I need to delete the "end of line" of the previous line when current line starts is not a number ^[!0-9], basically if match, append to the line before, I'm a sed & awk n00b, and really like them btw. thanks
edit:
$ cat file
1;1;1;text,1
2;4;;8;some;1;1;1;more
100;tex
t
broke
4564;1;1;"also
";12,2121;546465
$ "script" file
1;1;1;text,1
2;4;;8;some;1;1;1;more
100;text broke
4564;1;1;"also";12,2121;546465
You didn't post any sample input or expected output so this is a guess but it sounds like what you're asking for:
$ cat file
a
b
3
4
c
d
$ awk '{printf "%s%s",(NR>1 && /^[[:digit:]]/ ? ORS : ""),$0} END{print ""}' file
ab
3
4cd
On the OPs newly posted input:
$ awk '{printf "%s%s",(NR>1 && /^[[:digit:]]/ ? ORS : ""),$0} END{print ""}' file
1;1;1;text,1
2;4;;8;some;1;1;1;more
100;textbroke
4564;1;1;"also";12,2121;546465
This might work for you (GNU sed):
sed -r ':a;$!N;s/\n([^0-9]|$)/\1/;ta;P;D' file
Keep two lines in the pattern space and if the start of the second line is empty or does not start with an integer, remove the newline.
if you have Ruby on your system
array = File.open("file").readlines
array.each_with_index do |val,ind|
array[ind-1].chomp! if not val[/^\d/] # just chomp off the previous item's \n
end
puts array.join
output
# ruby test.rb
1;1;1;text,1
2;4;;8;some;1;1;1;more
100;textbroke
4564;1;1;"also";12,2121;546465
Related
I have a awk script from this example:
awk '/START/{if (x) print x; x="";}{x=(!x)?$0:x","$0;}END{print x;}' file
Here's a sample file with lines:
$ cat file
START
1
2
3
4
5
end
6
7
START
1
2
3
end
5
6
7
So I need to stop concatenating when destination string would contain end word, so the desired output is:
START,1,2,3,4,5,end
START,1,2,3,end
Short Awk solution (though it will check for /end/ pattern twice):
awk '/START/,/end/{ printf "%s%s",$0,(/^end/? ORS:",") }' file
The output:
START,1,2,3,4,5,end
START,1,2,3,end
/START/,/end/ - range pattern
A range pattern is made of two patterns separated by a comma, in the
form ‘begpat, endpat’. It is used to match ranges of consecutive
input records. The first pattern, begpat, controls where the range
begins, while endpat controls where the pattern ends.
/^end/? ORS:"," - set delimiter for the current item within a range
here is another awk
$ awk '/START/{ORS=","} /end/ && ORS=RS; ORS!=RS' file
START,1,2,3,4,5,end
START,1,2,3,end
Note that /end/ && ORS=RS; is shortened form of /end/{ORS=RS; print}
You can use this awk:
awk '/START/{p=1; x=""} p{x = x (x=="" ? "" : ",") $0} /end/{if (x) print x; p=0}' file
START,1,2,3,4,5,end
START,1,2,3,end
Another way, similar to answers in How to select lines between two patterns?
$ awk '/START/{ORS=","; f=1} /end/{ORS=RS; print; f=0} f' ip.txt
START,1,2,3,4,5,end
START,1,2,3,end
this doesn't need a buffer, but doesn't check if START had a corresponding end
/START/{ORS=","; f=1} set ORS as , and set a flag (which controls what lines to print)
/end/{ORS=RS; print; f=0} set ORS to newline on ending condition. Print the line and clear the flag
f print input record as long as this flag is set
Since we seem to have gone down the rabbit hole with ways to do this, here's a fairly reasonable approach with GNU awk for multi-char RS, RT, and gensub():
$ awk -v RS='end' -v OFS=',' 'RT{$0=gensub(/.*(START)/,"\\1",1); $NF=$NF OFS RT; print}' file
START,1,2,3,4,5,end
START,1,2,3,end
I have a file in this pattern:
Some text
---
## [Unreleased]
More text here
I need to replace the text between '---' and '## [Unreleased]' with something else in a shell script.
How can it be achieved using sed or awk?
Perl to the rescue!
perl -lne 'my #replacement = ("First line", "Second line");
if ($p = (/^---$/ .. /^## \[Unreleased\]/)) {
print $replacement[$p-1];
} else { print }'
The flip-flop operator .. tells you whether you're between the two strings, moreover, it returns the line number relative to the range.
This might work for you (GNU sed):
sed '/^---/,/^## \[Unreleased\]/c\something else' file
Change the lines between two regexp to the required string.
This example may help you.
$ cat f
Some text
---
## [Unreleased]
More text here
$ seq 1 5 >mydata.txt
$ cat mydata.txt
1
2
3
4
5
$ awk '/^---/{f=1; while(getline < c)print;close(c);next}/^## \[Unreleased\]/{f=0;next}!f' c="mydata.txt" f
Some text
1
2
3
4
5
More text here
awk -v RS="\0" 'gsub(/---\n\n## \[Unreleased\]\n/,"something")+1' file
give this line a try.
An awk solution that:
is portable (POSIX-compliant).
can deal with any number of lines between the start line and the end line of the block, and potentially with multiple blocks (although they'd all be replaced with the same text).
reads the file line by line (as opposed to reading the entire file at once).
awk -v new='something else' '
/^---$/ { f=1; next } # Block start: set flag, skip line
f && /^## \[Unreleased\]$/ { f=0; print new; next } # Block end: unset flag, print new txt
! f # Print line, if before or after block
' file
I'd like to know if I can remove a \n (newline) only if the current line has one ore more keywords from a list; for instance, I want to remove the \n if it contains the words hello or world.
Example:
this is an original
file with lines
containing words like hello
and world
this is the end of the file
And the result would be:
this is an original
file with lines
containing words like hello and world this is the end of the file
I'd like to use sed, or awk and, if needed, grep, wc or whatever commands work for this purpose. I want to be able to do this on a lot of files.
Using awk you can do:
awk '/hello|world/{printf "%s ", $0; next} 1' file
this is an original
file with lines
containing words like hello and world this is the end of the file
here is simple one using sed
sed -r ':a;$!{N;ba};s/((hello|world)[^\n]*)\n/\1 /g' file
Explanation
:a;$!{N;ba} read whole file into pattern, like this: this is an original\nfile with lines\ncontaining words like hell\
o\nand world\nthis is the end of the file$
s/((hello|world)[^\n]*)\n/\1 /g search the key words hello or world and remove the next \n,
g command in sed substitute stands to apply the replacement to all matches to the regexp, not just the first.
A non-regex approach:
awk '
BEGIN {
# define the word list
w["hello"]
w["world"]
}
{
printf "%s", $0
for (i=1; i<=NF; i++)
if ($i in w) {
printf " "
next
}
print ""
}
'
or a perl one-liner
perl -pe 'BEGIN {#w = qw(hello world)} s/\n/ / if grep {$_ ~~ #w} split'
To edit the file in-place, do:
awk '...' filename > tmpfile && mv tmpfile filename
perl -i -pe '...' filename
This might work for you (GNU sed):
sed -r ':a;/^.*(hello|world).*\'\''/M{$bb;N;ba};:b;s/\n/ /g' file
This checks if the last line, of a possible multi-line, contains the required string(s) and if so reads another line until end-of-file or such that the last line does not contain the/those string(s). Newlines are removed and the line printed.
$ awk '{ORS=(/hello|world/?FS:RS)}1' file
this is an original
file with lines
containing words like hello and world this is the end of the file
sed -n '
:beg
/hello/ b keep
/world/ b keep
H;s/.*//;x;s/\n/ /g;p;b
: keep
H;s/.*//
$ b beg
' YourFile
a bit harder due to check on current line that may include a previous hello or world already
principle:
on every pattern match, keep the string in hold buffer
other wise, load hold buffer and remove \n (use of swap and empty the current line due to limited buffer operation available) and print the content
Add a special case of pattern in last line (normaly hold so not printed otherwise)
I have a simple awk script named "script.awk" that contains:
/\/some_simple_string/ { print $0;}
I'm using it to parse some file that contains:
(by using: cat file | awk -f script.awk)
14 catcat one_two/some_thing
15 catcat one_three/one_more_some_simple_string
16 dogdog one_two/some_simple_string_again
17 dogdog one_four/some_simple_string
18 qweqwe firefire/ppp
I want the script to only print the stroke that fully reflect "/some_simple_string[END_OF_LINE]" but not 2 or 3.
Is there any simple way to do it?
I think, the most appropriate way is to add end-of-line sigh to the regular expression.
So it will parse only strokes that starting with "/some.." and have a new line at the end of "..string[END_OF_LINE]"
Desired output:
17 dogdog one_four/some_simple_string
Sorry for confusion, I was asking for END OF LINE sign in regular expressions.
The correct answer is:
/\/some_simple_string$/ { print $0;}
You can always use:
/\/some_simple_string$/ { print $0 }
I.e. match not only "some_simple_string" but match "/some_simple_string" followed by the end of the line ($ is end of line in regex)
grep '\some_simple_string$' file | tail -n 1 should do the trick.
Or if you really want to use awk do awk '/\/some_simple_string/{x = $0}END{print x}'
To return just the last of a group of matches, ...
Store the line in a variable and print it in the END block.
/some_simple_string/ { x = $0 }
END{ print x }
To print all the matches that end with the string /some_simple_string using regular expression you need to anchor to the the end of the line using $. The most suitable tool for this job is grep:
$ grep '/some_simple_string$' file
In awk the command is much the same:
$ awk '/[/]some_simple_string$/' file
To print all lines after the matching you would do:
$ awk 'print_flag{print;f=0} /[/]some_simple_string$/{print_flag=1}' file
Or just combine grep and tail if it makes it clearer using context option -A to print the following lines:
$ grep -A1 '/some_simple_string$' file | tail -n 1
I sometimes find that the input records can have a trailing carriage return (\r).
Yes, I deal with both Windows and Linux text files.
So I add the following 'pre-processor' to my awk scripts:
1 == 1 { # preprocess all records
res = gsub("\r", "") # remove unwanted trailing char
if(res>0 && NR<100) { print "(removed stuff)" > "/dev/stderr" } # optional
}
more optimally, let FS do the work instead of having it perform unnecessary and unrelated field splitting (adding the \r bit for Windows/DOS completeness):
mawk '!_<NF' FS='[/]some_simple_string[\r]?$'
17 dogdog one_four/some_simple_string
what would be the sed (or other tool) command to join lines together in a file that do not end w/ the character '0'?
I'll have lines like this
412|n|Leader Building Material||||||||||d|d|20||0
which need to be left alone, and then I'll have lines like this for example (which is 3 lines, but only one ends w/ 0)
107|n|Knot Tying Tools|||||Knot Tying Tools
|||||d|d|0||0
which need to be joined/combined into one line
107|n|Knot Tying Tools|||||Knot Tying Tools|||||d|d|0||0
sed ':a;/0$/{N;s/\n//;ba}'
In a loop (branch ba to label :a), if the current line ends in 0 (/0$/) append next line (N) and remove inner newline (s/\n//).
awk:
awk '{while(/0$/) { getline a; $0=$0 a; sub(/\n/,_) }; print}'
Perl:
perl -pe '$_.=<>,s/\n// while /0$/'
bash:
while read line; do
if [ ${line: -1:1} != "0" ] ; then
echo $line
else echo -n $line
fi
done
awk could be short too:
awk '!/0$/{printf $0}/0$/'
test:
kent$ cat t
#aasdfasdf
#asbbb0
#asf
#asdf0
#xxxxxx
#bar
kent$ awk '!/0$/{printf $0}/0$/' t
#aasdfasdf#asbbb0
#asf#asdf0
#xxxxxx#bar
The rating of this answer is surprising ;s (this surprised wink emoticon pun on sed substitution is intentional) given the OP specifications: sed join lines together.
This submission's last comment
"if that's the case check what #ninjalj submitted"
also suggests checking the same answer.
ie. Check using sed ':a;/0$/{N;s/\n//;ba}' verbatim
sed ':a;/0$/{N;s/\n//;ba}'
does
no one
ie. 0
people,
try
nothing,
ie. 0
things,
any more,
ie. 0
tests?
(^D aka eot 004 ctrl-D ␄ ... bash generate via: echo ^V^D)
which will not give (do the test ;):
does no one ie. 0
people, try nothing, ie. 0
things, any more, ie. 0
tests? (^D aka eot 004 ctrl-D ␄ ... bash generate via: echo ^V^D)
To get this use:
sed 'H;${z;x;s/\n//g;p;};/0$/!d;z;x;s/\n//g;'
or:
sed ':a;/0$/!{N;s/\n//;ba}'
not:
sed ':a;/0$/{N;s/\n//;ba}'
Notes:
sed 'H;${x;s/\n//g;p;};/0$/!d;z;x;s/\n//g;'
does not use branching and
is identical to:
sed '${H;z;x;s/\n//g;p;};/0$/!{H;d;};/0$/{H;z;x;s/\n//g;}'
H commences all sequences
d short circuits further script command execution on the current line and starts the next cycle so address selectors following /0$/! can only be /0$/!! so the address selector of
/0$/{H;z;x;s/\n//g;} is redundant and not needed.
if a line does not end with 0 save it in hold space
/0$/!{H;d;}
if a line does end with 0 save it too and then print flush (double entendre ie. purged and lines aligned)
/0$/{H;z;x;s/\n//g;}
NB ${H;z;x;s/\n//g;p;} uses /0$/ ... commands with an extra p to coerce the final print and with a now unnecessary z (to empty and reset pattern space like s/.*//)
A typically cryptic Perl one-liner:
perl -pe 'BEGIN{$/="0\n"}s/\n//g;$_.=$/'
This uses the sequence "0\n" as the record separator (by your question, I'm assuming that every line should end with a zero). Any record then should not have internal newlines, so those are removed, then print the line, appending the 0 and newline that were removed.
Another take to your question would be to ensure each line has 17 pipe-separated fields. This does not assume that the 17th field value must be zero.
awk -F \| '
NF == 17 {print; next}
prev {print prev $0; prev = ""}
{prev = $0}
'
if ends with 0 store, remove newline..
sed '/0$/!N;s/\n//'