How do you extract IP addresses from files using a

Question

How to extract a text part by regexp in linux shell? Lets say, I have a file where in every line is an IP address, but on a different position. What is the simplest way to extract those IP addresses using common unix command-line tools?

Answer 1

You can use sed. But if you know perl, that might be easier, and more useful to know in the long run:

perl -n '/(\d+\.\d+\.\d+\.\d+)/ && print "$1\n"' < file

Answer 2

Most of the examples here will match on 999.999.999.999 which is not technically a valid IP address.

The following will match on only valid IP addresses (including network and broadcast addresses).

grep -E -o '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)' file.txt

Omit the -o if you want to see the entire line that matched.

Answer 3

grep -E -o "([0-9]{1,3}[.]){3}[0-9]{1,3}"

Answer 4

for centos6.3

ifconfig eth0 | grep 'inet addr' | awk '{print $2}' | awk 'BEGIN {FS=":"} {print $2}'

Answer 5

I wrote a little script to see my log files better, it's nothing special, but might help a lot of the people who are learning perl. It does DNS lookups on the IP addresses after it extracts them.

Answer 6

Everyone here is using really long-handed regular expressions but actually understanding the regex of POSIX will allow you to use a small grep command like this for printing IP addresses.

grep -Eo "(([0-9]{1,3})\.){3}([0-9]{1,3})"

(Side note) This doesn't ignore invalid IPs but it is very simple.