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copied!<p>I don't know C# regex or what you are trying to do (ie: edit, validate,.. etc)<br> I know Perl, so I'll take a stab at it.<br> In regex's a period is a meta character. If you want a literal period, you have to escape<br> it with a '.' The meta period says match any character. If you want a literal period,<br> and to keep what you have, in Perl you either should not double quote the regex or escape<br> the escape. A single quote should work fine.<br> <code>'((www)\.([a-zA-Z0-9]{1,6}+\.+)*[a-zA-Z]{2,6})'</code> </p> <p>Now, unless you want multiple literal periods, you should get rid of the + quantifier<br> <code>'((www)\.([a-zA-Z0-9]{1,6}+\.)*[a-zA-Z]{2,6})'</code> </p> <p>As was said, {n,m} itself is a quantifier. Adding + after it is a double quantifier.<br> In Perl adding + after a quantifyer implies a possesive condition and is legal starting in version 5.10<br> So, getting rid of the extra + it is now<br> <code>'((www)\.([a-zA-Z0-9]{1,6}\.)*[a-zA-Z]{2,6})'</code> </p> <p>Finally, the * quantifyer implies 0 or more times. Why would you want to match 0 times?<br> Fixing that it is now<br> <code>'((www)\.([a-zA-Z0-9]{1,6}\.)+[a-zA-Z]{2,6})'</code></p> <p>As an extra, you have a main grouping around everything, a grouping around www, and one<br> around the middle ([a-zA-Z0-9]{1,6}.)+ which does no good in capture, just in grouping.<br> If you want to capture the beginning, middle, end, you should add appropriate capture<br> <code>'((www\.)(([a-zA-Z0-9]{1,6}\.)+)([a-zA-Z]{2,6}))'</code> </p> <p>Or, in Perl, that would be better written as<br> <code>'((www\.)((?:[a-zA-Z0-9]{1,6}\.)+)([a-zA-Z]{2,6}))'</code> </p>

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