CODEWITHSUNDEEP
regex
d82k
d82k

Reputation: 379

regex extract all valid IPv4 from string

From a similar string below I would like to extract only the valid IPv4 addresses

233.444.444.222 , 127.0.0.1 , 127.0.0.0.1 , 192.168.1.1 , 812.1.1.1.1

where I have some valid ip addresses and some random numbers separated by "." Using the below:

\b(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\b

I extract also 127.0.0.0 and 1.1.1.1 which are not good. Any suggestion please?

KR dk

Upvotes: 0

Views: 95

Answers (1)

blhsing
blhsing

Reputation: 106465

You can use negative lookarounds to avoid matching an IP-address led or trailed by a dot:

(?<!\.)\b(?:(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\.){3}(?:25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9]?[0-9])\b(?!\.)

Demo: https://regex101.com/r/6yHTnY/1

Upvotes: 1

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