How do you extract IP addresses from files using a regex in a linux shell?

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 e

Answer 1

All of the previous answers have one or more problems. The accepted answer allows ip numbers like 999.999.999.999. The currently second most upvoted answer requires prefixing with 0 such as 127.000.000.001 or 008.008.008.008 instead of 127.0.0.1 or 8.8.8.8. Apama has it almost right, but that expression requires that the ipnumber is the only thing on the line, no leading or trailing space allowed, nor can it select ip's from the middle of a line.

I think the correct regex can be found on http://www.regextester.com/22

So if you want to extract all ip-adresses from a file use:

grep -Eo "(([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.){3}([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])" file.txt

If you don't want duplicates use:

grep -Eo "(([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\.){3}([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])" file.txt | sort | uniq

Please comment if there still are problems in this regex. It easy to find many wrong regex for this problem, I hope this one has no real issues.