Bash IP from file

Question

I am trying to get the IPv4 addresses from a text file using the following command:

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 .

This will display the valid IP addresses but will also display the valid part from a longer invalid IP which I don't want.

Input example:

IP address one is 192.168.1.1, ip 2 is 192.168.1.2.
192.168.1.3  192.168.1.10.1
Output I get:
192.168.1.1
192.168.1.2
192.168.1.3
192.168.1.10
Output I want:
192.168.1.1
192.168.1.2
192.168.1.3

Answer 1

You need to exclude all matches where a <digit>. is present immediately before and a .<digit> immediately after the IP.

That is why you need a GNU grep with the following regex:

grep -Po '\b(?<!\d\.)(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b(?!\.\d)' file

The \\b(?<!\\d\\.) at the start matches a word boundary first (it means there must be no letter, digit or _ immediately to the left of the current location) and there must be no digit and . sequence there, either.

The \\b(?!\\.\\d) at the end matches a word boundary first (here, it means there must be no letter, digit or _ immediately to the right of the current location) and there must be no . + digit sequence there, either.

See an online grep demo :

s="IP address one is 192.168.1.1, ip 2 is 192.168.1.2.
192.168.1.3  192.168.1.10.1"
grep -Po '\b(?<!\d\.)(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\b(?!\.\d)' <<< "$s"

Output:

192.168.1.1
192.168.1.2
192.168.1.3

See the PCRE regex demo .