Question

Hey there, I’m trying to perform a backwards regular expression search on a string to divide it into groups of 3 digits. As far as i can see from the AS3 documentation, searching backwards is not possible in the reg ex engine.

The point of this exercise is to insert triplet commas into a number like so:

10000000 => 10,000,000 

I’m thinking of doing it like so:

string.replace(/(\d{3})/g, ',$1') 

But this is not correct due to the search not happening from the back and the replace $1 will only work for the first match.

I’m getting the feeling I would be better off performing this task using a loop.

UPDATE:

Due to AS3 not supporting lookahead this is how I have solved it.

public static function formatNumber(number:Number):String {     var numString:String = number.toString()     var result:String = ''      while (numString.length > 3)     {         var chunk:String = numString.substr(-3)         numString = numString.substr(0, numString.length - 3)         result = ',' + chunk + result     }      if (numString.length > 0)     {         result = numString + result     }      return result } 

Answer 1

If your language supports postive lookahead assertions, then I think the following regex will work:

(\d)(?=(\d{3})+$) 

Demonstrated in Java:

import static org.junit.Assert.assertEquals;  import org.junit.Test;  public class CommifyTest {      @Test     public void testCommify() {         String num0 = '1';         String num1 = '123456';         String num2 = '1234567';         String num3 = '12345678';         String num4 = '123456789';          String regex = '(\\d)(?=(\\d{3})+$)';          assertEquals('1', num0.replaceAll(regex, '$1,'));         assertEquals('123,456', num1.replaceAll(regex, '$1,'));         assertEquals('1,234,567', num2.replaceAll(regex, '$1,'));         assertEquals('12,345,678', num3.replaceAll(regex, '$1,'));         assertEquals('123,456,789', num4.replaceAll(regex, '$1,'));         }     }