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 $1 will only work for the first match.
I'm getting the feeling I would be better off performing this task using a loop.
UPDATE:
Due to AS3 not supporting lookahead this is how I have solved it.
public static function formatNumber(number:Number):String
{
var numString:String = number.toString()
var result:String = ''
while (numString.length > 3)
{
var chunk:String = numString.substr(-3)
numString = numString.substr(0, numString.length - 3)
result = ',' + chunk + result
}
if (numString.length > 0)
{
result = numString + result
}
return result
}
If your language supports postive lookahead assertions, then I think the following regex will work:
(\d)(?=(\d{3})+$)
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,"));
}
}