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.NET Regex: Howto extract IPv6 address parts

Answer 1

A:

Regex.Replace() may aid your cause. You could use it like this to return a mutated string with "::" replaced with ":0:" in a string

var myStr = "test::me";
Console.WriteLine( Regex.Replace(myStr, @"::", ":0:") );

I tested that out with LINQPad and it worked just fine for me.

jlafay 2010-05-28 21:27:07

The only problem is you don't know HOW MANY "0:" are replaced by ::

Quandary 2010-05-29 10:20:22

Ok sorry about that, I thought you were encountering "::" and need to replace it with ":0:"

jlafay 2010-05-30 02:44:20

Answer 2

+2 A:

    private static IEnumerable<string> Parse(string input)
    {
        const string partPattern = @"([A-Fa-f0-9]{1,4})";
        string longPattern = string.Format(@"{0} (?:\:{0})*", partPattern); //Group 1 -> 2
        string compactPattern = string.Format(@"{0} (?:\:{0})* \:\: {0} (?:\:{0})*", partPattern, RegexOptions.IgnorePatternWhitespace); //Groups 3 ->6
        string completePattern = string.Format(@"^{0}$ | ^{1}$", longPattern, compactPattern);
        var match = Regex.Match(input, completePattern, RegexOptions.IgnorePatternWhitespace);
        if (match.Success)
        {
            if (match.Groups[1].Success)
            {
                yield return match.Groups[1].Value.PadLeft(4, '0');
                for (int i = 0; i < match.Groups[2].Captures.Count; ++i)
                    yield return match.Groups[2].Captures[i].Value.PadLeft(4, '0');
            }
            else
            {
                var count = 6 - match.Groups[4].Captures.Count - match.Groups[6].Captures.Count;

                //First part
                yield return match.Groups[3].Value.PadLeft(4, '0');
                for (int i = 0; i < match.Groups[4].Captures.Count; ++i)
                    yield return match.Groups[4].Captures[i].Value.PadLeft(4, '0');

                //:: block
                for (int i = 0; i < count; ++i)
                    yield return "0000";

                //Second part
                yield return match.Groups[5].Value.PadLeft(4, '0');
                for (int i = 0; i < match.Groups[6].Captures.Count; ++i)
                    yield return match.Groups[6].Captures[i].Value.PadLeft(4, '0');
            }
        }
        else
            throw new Exception("No match");
    }

Something like this should work. The basic idea is just to split the case of the address containing :: and the rest into two different patterns. I used the assumption that :: can never be at the beginning or end so if that doesn't hold you'll have to modify the patterns a bit.

alun 2010-05-29 13:10:37

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.NET Regex: Howto extract IPv6 address parts

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