IPV6 address into compressed form in Java

问题:

I have used Inet6Address.getByName("2001:db8:0:0:0:0:2:1").toString() method to compress IPv6 address, and the output is 2001:db8:0:0:0:0:2:1 ,but i need 2001:db8::2:1 . , Basically the compression output should based on RFC 5952 standard , that is

1) Shorten as Much as Possible : For example, 2001:db8:0:0:0:0:2:1 must be shortened to
2001:db8::2:1.Likewise, 2001:db8::0:1 is not acceptable, because the symbol "::" could have been used to produce a shorter representation 2001:db8::1.

2) Handling One 16-Bit 0 Field : The symbol "::" MUST NOT be used to shorten just one 16-bit 0 field. For example, the representation 2001:db8:0:1:1:1:1:1 is correct, but 2001:db8::1:1:1:1:1 is not correct.

3) Choice in Placement of "::" : = When there is an alternative choice in the placement of a "::", the longest run of consecutive 16-bit 0 fields MUST be shortened (i.e., the sequence with three consecutive zero fields is shortened in 2001: 0:0:1:0:0:0:1). When the length of the consecutive 16-bit 0 fields are equal (i.e., 2001:db8:0:0:1:0:0:1), the first sequence of zero bits MUST be shortened. For example, 2001:db8::1:0:0:1 is correct representation.

I have also checked another post in Stack overflow, but there was no condition specified (example choice in placement of ::).

Is there any java library to handle this? Could anyone please help me?

Thanks in advance.

回答1:

How about this?

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

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?)

回答2:

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).

回答3:

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.

回答4:

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

回答5:

After performing some tests, I think the following captures all the different IPv6 scenarios:

"((?:(?::0|0:0?)\\b){2,}):?(?!\\S*\\b\\1:0\\b)(\\S*)" -> "::$2"

回答6:

The open-source IPAddress Java library can do as described, it provides numerous ways of producing strings for IPv4 and/or IPv6, including the canonical string which for IPv6 matches rfc 5952. Disclaimer: I am the project manager of that library.

Using the examples you list, sample code is:

    IPAddress addr = new IPAddressString("2001:db8:0:0:0:0:2:1").getAddress();
    System.out.println(addr.toCanonicalString());
    // 2001:db8::2:1
    addr = new IPAddressString("2001:db8:0:1:1:1:1:1").getAddress();
    System.out.println(addr.toCanonicalString());
    // 2001:db8:0:1:1:1:1:1
    addr = new IPAddressString("2001:0:0:1:0:0:0:1").getAddress();
    System.out.println(addr.toCanonicalString());
    // 2001:0:0:1::1
    addr = new IPAddressString("2001:db8:0:0:1:0:0:1").getAddress();
    System.out.println(addr.toCanonicalString());
    //2001:db8::1:0:0:1