]+) />\1\n\1/' file # or
perl -pe 's/^>(\S+) />$1\n$1/' file
sed is for simple substitutions on individual lines, that is all. For anything else you should be using awk, e.g.:
$ awk '{print $1 ORS substr($1,2) $2}' file
>AB
ABABABABA
>AC
ACACACACA
I would like to delete all "\n" (quotes, new line, quotes) in a text file.
I have tried:
sed 's/"\n"//g' < in > out
and also sed '/"\n"/d' < in > out but non of those seds worked.
What am I doing wrong?
This works with GNU sed on Linux: I don't have a Mac to test with.
sed '
# this reads the whole file into pattern space
:a; N; $ bb; ba; :b
# *now* make the replacement
s/"\n"//g
' <<END
one
two"
"three
four"
five
"six
END
one
twothree
four"
five
"six
This perl command accomplishes the same thing:
perl -0777 -pe 's/"\n"//g'
This awk-oneliner works here, you can give it a try:
awk -F'"\n"' -v RS='\0' -v ORS="" '{$1=$1;print}' file
a small test: tested with gawk
kent$ cat f
foo"
"bar"
"bla"
new line should be kept
this too
kent$ awk -F'"\n"' -v RS='\0' -v ORS="" '{$1=$1;print}' f
foo bar bla"
new line should be kept
this too
If you don't want to have the space between foo and bar blah .., add -v OFS="" to awk
Try this -- you need to escape the backslash to make it literal.
sed 's/"\\n"//g' < in > out
Verified on OSX.
The accepted answer was marked as such because of the Perl command it contains.
The sed command doesn't actually work on OSX, because it uses features specific to GNU sed, whereas OSX use BSD sed.
An equivalent answer requires only a few tweaks - note that this will work with both BSD and GNU sed:
Using multiple -e options:
sed -e ':a' -e '$!{N;ba' -e '}; s/"\n"//g' < in > out
Or, using an ANSI C-quoted string in Bash:
sed $':a\n$!{N;ba\n}; s/"\\n"//g' < in > out
Or, using a multi-line string literal:
sed ':a
$!{N;ba
}; s/"\n"//g' < in > out
BSD sed requires labels (e.g., :a) and branching commands (e.g., b) to be terminated with an actual newline (whereas in GNU sed a ; suffices), or, alternatively, for the script to be broken into multiple -e options, with each part ending where a newline is required.
For a detailed discussion of the differences between GNU and BSD sed, see https://stackoverflow.com/a/24276470/45375
$':a\n$!{N;ba\n}' is a common sed idiom for reading all input lines into the so-called pattern space (buffer on which (subsequent) commands operate):
:a is a label that can be branched to
$! matches every line but the last
{N;ba\n} keeps building the buffer by adding the current line (N) to it, then branching back to label :a to repeat the cycle.
Once the last line is reached, no branching is performed, and the buffer at that point contains all input lines, at which point the desired substitution (s/"\n"//g) is performed on the entire buffer.
As for why the OP's approach didn't work:
sed reads files line by line by default, so by default it can only operate on one line at a time.
In order to be able to replace newline chars. - i.e., to operate across multiple lines - you must explicitly read multiple/all lines first, as above.
instead of sed you could also use tr, I've tested it and for me it worked
tr -d '"\\n"' < input.txt > output.txt
We are receiving a file that is delimited into rows with the \(newline) and columns with the \(tab) character.
When there is a manual newline present in one of the "fields" of the file, it comes in as a special newline with two backslashes (\\newline).
To remove the special tabs \(tab), we are using this sed command, which works correctly:
sed "s/$(printf '\\\\\t')/ /g"
The corresponding command for newlines, however does not:
sed "s/$(printf '\\\\\n')/ /g"
It does not remove the \n, only the backslash before it. Is there special handling that needs to be done to remove \(newline)?
Clarification: normal newlines are formatted like this:
\(newline)
Wheras the special characters that need removal are
\\(newline)
Here you go:
echo -e 'hello\\\nthere' | perl -ne 's/\\\n/ /; print'
It would be difficult (but probably possible) to do this in sed, because sed processes input line by line, and your data is broken into multiple lines. This perl one-liner processes the input line by line, and since it treats the newline character as part of the line, it can perform a substitution with space, which I think has the effect that you want.
Or if you prefer awk:
echo -e 'hello\\\nthere' | awk '{ if (gsub(/\\$/, " ")) printf; else print }'
At first I suspected your "special newline" character is just the string \\n like in the output of this command:
echo 'hello\\nthere'
You can replace the string \\n with a space like this:
echo 'hello\\nthere' | sed -e 's/\\\\n/ /g'
You can use tr (translate) command as well to do this, like
tr '\n' ' ' < inputfile.txt
EdIT: In that case use it like
tr '\\\n' ' ' < inputfile.txt