Get string by regex bash script

Question

I work with bash script. I have lines

3    ACCEPT     tcp  --  0.0.0.0/0            0.0.0.0/0            state NEW tcp dpt:22
4    ACCEPT     tcp  --  0.0.0.0/0            0.0.0.0/0            state NEW tcp dpt:80

I want to get "22,80" from this lines.Can anybodey help me?

You can check here tldp.org/LDP/Bash-Beginners-Guide/html/… or you can do it easily in perl, python etc. — Netro
– Netro, Commented Jun 26, 2013 at 11:42

fedorqui · Accepted Answer · 2013-06-26 11:47:26Z

1

As the delimiter before 20 and 80 is :, you can mainly do it with cut:

$ cut -d: -f2 file
22
80

With bash:

$ while read line
do
  echo ${line#*:}
done < file
22
80

Even with awk:

$ awk -F: '{print $2}' file
22
80

And to complete it, with sed:

$ sed  's/.*://' file
22
80

edited Jun 26, 2013 at 11:47

answered Jun 26, 2013 at 11:41

fedorqui

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3 Comments

jaypal singh Over a year ago

perl -F: -lane 'print $F[1]' file ... +1 ;)

fedorqui Over a year ago

Jaypal and his new look to the rescue :) Thanks for improving my answer!

RARE Kpop Manifesto Over a year ago

@fedorqui @jaypal : dang perl is so verbose compared to awk -F: '$_=$2'

Nikolai Popov · Accepted Answer · 2013-06-26 11:42:39Z

0

awk 'BEGIN{FS=":"} {print $2}' file

answered Jun 26, 2013 at 11:42

Nikolai Popov

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