awk - How to filter on column value in bash -

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i'm facing problem while trying grep (filter) logfile on value of integer.

logfile.log:

2014-11-16 21:22:15 8 10.133.23.9 proxied ... 2014-11-16 21:22:15 1 163.104.40.133 authentication_failed denied ... 2014-11-16 21:22:15 15 163.104.40.134 authentication_failed denied ... 2014-11-16 21:22:16 9 163.104.124.209 proxied ...

i have int in column 3 : 8, 1, 15, 9.

i need grep if: value > 10.

something like: 

cat logfile.log | grep {$2>10} 2014-11-16 21:22:15 15 163.104.40.134 authentication_failed denied ...

this should make!

$ awk '$3>10' file 2014-11-16 21:22:15 15 163.104.40.134 authentication_failed denied ...

with $3 refer 3rd field, $3>10 means: lines having 3rd field bigger 10. if accomplished, condition true , hence awk performs default behaviour: print line.

you of course say: print specific field, saying awk '$3>10 {print $1}' file, example.


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