Is there a way to print all records separated by the OFS without typing out each column number.
#Desired style of syntax, undesired result [kbrandt@glade: ~] echo "1 2 3 4" | gawk 'BEGIN { OFS=" :-( "}; {print $0}' 1 2 3 4 #Desired result, undesired syntax [kbrandt@glade: ~] echo "1 2 3 4" | gawk 'BEGIN { OFS=" :-) "}; {print $1,$2,$3,$4}' 1 :-) 2 :-) 3 :-) 4
With space as the separator you could even generalize the solution; e.g., the following returns everything from the 3rd field: awk '{sub(/^[ ]*([^ ]+ +){2}/, ""); print $0}' It gets trickier with arbitrary field separators, though.
awk '{ print $2; }' prints the second field of each line. This field happens to be the process ID from the ps aux output. xargs kill -${2:-'TERM'} takes the process IDs from the selected sidekiq processes and feeds them as arguments to a kill command.
Any special characters in the field separator must be escaped appropriately. For example, to use a ' \ ' as the field separator on the command line, you would have to type: # same as FS = "\\" awk -F\\\\ '…' files … Because ' \ ' is used for quoting in the shell, awk sees ' -F\\ '.
This is a variation on the first style:
echo "1 2 3 4" | gawk 'BEGIN { OFS=" :-( "}; {$1=$1; print $0}'
Results:
1 :-( 2 :-( 3 :-( 4
Explanation:
the $1=$1
is to rebuild the record, using the current OFS (you can also see http://www.gnu.org/software/gawk/manual/gawk.html#Changing-Fields)
Update:
(suggested by @EdMorton and @steve) This is a briefer, equivalent version of the awk command, that sets OFS in the command line, and takes advantage of print $0
as the default action:
awk -v OFS=" :-( " '{$1=$1}1'
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