Regex: How can I match third IPv4 address? - regex

Question

I'm a regex noob and for the life of me I can't figure out how to match the third IPv4 address on line that contains three IPv4 addresses.
The line in question:
ip route 214.25.48.547 255.255.255.255 16.48.75.46 name Chicago-VPN
The regex I have so far that matches one IP:
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})
If I put a {3} at the end of it, it breaks. I think it has something to do with the spaces between the addresses but I can't figure out how to handle that. I need to capture the third address.
https://regex101.com/r/mN3cR6/1

Answer 1

You just need to add a multiline modifier to the code.
Your new code should be like this
/([0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3})/g
See this demo https://regex101.com/r/mN3cR6/2

Answer 2

Try
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\s?)+
This should match one, two, or three, or even more "IPs".
Or
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})\s([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})\s([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})
for exactly 3.
Or
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\s?){3}
for a shorter formula with some possible errors.
Note that the basic idea is problematic too, as it matches "999.999.999.999" when it is definitely not a valid IP address.

Answer 3

The following should match the third ip
(?:[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\s){2}([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})
It's possible to be more compact depending what language you're using - for instance in ruby
string.scan(/([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})/)[2]
would give you what you want. You could also collapse the multiple [0-9]{1,3}. instances using non matching groups and counts

Answer 4

The problem is, that the regex needs to not only contain the IPs but also the spaces between the IPs.
So adding a space into the repeated group should do the trick:
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3} ){3}
If you don't want tat space in the final match, you make it non-greedy, using ?? (or *?):
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3} ??){3}
Also note, that your regex matches more than just valid IPs. e.g. 999.999.999.999 would match nicely.

Answer 5

You are already matching all three IPs with that regex.
([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})
Match 1
214.25.48.547
Match 2
255.255.255.255
Match 3
16.48.75.46
You can test it here:
http://rubular.com/
The problem may be with how you are trying to access them.
In Ruby, your regex works perfectly:
regex = /([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})/
"ip route 214.25.48.547 255.255.255.255 16.48.75.46 name Chicago-VPN".scan(regex)
=> [["214.25.48.547"], ["255.255.255.255"], ["16.48.75.46"]]