This question already has answers here:
Editing/Replacing content in multiple files in Unix AIX without opening it
(2 answers)
Closed 6 years ago.
I have an output that looks as below
- 0.1-1
- 0.1-2
- 0.1-3
- 0.1-6
- 0.1-7
- 0.1-9
How to use grep or something else so as to remove "-" and a space from the beginning.
0.1-1
0.1-2
0.1-3
0.1-6
0.1-7
0.1-9
With sed:
sed -e 's/^- //' input.txt
Or with GNU grep:
grep -oP '^- \K.*' input.txt
You may use grep also,
grep -oE '[0-9].*' file
With awk:
awk '{print $2}' file
You can use cut to remove the first two columns of every line:
cut -c3- input.txt
Related
This question already has answers here:
How do I delete a matching line, the line above and the one below it, using sed?
(7 answers)
Closed 1 year ago.
I have a file named file.txt and it contains several lines containing string "NaN". How can I delete lines containing "Nan" and one line before and after it. I know sed －i '/pattern/d' file.txt can delete the matched line, but how can I delete neaby lines of the matched line.
Best regards
This is a most inelegant solution, but works. Perhaps, it will anger a UNIX guru to come and provide a real answer.
grep 'Nan' -n -C 1 target_file.txt | awk -F '[-:]' '{print $2}' | sed '2d' | paste -d, - - | sed 's/$/d/' > del_lines.sed && sed -f del_lines.sed target_file.txt > output_file.txt
fyi, The word "perticular" is misspelled, its "par-" like in golf!
This question already has answers here:
How to use sed/grep to extract text between two words?
(14 answers)
Closed 4 years ago.
how to print text between two specific words using awk, sed ?
$ ofed_info | awk '/MLNX_OFED_LINUX/{print}'
MLNX_OFED_LINUX-4.1-1.0.2.0 (OFED-4.1-1.0.2):
$
Output required:-
4.1-1.0.2.0
Following awk may help you here.(considering that your input to awk will be same as shown sample only)
your_command | awk '{sub(/[^-]*/,"");sub(/ .*/,"");sub(/-/,"");print}'
Solution 2nd: With sed solution now.
your_command | sed 's/\([^-]*\)-\([^ ]*\).*/\2/'
Solution 3rd: Using awk's match utility:
your_command | awk 'match($0,/[0-9]+\.[0-9]+\-[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+/){print substr($0,RSTART,RLENGTH)}'
You may use this sed:
echo 'MLNX_OFED_LINUX-4.1-1.0.2.0 (OFED-4.1-1.0.2):' |
sed -E 's/^[^-]*-| .*//g'
4.1-1.0.2.0
This sed command removes text till first hyphen from start or text starting with space towards end.
Try this:
ofed_info | sed -n 's/^MLNX_OFED_LINUX-\([^ ]\+\).*/\1/p'
The sed command only selects lines starting with the keyword and prints the version attached to it.
This question already has answers here:
Can grep show only words that match search pattern?
(15 answers)
Closed 8 years ago.
I have a bash file that has below line along with other lines.
var BUILD_VERSION = '2014.17.10_23';
I just want to extract 2014.17.10_23 and this value may change so something like grep for 2014* . However when I do that I get the whole line returned instead of the value 2014.17.10_23.
What would be the best way to achieve this?
Thanks
Using awk:
awk -F= '/BUILD_VERSION/{print $2}' input | tr -d "[' ;]"
And with sed:
sed -n "/BUILD_VERSION/s/.*'\([^']*\)'.*/\1/p" input
grep 'BUILD_VERSION' <your file> | sed -e 's/var BUILD_VERSION = //g'
Would get you '2014.17.10_23'; tweak the sed expression (or pipe it through a few more) to get rid of quotes.
It would be a 1 liner regex in Perl...
Here is another awk solution:
awk -F' = ' '/BUILD_VERSION/ {gsub(/\x27|;/,""); print $NF}'
You can use this awk
awk -F\' '/BUILD_VERSION/ {print $2}' file
2014.17.10_23
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Use grep to report back only line numbers
I only want to see the line number. I don't need to see the remaining output.
Pipe your grep -n output, which normally looks something like:
11: stuff that matched
43: more stuff that matched
through sed to strip out the matching parts:
grep -n pattern file | sed -e 's/:.*//g'
11
43
grep -n or --line-number option will do this for you. You can find this information in the grep help file, which you can find by using grep --help or grep --help | less to read it more carefully. Also consider using the manual page: man grep
You could use awk too.
grep -n word file | awk -F: '{ print $1 }'
As #Barmar pointed out you could just use an awk one-liner as such:
awk '/regex/ { print NR }' file
Since you don't have awk you could also use cut:
grep -n word file | cut -d: -f1
This question already has answers here:
sed unknown option to `s' in bash script [duplicate]
(4 answers)
Closed 2 years ago.
I am trying
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
and getting
unknown option to `s'
What am I doing wrong?
Edit:
As per the comments the code is actually correct. My full code resembled something like the following
grep searchterm myfile.csv | sed 's/replaceme/withthis/g'
# my comment
And it appears that for some reason my comment was being fed as input into sed. Very strange.
use the --expression option
grep searchterm myfile.csv | sed --expression='s/replaceme/withthis/g'
use "-e" to specify the sed-expression
cat input.txt | sed -e 's/foo/bar/g'
To make sed catch from stdin , instead of from a file, you should use -e.
Like this:
curl -k -u admin:admin https://$HOSTNAME:9070/api/tm/3.8/status/$HOSTNAME/statistics/traffic_ips/trafc_ip/ | sed -e 's/["{}]//g' |sed -e 's/[]]//g' |sed -e 's/[\[]//g' |awk 'BEGIN{FS=":"} {print $4}'
If you are trying to do an in-place update of text within a file, this is much easier to reason about in my mind.
grep -Rl text_to_find directory_to_search 2>/dev/null | while read line; do sed -i 's/text_to_find/replacement_text/g' $line; done
Open the file using vi myfile.csv
Press Escape
Type :%s/replaceme/withthis/
Type :wq and press Enter
Now you will have the new pattern in your file.