Regular Expression needed

Yeah thanks for ur comment vulpes 

Yeah, mine didn't deal with this one either:

    "0000,123.45"

However, Dorababu's expression seems to be dealing with all cases correctly, so I'd use that.

Excellent - thank you very much!

However, it is not perfect - for example 1234,567 (missing comma after the 1) is deemed valid by the Reg Exp.
Having said that, I think that the solution provided is "good enough"

Thanks
Gary

The following seems to be working but I'd test it thoroughly before use. I've assumed there should not be more than two digits after a decimal point, if there is one:

class Test
{
   static void Main()
   {
       string[] numbers = new string[]
       {
         "1.23",
         "100",
         "1,234",
         "1234",
         "1,234.56",
         "0.56",
         "1,234,567.89",
         "1.ab",
         "1,2345.67",
         "0,123.45",
         "1.24,687",
         "0.5" 
       };
       string regExp = @"^(?!0{1,3},)\d{1,3}(,?\d{3})*(.\d{1,2})?$";
       Regex r = new Regex(regExp);

       foreach(string number in numbers)
       {
         Console.WriteLine(r.IsMatch(number));
       }
       Console.ReadKey();
    }
} 

yes, 0.5 should be allowed.

Other examples of allowed numbers include:

1,234,567
1,234,567.89
12,345,678
12,345,678.9
123,456,789

and so on (basically any number that we as humans see as valid)