IPV6 address into compressed form in Java

I recently ran into the same problem and would like to (very slightly) improve on Tim's answer.

The following regular expression offers two advantages:

((?:(?:^|:)0+\\b){2,}):?(?!\\S*\\b\\1:0+\\b)(\\S*)

Firstly, it incorporates the change to match multiple zeroes. Secondly, it also correctly matches addresses where the longest chain of zeroes is at the beginning of the address (such as 0:0:0:0:0:0:0:1).

java-ipv6 is almost what you want. As of version 0.10 it does not check for the longest run of zeroes to shorten with :: - for instance 0:0:1:: is shortened to ::1:0:0:0:0:0. It is a very decent library for the handling of IPv6 addresses, though, and this problem should be fixed with version 0.11, such that the library is RFC 5952 compliant.

Guava's InetAddresses class has toAddrString() which formats according to RFC 5952.

How about this?

String resultString = subjectString.replaceAll("((?::0\\b){2,}):?(?!\\S*\\b\\1:0\\b)(\\S*)", "::$2").replaceFirst("^0::","::");

Explanation without Java double-backslash hell:

(       # Match and capture in backreference 1:
 (?:    #  Match this group:
  :0    #  :0
  \b    #  word boundary
 ){2,}  # twice or more
)       # End of capturing group 1
:?      # Match a : if present (not at the end of the address)
(?!     # Now assert that we can't match the following here:
 \S*    #  Any non-space character sequence
 \b     #  word boundary
 \1     #  the previous match
 :0     #  followed by another :0
 \b     #  word boundary
)       # End of lookahead. This ensures that there is not a longer
        # sequence of ":0"s in this address.
(\S*)   # Capture the rest of the address in backreference 2.
        # This is necessary to jump over any sequences of ":0"s
        # that are of the same length as the first one.

Input:

2001:db8:0:0:0:0:2:1
2001:db8:0:1:1:1:1:1
2001:0:0:1:0:0:0:1
2001:db8:0:0:1:0:0:1
2001:db8:0:0:1:0:0:0

Output:

2001:db8::2:1
2001:db8:0:1:1:1:1:1
2001:0:0:1::1
2001:db8::1:0:0:1
2001:db8:0:0:1::

(I hope the last example is correct - or is there another rule if the address ends in 0?)