Dart how to add commas to a string number - regex

Question

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?

Answer 1

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

Answer 2

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)

Answer 3

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.

Answer 4

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]},');

Answer 5

extension on int {
String get priceString {
final numberString = toString();
final numberDigits = List.from(numberString.split(''));
int index = numberDigits.length - 3;
while (index > 0) {
numberDigits.insert(index, ',');
index -= 3;
}
return numberDigits.join();
}
}

Answer 6

because in case of type double, the output will change based on the way, so check them.
If you need to format integer then any way works.
//1233.45677 => 1,233.4567
String num='1233.45677';
RegExp pattern = RegExp(r'(?<!\.\d*)(\d)(?=(?:\d{3})+(?:\.|$))');
String Function(Match) replace = (m) => '${m[1]},';
print(num..replaceAllMapped(pattern, replace));
//1233.45677 => 1,233.456,7
String num='1233.45677';
pattern = RegExp(r'(\d{1,3})(?=(\d{3})+(?!\d))');
String Function(Match) replace = (m) => '${m[1]},';
print(num..replaceAllMapped(pattern, replace));
//1233.45677 => 1,233.46
//after import intl package, to be able to use NumberFormat
String num='1233.45677';
var f = NumberFormat("###,###.0#", "en");
print(f.format(double.parse()));
if the number is in String type.
//in case of int data type
int.parse(num);
//in case of double data type
double.parse(num);