grep 2 elements in a line and print them - bash

Question

Here is my issue, i have a file with the entries, i would like to get just the date + the last command after the last "]: "
Aug 17 14:25:17 snaper[22134]: [ip:10.1.15.245 37985 10.1.15.18 22 uid:10000 sid:21680 tty: cwd:/data/www/hybris/hybris/bin/platform filename:/bin/ps]: /bin/ps -p 6763
How can i get it when i cat the file ?
I can get the date with:
awk '{print $1,$2,$3}'
and the last command with :
awk -F': ' '{print $NF}'
But how to combine them to get it in a single line ?
I'm not awk limited, any sed grep or other command is ok for me :)
Thanks in advance

Answer 1

Just remove everything between the date and the last command:
sed 's/^\(... .. ..:..:..\).*: /\1 /'

Answer 2

Simple solution using AWK
$awk '{print $1,$2,$3, $(NF-2), $(NF-1), $NF }' file
Aug 17 14:25:17 /bin/ps -p 6763

Answer 3

Using GNU grep
grep -oP '^.{15}|.*\]: \K.*' file | paste - -

Answer 4

Possible use tee for use 1 output to 2 commands:
echo 'Aug 17 14:25:17 snaper[22134]: [ip:10.1.15.245 37985 10.1.15.18 22 uid:10000 sid:21680 tty: cwd:/data/www/hybris/hybris/bin/platform filename:/bin/ps]: /bin/ps -p 6763' | tee >(awk -F': ' '{print $NF}') | awk '{print $1,$2,$3}' | tr '\n' ' '
and we have output as:
Aug 17 14:25:17 /bin/ps -p 6763

Answer 5

$ s="Aug 17 14:25:17 snaper[22134]: [ip:10.1.15.245 37985 10.1.15.18 22 uid:10000 sid:21680 tty: cwd:/data/www/hybris/hybris/bin/platform filename:/bin/ps]: /bin/ps -p 6763"
Use sed to achieve your goal,
$ sed -r 's/(^.*:[0-9]{2}) .*]:/\1/' <<< "$s"

Answer 6

Try following solutions too. Considering your Input_file will be having same data as shown sample.
Solution 1st: Using simple cut command.
cut -d" " -f1,2,3,14,15,16 Input_file
Solution 2nd: using awk command where I am making string snaper and ]: as field separators.
awk -F' snaper|]:' '{print $1,$4}' Input_file
Solution 3rd: making record separator as space and then printing only those lines which we need as per OP's request.
awk -v RS=" " 'NR<4||NR>13{printf("%s%s",$0,NR<3||NR<16?" ":"")}' Input_file
Solution 4th: Substituting everything from snap to till : and get whatever is OP's request.
awk '{sub(/snaper\[.*\]: /,"");print}' Input_file
Solution 5th: Using --re-interval here(as I have old version of awk) you could remove it if you latest awk version in your system too.
awk --re-interval '{match($0,/.*[0-9]{2}:[0-9]{2}:[0-9]{2}/);print substr($0,RSTART,RLENGTH),$(NF-2),$(NF-1),$NF}' Input_file
Solution 6th: using sed and substituting everything till snaper and then everything till colon and printing the match only.
sed 's/\(.[^s]*\)\(.*:\)\(.*\)/\1\3/' Input_file