I am trying to extract text between pattern1 (fixed) and pattern2 (this can be p2-1/p2-2).
can you please tell me how to achieve this in a single command?
A file starts with start and ends with either end or close
File1:
======
junktest
data
start
stackoverflow
sed
close
File2:
======
data2
start
stackoverflow
end
I can extract text from File1 with
sed -n "/start/,/close/p"
And from File2 with
sed -n "/start/,/end/p"
I need a single sed command to achieve both..
something like:
sed -n "/start/, /close or end /p"
Both GNU sed and BSD sed:
sed -nE '/start/,/close|end/p' file
This awk looks better
awk '/start/,/end|close/' file
sed -n -E "/Word1/,/Word2-1/p" | sed -n -E "/Word1/,/Word2-2/p"
Easy with awk:
$ awk '/start/{p=1}p{print}/end|close/{p=0}' file
Related
I am trying to change the column names to lowercase in a csv file. I found the code to do that online but I dont know how to replace the old column names(uppercase) with new column names(lowercase) in the original file. I did something like this:
$cat head -n1 xxx.csv | tr "[A-Z]" "[a-z]"
But it simply just prints out the column names in lowercase, which is not enough for me.
I tried to add sed -i but it did not do any good. Thanks!!
Using awk (readability winner) :
concise way:
awk 'NR==1{print tolower($0);next}1' file.csv
or using ternary operator:
awk '{print (NR==1) ? tolower($0): $0}' file.csv
or using if/else statements:
awk '{if (NR==1) {print tolower($0)} else {print $0}}' file.csv
To change the file for real:
awk 'NR==1{print tolower($0);next}1' file.csv | tee /tmp/temp
mv /tmp/temp file.csv
For your information, sed using the in place edit switch -i do the same: it use a temporary file under the hood.
You can check this by using :
strace -f -s 800 sed -i'' '...' file
Using perl:
perl -i -pe '$_=lc() if $.==1' file.csv
It replace the file on the fly with -i switch
You can use sed to tell it to replace the first line with all lower-case and then print the rest as-is:
sed '1s/.*/\L&/' ./xxx.csv
Redirect the output or use -i to do an in-place edit.
Proof of Concept
$ echo -e "COL1,COL2,COL3\nFoO,bAr,baZ" | sed '1s/.*/\L&/'
col1,col2,col3
FoO,bAr,baZ
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 file, with lines, contains ip with netmask
a.b.c.d/24
w.x.y.z/32
etc
How to delete delete specific row?
i'm using
sed -ie "s#a.b.c.d/24##g" %filname%
but after the removal is an empty string in file.
It should run inside a script, with ip as parameter and also work in freebsd under sh.
Sed solution
sed -i '/<pattern-to-match-with-proper-escape>/d' data.txt
-i option will change the original file.
Awk solution
awk '!/<pattern-to-match-with-proper-escape>/' data.txt
Using sed:
sed -i '\|a.b.c.d/24|d' file
Command line arg:
For the input being command line argument, say 1st argument($1):
sed -i "\|$1|d" file
Replace $1 with appropriate argument number as is your case.
You should use d (delete) not g. Also do not use s (replacement).
sed -ie '/a.b.c.d\/24/d' %filename%
In a script you should using it in this way
IP=$1
IPA=${IP////\\/}
sed -i /"${IPA}"/d %filename%
And the script parameter should be called in this way:
./script.sh a.b.c.d/24
perl -i -lne 'print unless(/a.b.c.d\/24/)' your_file
or in awk if you donot want to do inplace editing:
awk '$0!~/a.b.c.d\/24/' your_file
I am trying to find the occurance of tab in a file some_file and print those line with leading line number.
grep -nP "\t" some_file works well for me but I want sed or awk equivalent command for the same.
To emulate: grep -nP "\t" file.txt
Here's one way using GNU awk:
awk '/\t/ { print NR ":" $0 }' file.txt
Here's one way using GNU sed:
< file.txt sed -n '/\t/{ =;p }' | sed '{ N;s/\n/:/ }'
Well, you can always do it in sed:
cat -n test.txt | sed -n "/\t/p"
Unfortunately, sed can only print line numbers to stdout with a new line, so in any case, more than one command is necessary. A more lengthy (unnecessary so) version of the above, but one only using sed, would be:
sed = test.txt | sed -n "N;s/\n/ /;/\t/p"
but I like the one with cat more. CATS ARE NICE.
I want to add a line at top of file say f1 using awk.
Is there a better way than the following?
awk 'BEGIN{print "word"};{print $0}' f1 > aux;cp aux f1;\rm aux<br/>
Does awk has something like -i option in sed?
Why not use sed - it would make the solution more straightforward
$sed -i.bak '1i\
word
' <filename>
An alternate way to do this is:
sed -i '1s:^: Word1\nWord2 :' file