I need to print the string between these characters....
atob(' ')
I am using a = in the second part as an attempt to stop the code on an equal signs (which the base64 string I'm trying to get ends in.)
I use this script, but it prints the entire line containing the above characters. I need just the data in between.
sed -n '/atob/,${p;/==/q;}'
I appreciate any help. Thank you.
Does this work (tested for GNU sed 4.2.2)?
sed -n -e "s/atop('\(.*\)')/\1/p" b.txt
where b.txt is
atop('safdasdfasf')
or you can try awk
awk -F\' '/atop/ {print $2}' b.txt
(tested for gnu awk 4.0.2 and added the suggestion by Jotne)
And another working sed:
echo "atop('safdasdfasf')" | sed -r "/atop/ s/^[^']+'([^']+)'.*/\1/"
safdasdfasf
For example :
file1 contains :
This is an example.
This is Sed example.
This is an example.
I want to create a file like :
This is an example.
This is Sed example.
This is awk example.
This is an example.
and this should be everywhere in the file it finds the keyword: sed, it should change sed to awk and print the entire line with the change just below the original line in that file itself
thanks
This might work for you (GNU sed):
sed '/\bsed\b/I{p;s//awk/g}' file
This matches the word sed in any combination of case, prints the original line and then substitutes the word awk throughout the line and prints that line too.
The awk version:
awk '{IGNORECASE=1} $0 ~ /^.*sed.*$/ { print; gsub("sed", "awk" ); print; next}1' file.txt
Output:
This is an example.
This is Sed example.
This is awk example.
This is an example.
I have a file: users.txt that looks like the below:
Name
Jack
Name
Jackson
Name
Peter
I need to remove EVERYTHING ABOVE the last match found Name so that the file looks like:
Name
Peter
I have tried searching the net, but only find awk command's that can achieve this.
I need it to be sed in order to use the -i option to save. I have come across where a file is sometimes empty when redirecting > standard output using awk.
Your time and assistance is greatly appreciated.
You can use tac twice with awk:
tac file | awk '{print} /Name/{exit}' | tac
Name
Peter
Through sed,
tac file | sed -n '0,/Name/p' | tac
Using sed without tac:
sed -n '/Name/h; //!H; ${g;p}' file
When Name is seen, replace the hold space with the current line
When Name is NOT seen, append the current line to the hold space
At the last line, replace the pattern space with the hold space and print.
Since we don't have sed in a pipeline, you can use -i to replace the file.
All in awk
awk 'x{x=x"\n"$0}/Name/{x=$0}END{print x}' file
This might work for you (GNU sed):
sed 'H;/Name/h;$!d;x' file
Hold matches in the hold space and at the end of file print the hold space with the last match.
How to use sed to find lines with the word linux? As later display a first line 10 with the word linux?
EX.:
cat file | sed -e '/linux/!d' -e '10!d' ### I can not display the first 10 lines of the word linux
cat file | sed '/linux/!d' | sed '10!d' ### It is well
How to make it work with one sed?
cat file | sed -e '/linux/!d; ...?; 10!d'
...? - storing of the buffer linux? 10 later cut the lines?
Someone explain to me?
I would use awk:
awk '/linux/ && c<10 {print;c++} c==10 {exit}' file
This might work for you (GNU sed):
sed -nr '/linux/{p;G;/(.*\n){10}/q;h}' file
Print the line if it contains the required string. If the required number of lines has already been printed quit, otherwise store the line and previous lines in the hold space.
You could use perl:
perl -ne 'if (/linux/) {print; ++$a;}; last if $a==10' inputfile
Using GNU sed:
sed -rn "/linux/{p;x;s/^/P/;ta;:a;s/^P{10}$//;x;Tb;Q;:b}" filename
Thanks. You are great. All of the examples look very nice. Wooow :) It is a pity that I can not do that.
I have not seen for 'r' option in sed. I need to learn.
echo -e 'windows\nlinux\nwindows\nlinux\nlinux\nwindows' | sed -nr '/linux/{p;G;/(.*\n){2}/q;h}'
It works very well.
echo -e 'windows\nlinux\nwindows\nlinux\nlinux\nwindows' | sed -nr '/linux/{p;G;/(.*\n){2}/q;h}' | sed '2s/linux/debian/'
Can I ask you one more example? How to get a result at one sed?
I have lines in a file in linux, and i am trying print the line without the | and without some parameters
$cat file
2013-07-15,Provider 1.99,3|30000055|2347|0,12222,1,3,0,0,0,19,aaa,bbb
2013-07-15,Provider 1.99,3|30000055|2347|0,12222,44,12,0,0,0,33,aaa,bbb
and i need the output like:
2013-07-15,Provider,2347,12222,1,3,0,0,0,19,aaa,bbb
2013-07-15,Provider,2347,12222,44,12,0,0,0,33,aaa,bbb
and i am trying with awk, but i have some problems.
If your lines have similar pattern you would to retain then you can do:
awk 'BEGIN{FS=OFS=","}{$2="Provider";$3=2347}1' file
If you don't know what the patterns are then here is a more generic one:
awk 'BEGIN{FS=OFS=","}{split($2,a,/ /);split($3,b,/\|/);$2=a[1];$3=b[3]}1' file
If it doesn't solve your problem, I am pretty sure it would help you guide to get one.
Using sed:
sed 's/ [^|]*|[^|]*|\([^|]*\)|[^,]/,\1/' input
and some shorter version:
sed 's/ .*|\([^|]*\)|[^,]*/,\1/' input
and even shorter:
sed 's/ .*|\(.*\)|[^,]*/,\1/' input
Use awk, and let blank or comma or pipe be the field separators:
awk -F '[[:blank:],|]' -v OFS=, '{
print $1,$2,$6,$8,$9,$10,$11,$12,$13,$14,$15,$16
}' file
2013-07-15,Provider,2347,12222,1,3,0,0,0,19,aaa,bbb
2013-07-15,Provider,2347,12222,44,12,0,0,0,33,aaa,bbb