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foamy
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C# E-mail formatting

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Hi all,

I'm useless at regex, but for the task I'm currently working on it seems like the correct approach. I have a Windows Service (.NET 2.0) which needs to filter e-mail addresses.
It will receive input in a format like these:

library1@domain.com
library2@domain.dk
library3@subdomain.domain.com
etc..


What i need to do is format them into this:

library1.domain@predefined-domain.dk
library2.domain@predefined-domain.dk
library3.subdomain.domain@predefined-domain.dk


If anyone can point (or throw) me in the right direction I'd appreciate it :)


EDIT: I came up with this:

public static string FormatEmail(string inputAddress, string newDomain)
{
    string result = inputAddress.Replace("@", ".");
    foreach (string domain in topleveldomains)
    {
        if (result.LastIndexOf(domain) == result.Length - domain.Length)
        {
            result = result.Substring(0, result.LastIndexOf(domain));
            result += "@" + newDomain;
            break;
        }
    }
    return result;
}

c# email regex format
  • asked 4 months ago
  • edited 4 months ago
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last answered 4 months ago

1 answers

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up1down
tick
link

Well, you could use regex but, if I've understood the pattern correctly, I think you'll find it easier and quicker to use ordinary string parsing here:

using System;

class Test
{
    static void Main()
    {
       string[] addresses =
       {
          "library1@domain.com",
          "library2@domain.dk",
          "library3@subdomain.domain.com"
       };

       foreach(string address in addresses)
       {
          Console.WriteLine(Reformat(address));
       }

       Console.ReadKey();   
    }

    static string Reformat(string address)
    {
        string[] items = address.Split('@');
        if (items.Length != 2) return ""; // or whatever 
        int index = items[1].LastIndexOf('.');
        return items[0] + "." + items[1].Substring(0, index) + "@predefined-domain.dk";
    }
}

  • answered 4 months ago by:

    vulpes
    12813
(x)
foamy
2135

That's somewhat simpler than the solution I came up with. They both seem to produce the same result though :) Thanks a bunch

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