URL regular expression in C#

Question

I am trying to find a regular expression that will validate as many urls as possible. I'm using it in an input field in MVC3:

[RegularExpression(@"expression...")]

I found this regular expression - "diegoperini", and I loved it but I dont know how to convert it to .net version from php:

_^(?:(?:https?|ftp)://)(?:\S+(?::\S*)?@)?(?:(?!10(?:\.\d{1,3}){3})
(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})
(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])
(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))
|(?:(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)(?:\.
(?:[a-z\x{00a1}-\x{ffff}0-9]+-?)*[a-z\x{00a1}-\x{ffff}0-9]+)*
(?:\.(?:[a-z\x{00a1}-\x{ffff}]{2,})))(?::\d{2,5})?(?:/[^\s]*)?$_iuS

How would it look in net?

Answer 1

You have to remove the PHP-delimiters and change the \x{0000}-patterns with \u0000.
So the regex should look like this.

String sourcestring = "source string to match with pattern";
Regex re = new Regex(@"^(?:(?:https?|ftp)://)(?:\S+(?::\S*)?@)?(?:(?!10(?:\.\d{1,3}){3})(?!127(?:\.\d{1,3}){3})(?!169\.254(?:\.\d{1,3}){2})(?!192\.168(?:\.\d{1,3}){2})(?!172\.(?:1[6-9]|2\d|3[0-1])(?:\.\d{1,3}){2})(?:[1-9]\d?|1\d\d|2[01]\d|22[0-3])(?:\.(?:1?\d{1,2}|2[0-4]\d|25[0-5])){2}(?:\.(?:[1-9]\d?|1\d\d|2[0-4]\d|25[0-4]))|(?:(?:[a-z\u00a1-\uffff0-9]+-?)*[a-z\u00a1-\uffff0-9]+)(?:\.(?:[a-z\u00a1-\uffff0-9]+-?)*[a-z\u00a1-\uffff0-9]+)*(?:\.(?:[a-z\u00a1-\uffff]{2,})))(?::\d{2,5})?(?:/[^\s]*)?$",RegexOptions.IgnoreCase | RegexOptions.Multiline);
Match m = re.Match(sourcestring);
for (int gIdx = 0; gIdx < m.Groups.Count; gIdx++)
{
   Console.WriteLine("[{0}] = {1}", re.GetGroupNames()[gIdx], m.Groups[gIdx].Value);
}

If you want, you can see a live example over here.

Just a short note: This pattern will match a full URL (including username, password, port, path, query, fragment), but it only does detailed validation of the domain part; the other parts are scantily validated. (Thanks to @nhahtdh for clue)