I want to grep a part of string that has numbers and dots(.) in it.
For example:
/home/xar/11.1.0/hez
/uaa/14.0.2.5/grd/pc
What i want is only this part of line:
14.0.2.5
11.1.0
I realized that cut command is not enough for this problem.
You can use this grep:
$ grep -o "[0-9.]*" file
11.1.0
14.0.2.5
-o is for "print just the matched part".
"[0-9.]*" matches any combination of numbers and dots.
sed version:
sed -n 's!.*/\([0-9.]*\)/.*!\1!p' input
You can use awk, but here grep is the correct tool.
awk -F/ '{for (i=1;i<=NF;i++) if ($i~/[0-9.]+/) print $i}' file
11.1.0
14.0.2.5
Related
I need to grep some pattern and further i need to print some output within that. Currently I am using the below command which is working fine. But I like to eliminate using multiple pipe and want to use single awk command to achieve the same output. Is there a way to do it using awk?
root#Server1 # cat file
Jenny:Mon,Tue,Wed:Morning
David:Thu,Fri,Sat:Evening
root#Server1 # awk '/Jenny/ {print $0}' file | awk -F ":" '{ print $2 }' | awk -F "," '{ print $1 }'
Mon
I want to get this output using single awk command. Any help?
You can try something like:
awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file
Try this
awk -F'[:,]+' '/Jenny/{print $2}' file.txt
It is using muliple -F value inside the [ ]
The + means one or more since it is treated as a regex.
For this particular job, I find grep to be slightly more robust.
Unless your company has a policy not to hire people named Eve.
(Try it out if you don't understand.)
grep -oP '^[^:]*Jenny[^:]*:\K[^,:]+' file
Or to do a whole-word match:
grep -oP '^[^:]*\bJenny\b[^:]*:\K[^,:]+' file
Or when you are confident that "Jenny" is the full name:
grep -oP '^Jenny:\K[^,:]+' file
Output:
Mon
Explanation:
The stuff up until \K speaks for itself: it selects the line(s) with the desired name.
[^,:]+ captures the day of week (in this case Mon).
\K cuts off everything preceding Mon.
-o cuts off anything following Mon.
I need to print the string between these characters....
atob(' ')
I am using a = in the second part as an attempt to stop the code on an equal signs (which the base64 string I'm trying to get ends in.)
I use this script, but it prints the entire line containing the above characters. I need just the data in between.
sed -n '/atob/,${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
I have a file text.txt which contains the below words.
1. moon,one
2. sun,two
3. well,three
4. doll,four
if i grep this file using sun
grep -i sun text.txt
I will get the output
sun,two
But, my requirement is I need to grep with the word which is starting with sun not exactly sun.
grep -i sunlight text.txt
Here I need the same output for grep -i sun text.txt.
You don't need awk or gawk, nor sed. Just do
grep -o 'sun.*'
Other more complex / elegant solutions may be available depending on the system you are using.
What you are looking for are regular expressions.
In your case, it would be
grep -i 'sun.*' text.txt
Try using -o, as showed in the documentation.
The -o make grep return only the matched part. You can also use regular expressions.
grep -io sun text.txt
Is this what you're looking for?
awk -F ',' '/^[SsuUnN]/ {print $0}' test.txt
or if you want to search the pattern "sun" in general from the input_file, then use this:
awk -F ',' 'BEGIN{IGNORECASE=1} /sun/ {print $0}' test.txt
I have a file containing consecutive symbols (as pipe "|") like
ANKRD54,LIAR,allergy,|||
ANKRD54,LIAR,asthma,||20447076||
ANKRD54,LIAR,autism,||||
ANKRD54,LIAR,cancer,|||
ANKRD54,LIAR,chronic_obstructive_pulmonary_disease,|||
ANKRD54,LIAR,dental_caries,||||
Now using shell or a sed command in shell is it possible to replace multiple pipe with one pipe like
ANKRD54,LIAR,allergy,|
ANKRD54,LIAR,asthma,|20447076|
ANKRD54,LIAR,autism,|
ANKRD54,LIAR,cancer,|
ANKRD54,LIAR,chronic_obstructive_pulmonary_disease,|
ANKRD54,LIAR,dental_caries,|
I guess the easiest way is use built-in commands: cat your_file | tr -s '|'
Pass your text to sed (e.g. via a pipe)
cat your_file | sed "s/|\+/|/g"
You can do that with a simple awk gsub as:-
awk -F"," -v OFS="," '{gsub(/[|]+/,"|",$4)}1' file
See it in action:-
$ cat file
ANKRD54,LIAR,allergy,|||
ANKRD54,LIAR,asthma,||20447076||
ANKRD54,LIAR,autism,||||
ANKRD54,LIAR,cancer,|||
ANKRD54,LIAR,chronic_obstructive_pulmonary_disease,|||
ANKRD54,LIAR,dental_caries,||||
$ awk -F"," -v OFS="," '{gsub(/[|]+/,"|",$4)}1' file
NKRD54,LIAR,allergy,|
ANKRD54,LIAR,asthma,|20447076|
ANKRD54,LIAR,autism,|
ANKRD54,LIAR,cancer,|
ANKRD54,LIAR,chronic_obstructive_pulmonary_disease,|
ANKRD54,LIAR,dental_caries,|
In Linux (Cento OS) I have a file that contains a set of additional information that I want to removed. I want to generate a new file with all characters until to the first |.
The file has the following information:
ALFA12345|7890
Beta0-XPTO-2|30452|90 385|29
ZETA2334423 435; 2|2|90dd5|dddd29|dqe3
The output expected will be:
ALFA12345
Beta0 XPTO-2
ZETA2334423 435; 2
That is removed all characters after the character | (inclusive).
Any suggestion for a script that reads File1 and generates File2 with this specific requirement?
Try
cut -d'|' -f1 oldfile > newfile
And, to round out the "big 3", here's the awk version:
awk -F\| '{print $1}' in.dat
You can use a simple sed script.
sed 's/^\([^|]*\).*/\1/g' in.dat
ALFA12345
Beta0-XPTO-2
ZETA2334423 435; 2
Redirect to a file to capture the output.
sed 's/^\([^|]*\).*/\1/g' in.dat > out.dat
And with grep:
$ grep -o '^[^|]*' file1
ALFA12345
Beta0-XPTO-2
ZETA2334423 435; 2
$ grep -o '^[^|]*' file1 > file2