Question

I need a Regex for numbers that considers thousands and decimal places. The one I came up so far is:

\d{1,3}(\.\d{3})*(,\d*)?

It will catch 3.000,00 as well as 3,00 and 3.000. Sweet.

The problem is that I need it to be very greedy, and capture larger numbers first, for instance, for this input:

125,45.124.890,45,32,67.456

I’d need it to capture 45.124.890,45 as a number. It’s important to me because it’s the biggest one. My regex just won’t work, because it will capture 123,45 and then 45 will not be considered for the next match. So the next number it captures is 124.890,45 which is not good for me.

Any thoughts?

PS As I’m brazillian the separators are inverted

Answer 1

Use regex pattern

(?!0\d)\d{1,3}(?:\.\d{3})*(?:\,\d+(?![\d\.]))?

C# demo:

using System;
using System.Text.RegularExpressions;

public class Test
{
  public static void Main()
  {
    const string line = @"125,45.124.890,45,32,67.456";
    MatchCollection matches = Regex.Matches(line, 
      @"(?!0\d)\d{1,3}(?:\.\d{3})*(?:\,\d+(?![\d\.]))?");
    foreach(Match match in matches)
    {
      foreach (Capture capture in match.Captures)
      {
        Console.WriteLine(capture.Value);
      }
    }
  }
}

Output:

125
45.124.890,45
32
67.456

Test this demo here.