CODEWITHSUNDEEP
regex
hoju
hoju

Reputation: 29472

regular expression to parse IP

I am after a regular expression to parse IP addresses and extract their host, port, username, and password.

Here are the formats I am interested in:

Upvotes: 2

Views: 9421

Answers (6)

Alban
Alban

Reputation: 3215

for match exclusively a valid IP adress use

(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]?)){3}

instead of

([01]?[0-9][0-9]?|2[0-4][0-9]|25[0-5])(\.([01]?[0-9][0-9]?|2[0-4][0-9]|25[0-5])){3}

because many regex engine match the first possibility in the OR sequence

you can try your regex engine : 10.48.0.200

Upvotes: 0

Philar
Philar

Reputation: 3897

Here is a small script whipped up in perl that does the following things a) Strips out username and password after checking that the former starts with a character b) Validates ip address c) validated port

#!/usr/bin/perl

    while (<>) {
    chomp;
            if (/(?:(?:([a-zA-z]\w+)\:(\w+))@)?((\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3}))(?:\:(\d{1,5}))?/) {
                    print "username=$1\n";
                    print "password=$2\n";
                    print "ip address=$3\n";
                    print "port=$8\n";
                    print "Warning: IP Address invalid\n" if ($4>255||$5>255||$6>255||$7>255);
                    print "Warning: Port Address invalid\n" if ($8>65535);
            }
    }

EDIT: Recommendation from tchrist below

Upvotes: 2

Merlyn Morgan-Graham
Merlyn Morgan-Graham

Reputation: 59131

Doing the match this way may not be a best practice. It might be better to plug into some sort of code with real smarts in it, that can do general-purpose URI parsing. If you have limited needs, though, and can comment/document thoroughly that your code will break if you demand more of it, then maybe it makes sense to go down this path.

The simplest way is to match four sets of 1 to 3 digits, with:

  • optionally, one-or-more not-:, plus :, plus one-or-more not-@, plus @
  • optionally, :, plus 1 to 5 digits

Something like:

([^:]+:[^@]+@)?(\d{1,3}\.){3}\d{1,3}(:\d{1,5})?

But this would accept silly stuff, like "999.999.999.999:99999"

If you only want to accept valid IP addresses, and don't care that it happens to be part of a URI, or don't care what other garbage exists in the string, here is an example:

http://www.regular-expressions.info/examples.html

It basically matches four sets of:

  • 2, plus 0-4, plus 0-9
  • or 2, plus 5, plus 0-5
  • or 1, plus 0-9, plus 0-9
  • or 1-9, plus 0-9
  • or 0-9

That should get you started.

  • optionally, one-or-more not-:, plus :, plus one-or-more not-@, plus @ (max lengths may be interesting, here)
  • optionally, :, plus 0-65535 (this I'll leave up to you, based on the 0-255 rules above)

There are other range-based rules for matching IP addresses that you might want to avoid (stuff like 0.0.0.0, and reserved ranges), but it may be easier to do subsequent matching for these.

Basically, I'd suggest you use the very-simple example, or plug into a library.

Upvotes: 4

Anton Hansson
Anton Hansson

Reputation: 2171

Try something like this

(http://(\w+:\w+@)?)?(\d{1,3}\.){3}\d{1,3}(:\d{1,5})?

Explanation:

(http://(\w+:\w+@)?)? - optional group of http:// followed by optional user:pass@
(\d{1,3}\.){3} - three groups of one to three digits followed by a dot
\d{1,3} - one to three digits
(:\d{1,5})? - optional group of colon followed by one to five digits

Upvotes: 7

Antoine Pelisse
Antoine Pelisse

Reputation: 13129

You can start with that (python):

import re

pattern = "((?P<login>\w+):(?P<password>\w+)@)?(?P<ip>\d{1,3}.\d{1,3}.\d{1,3}.\d{1,3})(:(?P<port>\d+))?"

re.match(pattern, "12.34.56.789").groupdict()
re.match(pattern, "12.34.56.789:80").groupdict()
re.match(pattern, "john:[email protected]:80").groupdict()

And obviously, the IP you specified is not valid (as Matt says ...)

Upvotes: 3

Chathuranga Chandrasekara
Chathuranga Chandrasekara

Reputation: 20956

Regexlib would be a helpful resource for your question. You can find many solutions (May be you will need to combine some)

Upvotes: 0

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