Reverse : Regular Expression « Regular Expression « C# / CSharp Tutorial






using System;
using System.Text;
using System.Text.RegularExpressions;

public class MainClass
{
    static void Main( string[] args ) {
        // Create regex to search for IP address pattern.
        string pattern = @"(?<part1>[01]?\d\d?|2[0-4]\d|25[0-5])\." +
                         @"(?<part2>[01]?\d\d?|2[0-4]\d|25[0-5])\." +
                         @"(?<part3>[01]?\d\d?|2[0-4]\d|25[0-5])\." +
                         @"(?<part4>[01]?\d\d?|2[0-4]\d|25[0-5])";
        Regex regex = new Regex( pattern );
        Match match = regex.Match( "192.168.123.1" );

        string replace = @"${part4}.${part3}.${part2}.${part1}" +
                         @" (the reverse of $&)";
        Console.WriteLine( regex.Replace("192.168.123.1", replace) );
    }
}
1.123.168.192 (the reverse of 192.168.123.1)








17.1.Regular Expression
17.1.1.Use Regular Expressions to split string
17.1.2.Do a replace using the Regex
17.1.3.Reverse
17.1.4.Regular Expresion Part
17.1.5.new Regex('^\\d+') 1
17.1.6.new Regex('\d+$') 2
17.1.7.new Regex('^\\d+$') 3
17.1.8.new Regex('(abc)|(xyz)*') 1
17.1.9.new Regex('((abc)|(xyz))*') 2
17.1.10.new Regex('((?:abc)|(?:xyz))*') 3
17.1.11.new Regex(?(^\d)^\d+$|^\D+$)
17.1.12.new Regex((abc)*)x(\1)
17.1.13.new Regex(^\d+$\n+, RegexOptions.Multiline)
17.1.14.new Regex((abc)*abc((abcd)|z)bc)
17.1.15.new Regex(<[^>]+>[^<]*]+>): for XML tag
17.1.16.new Regex(<([^>]+)>[^<]*: for XML tag
17.1.17.Using RegEx