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I'm trying to adapt this: Insert commas into number string to work in dart, but no luck.

either one of these don't work:

print("1000200".replaceAllMapped(new RegExp(r'/(\d)(?=(\d{3})+$)'), (match m) => "${m},"));
print("1000300".replaceAll(new RegExp(r'/\d{1,3}(?=(\d{3})+(?!\d))/g'), (match m) => "$m,"));

Is there a simpler/working way to add commas to a string number?

Community
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Alex L.
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  • In first adaptation you miss `$&,` replacer. It's for JS only, without this replacer expression doesn't work. We should change expression to a little, look at my answer bellow. – kelegorm Aug 11 '15 at 06:54

4 Answers4

51

You just forgot get first digits into group. Use this short one:

'12345kWh'.replaceAllMapped(RegExp(r'(\d{1,3})(?=(\d{3})+(?!\d))'), (Match m) => '${m[1]},')

Look at the readable version. In last part of expression I added checking to any not digit char including string end so you can use it with '12 Watt' too.

RegExp reg = RegExp(r'(\d{1,3})(?=(\d{3})+(?!\d))');
String Function(Match) mathFunc = (Match match) => '${match[1]},';

List<String> tests = [
  '0',
  '10',
  '123',
  '1230',
  '12300',
  '123040',
  '12k',
  '12 ',
];

for (String test in tests) {    
  String result = test.replaceAllMapped(reg, mathFunc);
  print('$test -> $result');
}

It works perfectly:

0 -> 0
10 -> 10
123 -> 123
1230 -> 1,230
12300 -> 12,300
123040 -> 123,040
12k -> 12k
12  -> 12
jamesdlin
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kelegorm
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    Kelegorm, AWESOME! thank you so much! that worked. Here is my working code: `print("1000200".replaceAllMapped(new RegExp(r'(\d{1,3})(?=(\d{3})+(?!\d))'), (Match m) => "${m[1]},"));` – Alex L. Aug 11 '15 at 16:43
  • Thanks! Can you please explain how it works? – batuhankrbb Aug 09 '21 at 03:34
  • It's nothing special, usual Regular Expression stuff. You can read any book about that. I have read that one: https://www.oreilly.com/library/view/mastering-regular-expressions/0596528124/ – kelegorm Sep 27 '21 at 09:37
4
import 'package:intl/intl.dart';

var f = NumberFormat("###,###.0#", "en_US");
print(f.format(int.parse("1000300")));

prints 1,000,300.0 check dart's NumberFormat here

The format is specified as a pattern using a subset of the ICU formatting patterns.

  • 0 A single digit
  • # A single digit, omitted if the value is zero
  • . Decimal separator
  • - Minus sign
  • , Grouping separator
  • E Separates mantissa and expontent
  • + - Before an exponent, to say it should be prefixed with a plus sign.
  • % - In prefix or suffix, multiply by 100 and show as percentage
  • ‰ (\u2030) In prefix or suffix, multiply by 1000 and show as per mille
  • ¤ (\u00A4) Currency sign, replaced by currency name
  • ' Used to quote special characters
  • ; Used to separate the positive and negative patterns (if both present)
Clyde Santos
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2

Try the following regex: (\d{1,3})(?=(\d{3})+$)

This will provide two backreferences, and replacing your number using them like $1,$2, will add commas where they are supposed to be.

Dungeonfire
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    Wouldn't this produce something like `12,423,`, adding a comma at the end? – Flipybitz Aug 11 '15 at 06:38
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    No, it doesn't add an extra comma at the end of the new number – Dungeonfire Aug 11 '15 at 06:39
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    Ah okay, and yeah I just tested it and you're right. GL OP! – Flipybitz Aug 11 '15 at 06:40
  • Dungeonfire >> I tried adapting your suggestion, but no dice. Maybe is my limited experience w/replaceAll. `print("100030".replaceAll(new RegExp(r'(\d{1,3})(?=(\d{3})+$)'), '$1,$2');` That didn't work..neither using replaceAllMapped. – Alex L. Aug 11 '15 at 16:41
1

Let's take the example amount 12000. now our expected amount should be 12,000.00

so, the solution is

double rawAmount = 12000;
String amount = rawAmount.toStringAsFixed(2).replaceAllMapped(RegExp(r'(\d{1,3})(?=(\d{3})+(?!\d))'), (Match m) => '${m[1]},');

or if you don't want to add .00 then, we just need to use toString() instead of toStringAsFixed().

String amount = rawAmount.toString().replaceAllMapped(RegExp(r'(\d{1,3})(?=(\d{3})+(?!\d))'), (Match m) => '${m[1]},');
Rushi Dave
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