I need help with using sed to comment a matching lines and 4 lines which follows it.
in a text file.
my text file is like this:
[myprocess-a]
property1=1
property2=2
property3=3
property4=4
[anotherprocess-b]
property1=gffgg
property3=gjdl
property2=red
property4=djfjf
[myprocess-b]
property1=1
property4=4
property2=2
property3=3
I want to prefix # to all the lines having text '[myprocess' and 4 lines that follows it
expected output:
#[myprocess-a]
#property1=1
#property2=2
#property3=3
#property4=4
[anotherprocess-b]
property1=gffgg
property3=gjdl
property2=red
property4=djfjf
#[myprocess-b]
#property1=1
#property4=4
#property2=2
#property3=3
Greatly appreciate your help on this.
You can do this by applying a regular expression to a set of lines:
sed -e '/myprocess/,+4 s/^/#/'
This matches lines with 'myprocess' and the 4 lines after them. For those 4 lines it then inserts a '#' at the beginning of the line.
(I think this might be a GNU extension - it's not in any of the "sed one liner" cheatsheets I know)
sed '/\[myprocess/ { N;N;N;N; s/^/#/gm }' input_file
Using string concatenation and default action in awk.
http://www.gnu.org/software/gawk/manual/html_node/Concatenation.html
awk '/myprocess/{f=1} f>5{f=0} f{f++; $0="#" $0} 1' foo.txt
or if the block always ends with empty line
awk '/myprocess/{f=1} !NF{f=0} f{$0="#" $0} 1' foo.txt
Related
I have a file with text as follows:
###interest1 moreinterest1### sometext ###interest2###
not-interesting-line
sometext ###interest3###
sometext ###interest4### sometext othertext ###interest5### sometext ###interest6###
I want to extract all strings between ### .
My desired output would be something like this:
interest1 moreinterest1
interest2
interest3
interest4
interest5
interest6
I have tried the following:
grep '###' file.txt | sed -e 's/.*###\(.*\)###.*/\1/g'
This almost works but only seems to grab the first instance per line, so the first line in my output only grabs
interest1 moreinterest1
rather than
interest1 moreinterest1
interest2
Here is a single awk command to achieve this that makes ### field separator and prints each even numbered field:
awk -F '###' '{for (i=2; i<NF; i+=2) print $i}' file
interest1 moreinterest1
interest2
interest3
interest4
interest5
interest6
Here is an alternative grep + sed solution:
grep -oE '###[^#]*###' file | sed -E 's/^###|###$//g'
This assumes there are no # characters in between ### markers.
With GNU awk for multi-char RS:
$ awk -v RS='###' '!(NR%2)' file
interest1 moreinterest1
interest2
interest3
interest4
interest5
interest6
You can use pcregrep:
pcregrep -o1 '###(.*?)###' file
The regex - ###(.*?)### - matches ###, then captures into Group 1 any zero o more chars other than line break chars, as few as possible, and ### then matches ###.
o1 option will output Group 1 value only.
See the regex demo online.
sed 't x
s/###/\
/;D; :x
s//\
/;t y
D;:y
P;D' file
Replacing "###" with newline, D, then conditionally branching to P if a second replacement of "###" is successful.
This might work for you (GNU sed):
sed -n 's/###/\n/g;/[^\n]*\n/{s///;P;D}' file
Replace all occurrences of ###'s by newlines.
If a line contains a newline, remove any characters before and including the first newline, print the details up to and including the following newline, delete those details and repeat.
To delete/comment 3 lines befor a pattern (including the line with the pattern):
how can i achive it through sed command
Ref:
sed or awk: delete n lines following a pattern
the above ref blog help to achive the this with after a pattern match but i need to know before match
define host{
use xxx;
host_name pattern;
alias yyy;
address zzz;
}
the below sed command will comment the '#' after the pattern match for example
sed -e '/pattern/,+3 s/^/#/' file.cfg
define host{
use xxx;
#host_name pattern;
#alias yyy;
#address zzz;
#}
like this how can i do this for the before pattern?
can any one help me to resolve this
If tac is allowed :
tac|sed -e '/pattern/,+3 s/^/#/'|tac
If tac isn't allowed :
sed -e '1!G;h;$!d'|sed -e '/pattern/,+3 s/^/#/'|sed -e '1!G;h;$!d'
(source : http://sed.sourceforge.net/sed1line.txt)
Reverse the file, comment the 3 lines after, then re-reverse the file.
tac file | sed '/pattern/ {s/^/#/; N; N; s/\n/&#/g;}' | tac
#define host{
#use xxx;
#host_name pattern;
alias yyy;
address zzz;
}
Although I think awk is a little easier to read:
tac file | awk '/pattern/ {c=3} c-- > 0 {$0 = "#" $0} 1' | tac
This might work for you (GNU sed):
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/^/#/mg;P;D' file
Gather up 4 lines in the pattern space and if the last line contains pattern insert # at the beginning of each line in the pattern space.
To delete those 4 lines, use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/d;P;D' file
To delete the 3 lines before pattern but not the line containing pattern use:
sed ':a;N;s/\n/&/3;Ta;/pattern[^\n]*$/s/.*\n//;P;D'
I want to read from the file /etc/lvm/lvm.conf and check for the below pattern that could span across multiple lines.
tags {
hosttags = 1
}
There could be as many white spaces between tags and {, { and hosttags and so forth. Also { could follow tags on the next line instead of being on the same line with it.
I'm planning to use awk and sed to do this.
While reading the file lvm.conf, it should skip empty lines and comments.
That I'm doing using.
data=$(awk < cat `cat /etc/lvm/lvm.conf`
/^#/ { next }
/^[[:space:]]*#/ { next }
/^[[:space:]]*$/ { next }
.
.
How can I use sed to find the pattern I described above?
Are you looking for something like this
sed -n '/{/,/}/p' input
i.e. print lines between tokens (inclusive)?
To delete lines containing # and empty lines or lines containing only whitespace, use
sed -n '/{/,/}/p' input | sed '/#/d' | sed '/^[ ]*$/d'
space and a tab--^
update
If empty lines are just empty lines (no ws), the above can be shortened to
sed -e '/#/d' -e '/^$/d' input
update2
To check if the pattern tags {... is present in file, use
$ tr -d '\n' < input | grep -o 'tags\s*{[^}]*}'
tags { hosttags = 1# this is a comment}
The tr part above removes all newlines, i.e. makes everything into one single line (will work great if the file isn't to large) and then search for the tags pattern and outputs all matches.
The return code from grep will be 0 is pattern was found, 1 if not.
Return code is stored in variable $?. Or pipe the above to wc -l to get the number of matches found.
update3
regex for searcing for tags { hosttags=1 } with any number of ws anywhere
'tags\s*{\s*hosttags\s*=\s*1*[^}]*}'
try this line:
awk '/^\s*#|^\s*$/{next}1' /etc/lvm/lvm.conf
One could try preprocessing the file first, removing commments and empty lines and introducing empty lines behind the closing curly brace for easy processing with the second awk.
awk 'NF && $1!~/^#/{print; if(/}/) print x}' file | awk '/pattern/' RS=
I have a file split in blocks like the following:
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGG
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTT
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTA
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
TATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
GGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
I've truncated/wrapped the lines for clarity's sake, but imagine very long lines. The point of my question is that I want a final file that looks like this:
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGGAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTTAGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGTTATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
Where this new block has:
the same number of lines as the initial blocks,
each of the lines of the resulting block is a concatenation of the lines with the same line-number in the initial blocks.
this concatenation should be in-order (i.e. "1st line of 1st block" + "1st line of 2nd block", etc
Is it possible to achieve this final block using sed and/or awk, could you show me how it could be done?
In bash with paste:
$ paste <(head -4 file) <(tail -4 file) | tr -d '\t'
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGGAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTTAGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGTTATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
try this:
awk -vOFS="" '$0{a[NR]=$0}END{for(i=1;i<=NR/2;i++)print a[i],a[i+5]}' file
test with your example:
kent$ cat tmp.txt
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGG
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTT
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTA
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
TATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
GGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
kent$ awk -vOFS="" '$0{a[NR]=$0}END{for(i=1;i<=NR/2;i++)print a[i],a[i+5]}' tmp.txt
AGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTGGGGAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTT
AGGTAGTTATTATTTTTTTGGTTTTTAGTATTTAATTGAGTGTTTAGTTTTTTTTTATTTGTCGGGATATTTTAGTTGATTTTAGATTGC
ATGTAGGTGTTTATGTATTAGTTTTTTTTAGGTTTAGGGTGTTGTTATATTTTTAGTTTCGATTCGTCGTAAGTTTTATTTTTTTTTAAT
ATTTAGGTTTTGTGTTTTGTGTATTATTGAATTTAATTAAAGTTAGGATAGGTTTTGGTGTTTGAGGTTAATTTTGTTTTATTTTTTTTT
awk -F'\n' -v RS= '{for (i=1;i<=NF;i++) {str[i] = str[i] $i} END {for (i=1;i<=NF;i++) print str[i]}' file
How would like to join two lines usung awk or sed?
For example, I have data like below:
abcd
12:12:12:12:12:12:12:12
efgh001_01
45:45:45:45:45:45:45:45
ijkl7464746
78:78:78:78:78:78:78:78
and I need output like below:
abcd 12:12:12:12:12:12:12:12
efgh001_01 45:45:45:45:45:45:45:45
ijkl7464746 78:78:78:78:78:78:78:78
Running this almost works, but I need the space or tab:
awk '!(NR%2){print$0p}{p=$0}'
You're almost there:
awk '(NR % 2 == 0) {print p, $0} {p = $0}'
With sed you can do that as follows:
sed -n 'N;s/\n/ /p' file
where:
N reads next line
s replaces the new line character with a space to join both lines properly
p prints the result
This might work for you:
sed '$!N;s/\n/ /' file
or this:
paste -sd' \n' file