Not really.

You can:

strmemo = strmemo.Replace("@@Name",
value1.ToString())Replace("@@Surname",
value2.ToString()).Replace("@@Address", value3.ToString());

but it is still 3 replaces.

But why not:

strmemo = String.Format("new Data for user;  Name:{0}, Surname:{1},
Address:{2}", value1.ToString(), value2.ToString(), value3.ToString());

it looks much better.

Arne

How about using a custom evaluator? Now all you need to do is add things
to the dictionary and they should work... For the record, you might also
want to handle escaping (i.e. how do yuo write @@Foo as a literal?), and
termintated tokens (i.e. you do you write a literal immediately after a
token - i.e. "@@Size" + "mm" (if you see what I mean) - @Size@mm would
be easier... (i.e. @Foo@ is a token for the argument Foo)

Marc

using System.Collections.Generic;
using System.Text.RegularExpressions;
static class Program
{
    static readonly Regex regex = new Regex("(@@)([a-zA-z]+)",
RegexOptions.Compiled);
    static void Main()
    {
        Dictionary<string, string> args = new Dictionary<string, string>();
        args.Add("Surname", "Smith");
        args.Add("Name", "John");
        args.Add("Address", "Somewhere");

        string input = "new Data for user; Missing:@@Missing,
Name:@@Name , Surname:@@Surname, Address:@@Address";

        string output = regex.Replace(input, delegate(Match match)
        {
            string value;
            args.TryGetValue(match.Groups[2].Value, out value);
            return value;
        });
    }
}