Regex ignore part of a String - Java

Question

I am parsing an IP, and I don't care about anything, but the IP. Here is what I have, but I don't care what follows after the '10', and just want to know if the String matches the IP:

[0-9]{1,3}\\.[0-9]{1,3}\\.(16|249)\\.10

What can I add into this to make it ignore everything else? This IP will be at the very beginning of the String every time as well.

Add .* after the 10? What does your string look like? How do you use the regex? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jun 14, 2016 at 22:17
Might want to use a better regex. safaribooksonline.com/library/view/regular-expressions-cookbook/… — OneCricketeer
– OneCricketeer, Commented Jun 14, 2016 at 22:18

Wiktor Stribiżew · Accepted Answer · 2016-06-14 22:22:30Z

1

If your string starts with a specific IP pattern, and you are using String#matches(), just append a word boundary after 10 and use .* after it:

"(?s)\\d{1,3}\\.\\d{1,3}\\.(?:16|249)\\.10\\b.*"

The (?s) is added to make sure you will match the whole string that can contain newlines.

Instead of a \b you may use (?!\d) ("(?!\\d)") to disallow matching IPs ending with 100 rather than 10.

NOTE that the first and second parts (\\d{1,3}) can be enhanced by replacing them with (?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?) and the regex would look like "(?s)(?: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]?)\\.(?:16|249)\\.10\\b.*".

answered Jun 14, 2016 at 22:22

Wiktor Stribiżew

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