I have a file in which some lines contain a json object on a single line, and I want to extract the value of the window_indicator property.
A normal regular expression is: "window_indicator":\s*([\-\d\.]+) in which I want the value of the fist match group.
Here it is working perfectly well: https://regex101.com/r/w9Iuch/1
I've settled on sed because it seems that grep has to print the whole line and can't limit to the match group value, and perl is overkill.
Unfortunately, sed isn't actually capable of doing this, is it?
# sed 's/("window_indicator:)/\1/' in.txt
sed: -e expression #1, char 26: invalid reference \1 on `s' command's RHS
# sed -E 's/("window_indicator":)/\1/p' in.txt
prints out every line of the file
# sed -rn 's/("window_indicator":)/\1/p' in.txt
prints the whole line
# sed -rn 's/("window_indicator":)/\1/' in.txt
nothing
With sed, you need to match the whole line, capture what you need, replace the whole match with Group 1 placeholder, and make sure you suppress the default line output and only print the new text after successful substitution:
sed -nE 's/.*"window_indicator":[[:space:]]*([-0-9.]+).*/\1/p' in.txt
If the first match is to be retrieved, add q to quit:
sed -nE 's/.*"window_indicator":[[:space:]]*([-0-9.]+).*/\1/p;q' in.txt
Note that \d is not supported in POSIX regex, it is replaced with 0-9 range in the bracket expression here.
Details
n - suppress default line output
E - enables POSIX ERE flavor
.*"window_indicator":[[:space:]]*([-0-9.]+).* - finds
.* - any text
"window_indicator": - a fixed string
[[:space:]]* - zero or more whitespaces (GNU sed supports \s, too)
([-0-9.]+) - Group 1: one or more digits, - or .
.* - any text
\1 - replaces with Group 1 value
p - prints the result upon successful replacement
q - quits processing the stream.
With GNU grep, it is even easier:
grep -oP '"window_indicator":\s*\K[-\d.]+' in.txt
To get the first match,
grep -oP '"window_indicator":\s*\K[-\d.]+' in.txt | head -1
Here,
o - outputs matched texts only
P - enables the PCRE regex engine
"window_indicator":\s*\K[-\d.]+ - matches
"window_indicator": - a fixed string
\s* - zero or more whitespaces
\K - removes the text matched so far from the match value
[-\d.]+ - matches one or more -, . or digits.
1st solution: With your shown samples please try following awk code. Though its always advised to use json parsers like: jq. Simple explanation would be, using match function of awk here, where using regex "window_indicator":[0-9]+} in it to match needed value. If regex is successfully matched then creating variable val which has sub-string of matched regex in current line. Then substituting "window_indicator": and } with NULL in val and printing val which will give needed value.
awk '
match($0,/"window_indicator":[0-9]+}/){
val=substr($0,RSTART,RLENGTH)
gsub(/"window_indicator":|}/,"",val)
print val
}
' Input_file
2nd solution: Using GNU grep where using positive look ahead and positive look behind mechanism and getting the expected output as per requirement.
grep -oP '(?<="window_indicator":)\d+(?=})' Input_file
Using sed
$ sed -E 's/.*window_indicator":([0-9]+).*/\1/' input_file
0
Using grep
$ grep -Po '.*window_indicator":\K\d+' input_file
0
Using awk
$ awk '{match($0,/.*window_indicator":([0-9]+)/,arr);print arr[1]}' input_file
0
I have a huge file that contains lines that follow this format:
New-England-Center-For-Children-L0000392290
Southboro-Housing-Authority-L0000392464
Crew-Star-Inc-L0000391998
Saxony-Ii-Barber-Shop-L0000392491
Test-L0000392334
What I'm trying to do is narrow it down to just this:
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Test
Can anyone help with this?
Using GNU awk:
awk -F\- 'NF--' OFS=\- file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Set the input and output field separator to -.
NF contains number of fields. Reduce it by 1 to remove the last field.
Using sed:
sed 's/\(.*\)-.*/\1/' file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Simple greedy regex to match up to the last hyphen.
In replacement use the captured group and discard the rest.
Version 1 of the Question
The first version of the input was in the form of HTML and parts had to be removed both before and after the desired text:
$ sed -r 's|.*[A-Z]/([a-zA-Z-]+)-L0.*|\1|' input
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
Version 2 of the Question
In the revised question, it is only necessary to remove the text that starts with -L00:
$ sed 's|-L00.*||' input2
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
Both of these commands use a single "substitute" command. The command has the form s|old|new|.
The perl code for this would be: perl -nle'print $1 if(m{-.*?/(.*?-.*?)-})
We can break the Regex down to matching the following:
- for that's between the city and state
.*? match the smallest set of character(s) that makes the Regex work, i.e. the State
/ matches the slash between the State and the data you want
( starts the capture of the data you are interested in
.*?-.*? will match the data you care about
) will close out the capture
- will match the dash before the L####### to give the regex something to match after your data. This will prevent the minimal Regex from matching 0 characters.
Then the print statement will print out what was captured (your data).
awk likes these things:
$ awk -F[/-] -v OFS="-" '{print $(NF-3), $(NF-2)}' file
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
This sets / and - as possible field separators. Based on them, it prints the last_field-3 and last_field-2 separated by the delimiter -. Note that $NF stands for last parameter, hence $(NF-1) is the penultimate, etc.
This sed is also helpful:
$ sed -r 's#.*/(\w*-\w*)-\w*\.\w*</loc>$#\1#' file
Special-Restaurant
Eliot-Cleaning
Kennedy-Plumbing
It selects the block word-word after a slash / and followed with word.word</loc> + end_of_line. Then, it prints back this block.
Update
Based on your new input, this can make it:
$ sed -r 's/(.*)-L\w*$/\1/' file
New-England-Center-For-Children
Southboro-Housing-Authority
Crew-Star-Inc
Saxony-Ii-Barber-Shop
Test
It selects everything up to the block -L + something + end of line, and prints it back.
You can use also another trick:
rev file | cut -d- -f2- | rev
As what you want is every slice of - separated fields, let's get all of them but last one. How? By reversing the line, getting all of them from the 2nd one and then reversing back.
Here's how I'd do it with Perl:
perl -nle 'm{example[.]com/bp/(.*?)/(.*?)-L\d+[.]htm} && print $2' filename
Note: the original question was matching input lines like this:
<loc>http://www.example.com/bp/Lowell-MA/Special-Restaurant-L0000423916.htm</loc>
<loc>http://www.example.com/bp/Houston-TX/Eliot-Cleaning-L0000422797.htm</loc>
<loc>http://www.example.com/bp/New-Orleans-LA/Kennedy-Plumbing-L0000423121.htm</loc>
The -n option tells Perl to loop over every line of the file (but not print them out).
The -l option adds a newline onto the end of every print
The -e 'perl-code' option executes perl-code for each line of input
The pattern:
/regex/ && print
Will only print if the regex matches. If the regex contains capture parentheses you can refer to the first captured section as $1, the second as $2 etc.
If your regex contains slashes, it may be cleaner to use a different regex delimiter ('m' stands for 'match'):
m{regex} && print
If you have a modern Perl, you can use -E to enable modern feature and use say instead of print to print with a newline appended:
perl -nE 'm{example[.]com/bp/(.*?)/(.*?)-L\d+[.]htm} && say $2' filename
This is very concise in Perl
perl -i.bak -lpe's/-[^-]+$//' myfile
Note that this will modify the input file in-place but will keep a backup of the original data in called myfile.bak
I work with strings like
abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf
and I need to get a new one where I remove in the original string everything from the beginning till the last appearance of "_" and the next characters (can be 3, 4, or whatever number)
so in this case I would get
_adf
How could I do it with "sed" or another bash tool?
Regular expression pattern matching is greedy. Hence ^.*_ will match all characters up to and including the last _. Then just put the underscore back in:
echo abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf | sed 's/^.*_/_/'
sed 's/^(.*)_([^_]*)$/_\2/' < input.txt
Do you need to modify the string, or just find everything after the last underscore? The regex to find the last _{anything} would be /(_[^_]+)$/ ($ matches the end of the string), or if you also want to match a trailing underscore with nothing after it, /(_[^_]*)$/.
Unless you really need to modify the string in place instead of just finding this piece, or you really want to do this from the command line instead of a script, this regex is a bit simpler (you tagged this with perl, so I wasn't sure quite how committed to using just the command line as opposed to a simple script you were).
If you do need to modify the string in place, sed -i 's/(_[^_]+)$/\1/' myfile or sed -i 's/(_[^_]+)$/\1/g' myfile. The -i (edit: I decided not to be lazy and look up the proper syntax...) the -i flag will just overwrite the old file with the new one. If you want to create a new file and not clobber the old one, sed -e 's/.../.../g' oldfile > newfile. The g after the s/// will do this for all instances in the file you pass into sed; leaving it out just replaces the first instance.
If the string is not by itself at the end of the line, but rather embedded in other text. but just separated by whitespace, replace the $ with \s, which will match a whitespace character (the end of a word).
If you have strings like these in bash variables (I don't see that specified in the question), you can use parameter expansion:
s="abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf"
t="_${s##*_}"
echo "$t" # ==> _adf
In Perl, you could do this:
my $string = "abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf";
if ( $string =~ m/(_[^_]+)$/ ) {
print $1;
}
[Edit]
A Perl one liner approach (ie, can be run from bash directly):
perl -lne 'm/(_[^_]+)$/ && print $1;' infile > outfile
Or using substitution:
perl -pe 's/.*(_[^_]+)$/$1/' infile > outfile
Just group the last non-underscore characters preceded by the last underscore with \(_[^_]*\), then reference this group with \1:
sed 's/^.*\(_[^_]*\)$/\1/'
Result:
$ echo abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf | sed 's/^.*\(_[^_]*\)$/\1/'
_adf
A Perl way:
echo 'abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf' | \
perl -e 'print ((split/(_)/,<>)[-2..-1])'
output:
_adf
Just for fun:
echo abc_dsdsds_ss_gsgsdsfsdf_ewew_wewewewewew_adf | tr _ '\n' | tail -n 1 | rev | tr '\n' _ | rev
I've got a file called 'res' that's 29374 characters of http data in a one-line string. Inside it, there are several http links, but I only want to be display those that end in '/idNNNNNNNNN' where N is a digit. In fact I'm only interested in the string 'idNNNNNNNNN'.
I've tried with:
cat res | sed -n '0,/.*\(id[0-9]*\).*/s//\1/p'
but I get the whole file.
Do you know a way to do it?
perl -n -E 'say $1 while m!/id(\d{9})!g' input-file
should work. That assumes exactly 9 digits; that's the {9} in the above. You can match 8 or 9 ({8,9}), 8 or more ({8,}), up to 9 ({0,9}), etc.
Example of this working:
$ echo -n 'junk jumk http://foo/id231313 junk lalala http://bar/id23123 asda' | perl -n -E 'say $1 while m!id(\d{0,9})!g'
231313
23123
That's with the 0 to 9 variant, of course.
If you're stuck with a pre-5.10 perl, use -e instead of -E and print "$1\n" instead of say $1.
How it works
First is the two command-line arguments to Perl. -n tells Perl to read input from standard input or files given on the command line, line by line, setting $_ to each line. $_ is perl's default target for a lot of things, including regular expression matches. -E merely tells Perl that the next argument is a Perl one-liner, using the new language features (vs. -e which does not use the 5.10 extensions).
So, looking at the one liner: say means to print out some value, followed by a newline. $1 is the first regular expression capture (captures are made by parentheses in regular expressions). while is a looping construct, which you're probably familiar with. m is the match operator, the ! after it is the regular expression delimiter (normally, you see / here, but since the pattern contains / it's easier to use something else, so you don't have to escape the / as \/). /id(\d{9}) is the regular expression to match. Keep in mind that the delimiter is !, so the / is not special, it just matches a literal /. The parentheses form a capture group, so $1 will be the number. The ! is the delimiter, followed by g which means to match as many times as possible (as opposed to once). This is what makes it pick up all the URLs in the line, not just the first. As long as there is a match, the m operator will return a true value, so the loop will continue (and run that say $1, printing out the match).
Two-sed solution
I think this is one way to do this with only sed. Much more complicated!
echo 'junk jumk http://foo/id231313 junk lalala http://bar/id23123 asda' | \
sed 's!http://!\nhttp://!g' | \
sed 's!^.*/id\([0-9]*\).*$!\1!'
cat res | perl -ne 'chomp; print "$1\n" if m/\/(id\d*)/'
The trouble is that sed and grep and awk work on lines, and you've only got one line. So, you probably need to split things up so you have more than one line -- then you can make the normal tools work.
tr ':' '\012' < res |
sed -n 's%.*/\(id[0-9][0-9]*\).*%\1%p'
This takes advantage of URLs containing colons and maps colons to newlines with tr, then uses sed to pick up anything up to a slash, followed by id and one or more digits, followed by anything, and prints out the id and digit string (only). Since these only occur in URLs, they will only appear one per line and relatively near the start of the line too.
Here's a solution using only one invocation of sed:
sed -n 's| |\n|g;/^http/{s|http://[^/]*/id\([0-9]*\)|\1|;P};D' inputfile
Explanation:
s| |\n|g; - Divide and conquer
/^http/{ - If pattern space begins with "http"
s|http://[^/]*/id\([0-9]*\)|\1|; - capture the id
P - Print the string preceding the first newline
}; - end if
D - Delete the string preceding the first newline regardless of whether it contains "http"
Edit:
This version uses the same technique but is more selective.
sed -n 's|http://|\n&|g;/^\n*http/{s|\n*http://[^/]*/id\([0-9]*\)|\1\n|;P};D' inputfile
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' $'\x01'|sed -e 's/\x01/===/g'
first line
second line
3rd line
EOF
first line===second line===3rd line===PROMPT~$