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.
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:
Delete empty lines using sed
(17 answers)
Closed 6 years ago.
I tried to use 'sed' command to remove the empty lines.
sed -i '/^$/d' file.txt
My sample txt file looks likes this. The second line has space characters. sed command only removes the empty lines but not the lines with white space.
Sample text
Sample text
So is there away to accomplish this via bash.
My intended out put is
Sample text
Sample text
Use character class [:blank:] to indicate space or tab:
With sed:
sed -i '/^[[:blank:]]*$/ d' file.txt
With perl:
perl -ne 'print if !/^[[:blank:]]*$/' file.txt
With awk:
awk '!/^[[:blank:]]*$/' file.txt
With grep:
grep -v '^[[:blank:]]*$' file.txt
If the tool does not support editing in-place, leverage a temporary file e.g. for grep:
grep -v '^[[:blank:]]*$' file.txt >file.txt.tmp && mv file.txt{.tmp,}
sed -i '/^ *$/d' file.txt
or to also match other white space characters such as tabs, etc:
sed -i '/^[[:space:]]*$/d' file.txt
the * character matches 0 or more instances of preceding character
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
This question already has answers here:
Print text between delimiters using sed
(2 answers)
Closed 2 years ago.
I have a line of text which looks like hh^ay-pau+h#ow, I want to extract the text between - and + which in this case is pau. This should be done in bash. Any help would be appreciated.
EDIT: I want to extract the text between the first occurence of the tokens
PS: My google search didn't take me anywhere. I apologize if this question is already asked.
The way to do this in pure bash, is by using parameter expansions in bash
$ a=hh^ay-pau+h#ow
$ b=${a%%+*}
$ c=${b#*-}
$ echo $c
pau
b: remove everything including and behind the first + occurence
c: remove everything excluding and before the first - ocurrence
More info about substring removing in bash parameter expansion
Try
grep -Po "(?<=\-).*?(?=\+)"
For example,
echo "hh^ay-pau+h#ow" | grep -Po "(?<=\-).*?(?=\+)"
If you have only one occurence of - and + you can use cut:
$ echo "hh^ay-pau+h#ow" | cut -d "-" -f 2 | cut -d "+" -f 1
pau
Assuming one occurence of + and -, you can stick to bash
IFS=+- read -r _ x _ <<<'hh^ay-pau+h#ow'
echo $x
pau
If you're guarenteed to only have one - and one + .
% echo "hh^ay-pau+h#ow" | sed -e 's/.*-//' -e 's/+.*//'
pau
echo "hh^ay-pau+h#ow" | awk -F'-' '{print $2}' |awk -F'+' '{print $1}'
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