How to remove trailing zeros using Dart

Question

I would like the optimal solution for removing trailing zeros using Dart. If I have a double that is 12.0 it should output 12. If I have a double that is 12.5 it should output 12.5

score 60 · Accepted Answer · edited Jan 11 '22 at 08:55

60

I made regular expression pattern for that feature.

double num = 12.50; // 12.5
double num2 = 12.0; // 12
double num3 = 1000; // 1000

RegExp regex = RegExp(r'([.]*0)(?!.*\d)');

String s = num.toString().replaceAll(regex, '');

edited Jan 11 '22 at 08:55

Tumist

119
3
11

answered Mar 14 '19 at 23:55

Yuna

1,323
10
8

3

This works for all scenarios. The toStringAsFixed(n.truncateToDouble() method in my answer auto rounds decimal place based values to 1 when they are less than 1 (i.e. 0.95) which is an unwanted side effect. – John Mar 15 '19 at 23:12
Why didn't you use the regex you've declared and assiged? It's this part: `RegExp regex = RegExp(r"([.]*0)(?!.*\d)");` Instead, you've given a new RegExp into `replaceAll`. – SametSahin Dec 30 '20 at 18:16
2

`double numx = 12.500; ` return value is 12.50; – alexwan02 Jan 05 '21 at 08:00
1

RegExp(r'\.0') is better for me – temirbek May 11 '21 at 11:39
8

@alexwan02 use `RegExp(r"([.]*0+)(?!.*\d)")` to remove all trailing zeroes after the decimal point. – Oleksii Shliama May 30 '21 at 02:26
6

@Yuna Could you check your answer again? When you pass 1000, you'll get 100. Why? – RuslanBek Jan 19 '22 at 09:38
2

@Ruslanbek0809, try RegExp(r'(\.[0]+)') – Fəqan Çələbizadə Feb 04 '22 at 14:34
6

Be careful - I confirmed that the REGEX in the accepted answer turns 1000 into 100 despite the commented code saying otherwise. The Regex above my comment does not work properly either @FəqanÇələbizadə - it turns 1000.05 into 10005 – JakeD Apr 05 '22 at 21:15
1

This approach has flaws, don't use for prices (1000 -> 100) – Kirill Karmazin Apr 19 '22 at 12:11
This writes 1 instead of 10 – gurnisht Jul 21 '22 at 07:38
This is incorrect answer, please down vote this. Following this resulted into major problem in my app. – Ashish Khurange Aug 22 '22 at 06:37
This converts 10.0 to 1. So don't use it – Muhammad Umair Saqib Aug 06 '23 at 07:22

John · Answer 2 · 2019-03-14T20:35:33.360

23

UPDATE
A better approach, just use this method:

String removeDecimalZeroFormat(double n) {
    return n.toStringAsFixed(n.truncateToDouble() == n ? 0 : 1);
}

OLD
This meets the requirements:

double x = 12.0;
double y = 12.5;

print(x.toString().replaceAll(RegExp(r'.0'), ''));
print(y.toString().replaceAll(RegExp(r'.0'), ''));

X Output: 12
Y Output: 12.5

edited Mar 14 '19 at 20:35

answered Mar 14 '19 at 17:41

John

2,551
3
17
29

it does not work if the variable x/y not in double. – Muhammad Tameem Rafay Feb 19 '21 at 13:31
2

it is not working for 99.96, it giving me 100. – Mohit Modh Jun 22 '21 at 09:50

score 19 · Answer 3 · answered Sep 01 '19 at 19:36

19

Use NumberFormat:

String formatQuantity(double v) {
  if (v == null) return '';

  NumberFormat formatter = NumberFormat();
  formatter.minimumFractionDigits = 0;
  formatter.maximumFractionDigits = 2;
  return formatter.format(v);
}

answered Sep 01 '19 at 19:36

greensuisse

1,727
16
18

1

be careful because this approach from`4.003000` will give you `4` – Kirill Karmazin Apr 19 '22 at 12:02
i love you man <3 – wesleyhodaj Sep 22 '22 at 13:28

August Kimo · Answer 4 · 2021-06-16T19:13:26.973

13

Lots of the answers don't work for numbers with many decimal points and are centered around monetary values.

To remove all trailing zeros regardless of length:

removeTrailingZeros(String n) {
  return n.replaceAll(RegExp(r"([.]*0+)(?!.*\d)"), "");
}

Input: 12.00100003000

Output: 12.00100003

If you only want to remove trailing 0's that come after a decimal point, use this instead:

removeTrailingZerosAndNumberfy(String n) {
    if(n.contains('.')){
      return double.parse(
        n.replaceAll(RegExp(r"([.]*0+)(?!.*\d)"), "") //remove all trailing 0's and extra decimals at end if any
      );
    }
    else{
      return double.parse(
        n
      );
    }
  }

edited Jun 16 '21 at 19:13

answered Jun 16 '21 at 18:34

August Kimo

1,503
8
17

so far this answer with `removeTrailingZerosAndNumberfy()` method is the best approach if you need to show prices on UI – Kirill Karmazin Apr 19 '22 at 12:24
this one worked for me after having explored the above solutions. – MetaStack Jan 03 '23 at 23:54

score 10 · Answer 5 · answered Jan 29 '20 at 09:15

10

If what you want is to convert a double without decimals to an int but keep it as a double if it has decimals, I use this method:

num doubleWithoutDecimalToInt(double val) {
  return val % 1 == 0 ? val.toInt() : val;
}

answered Jan 29 '20 at 09:15

SimonC

365
3
11

Great! Cheers SimonC – Sam Jun 29 '21 at 10:41

score 6 · Answer 6 · answered Jul 24 '20 at 12:56

6

I found another solution, to use num instead of double. In my case I'm parsing String to num:

void main() {
 print(num.parse('50.05').toString()); //prints 50.05
 print(num.parse('50.0').toString()); //prints 50
}

answered Jul 24 '20 at 12:56

dudeck

393
5
13

2

but num.parse('50.00') returns 50.0 – Fəqan Çələbizadə Feb 03 '22 at 17:04
1

Does not work for me when running dart from terminal. `num.parse('10.0').toString()` returns 10.0. I tried in DartPad and then it returned 10 however. – Simon Bengtsson Aug 20 '22 at 14:37

score 4 · Answer 7 · answered Oct 23 '20 at 11:58

Here is what I've come up with:

extension DoubleExtensions on double {
  String toStringWithoutTrailingZeros() {
    if (this == null) return null;
    return truncateToDouble() == this ? toInt().toString() : toString();
  }
}


void main() {
  group('DoubleExtensions', () {
    test("toStringWithoutTrailingZeros's result matches the expected value for a given double",
        () async {
      // Arrange
      final _initialAndExpectedValueMap = <double, String>{
        0: '0',
        35: '35',
        -45: '-45',
        100.0: '100',
        0.19: '0.19',
        18.8: '18.8',
        0.20: '0.2',
        123.32432400: '123.324324',
        -23.400: '-23.4',
        null: null
      };

      _initialAndExpectedValueMap.forEach((key, value) {
        final initialValue = key;
        final expectedValue = value;

        // Act
        final actualValue = initialValue.toStringWithoutTrailingZeros();

        // Assert
        expect(actualValue, expectedValue);
      });
    });
  });
}

great answer, it works even if there are more than one trailing zeros. Thanks. — Mol0ko, Mar 09 '21 at 07:11

score 3 · Answer 8 · answered May 08 '20 at 20:20

String removeTrailingZero(String string) {
  if (!string.contains('.')) {
    return string;
  }
  string = string.replaceAll(RegExp(r'0*$'), '');
  if (string.endsWith('.')) {
    string = string.substring(0, string.length - 1);
  }
  return string;
}

======= testcase below =======

000 -> 000
1230 -> 1230
123.00 -> 123
123.001 -> 123.001
123.00100 -> 123.001
abc000 -> abc000
abc000.0000 -> abc000
abc000.001 -> abc000.001

score 2 · Answer 9 · answered Sep 06 '20 at 16:29

2

To improve on What @John's answer: here is a shorter version.

String formatNumber(double n) {
 return n.toStringAsFixed(0) //removes all trailing numbers after the decimal. 
 }

answered Sep 06 '20 at 16:29

ombiro

869
1
12
19

`print((1.0 / 2).toStringAsFixed(0));` displays `1` instead of `0.5` – Udesh Apr 20 '23 at 05:43

Nick Hermann · Answer 10 · 2022-08-29T10:36:45.760

This function removes all trailing commas. It also makes it possible to specify a maximum number of digits after the comma.

   extension ToString on double {
      String toStringWithMaxPrecision({int? maxDigits}) {
        if (round() == this) {
          return round().toString();
        } else {
          if (maxDigits== null) {
            return toString().replaceAll(RegExp(r'([.]*0)(?!.*\d)'), "");
          } else {
            return toStringAsFixed(maxDigits)
                .replaceAll(RegExp(r'([.]*0)(?!.*\d)'), "");
          }
        }
      }
    }
//output without maxDigits: 
// 1.0 -> 1
// 1.0000 -> 1
// 0.99990 -> 0.9999
// 0.103 -> 0.103
//
////output with maxDigits of 2: 
// 1.0 -> 1
// 1.0000 -> 1
// 0.99990 -> 0.99
// 0.103 -> 0.1

batman2020 · Answer 11 · 2020-07-04T17:49:18.150

1

Here is a very simple way. Using if else I will check if the number equals its integer or it is a fraction and take action accordingly

num x = 24/2; // returns 12.0
num y = 25/2; // returns 12.5

if (x == x.truncate()) {
// it is true in this case so i will do something like
x = x.toInt();
}

edited Jul 04 '20 at 17:49

answered Jul 04 '20 at 11:50

batman2020

31
6

score 1 · Answer 12 · answered Apr 13 '23 at 14:45

"Production ready"

extension MeineVer on double {
  String get toMoney => '$removeTrailingZeros₺';
  String get removeTrailingZeros {
    // return if complies to int
    if (this % 1 == 0) return toInt().toString();
    // remove trailing zeroes
    String str = '$this'.replaceAll(RegExp(r'0*$'), '');
    // reduce fraction max length to 2
    if (str.contains('.')) {
      final fr = str.split('.');
      if (2 < fr[1].length) {
        str = '${fr[0]}.${fr[1][0]}${fr[1][1]}';
      }
    }
    return str;
  }
}

23.1250 => 23.12
23.0130 => 23.01
23.1300 => 23.13
23.2000 => 23.2
23.0000 => 23

print(23.1300.removeTrailingZeros) => 23.13
print(23.1300.toMoney) => 23.13₺

score 0 · Answer 13 · answered Oct 20 '20 at 07:10

0

user3044484's version with Dart extension:

extension StringRegEx on String {
  String removeTrailingZero() {
    if (!this.contains('.')) {
      return this;
    }

    String trimmed = this.replaceAll(RegExp(r'0*$'), '');
    if (!trimmed.endsWith('.')) {
      return trimmed;
    }

    return trimmed.substring(0, this.length - 1);
  }
}

answered Oct 20 '20 at 07:10

Walter

1
1

Last line should be `return trimmed.substring(0, trimmed.length - 1);` – tudorprodan Nov 04 '20 at 14:29

score 0 · Answer 14 · answered Mar 05 '21 at 05:36

// The syntax is same as toStringAsFixed but this one removes trailing zeros
// 1st toStringAsFixed() is executed to limit the digits to your liking
// 2nd toString() is executed to remove trailing zeros

extension Ex on double {
  String toStringAsFixedNoZero(int n) =>            
  double.parse(this.toStringAsFixed(n)).toString(); 
}

// It works in all scenarios. Usage

void main() {

  double length1 = 25.001; 
  double length2 = 25.5487000; 
  double length3 = 25.10000;
  double length4 = 25.0000;
  double length5 = 0.9;

  print('\nlength1= ' + length1.toStringAsFixedNoZero(3));
  print('\nlength2= ' + length2.toStringAsFixedNoZero(3));
  print('\nlenght3= ' + length3.toStringAsFixedNoZero(3));
  print('\nlenght4= ' + length4.toStringAsFixedNoZero(3));
  print('\nlenght5= ' + length5.toStringAsFixedNoZero(0));

}

// output:

// length1= 25.001
// length2= 25.549
// lenght3= 25.1
// lenght4= 25
// lenght5= 1

Mohammed Adel · Answer 15 · 2022-02-22T14:43:32.047

0

you can do a simple extension on the double class and add a function which in my case i called it neglectFractionZero()

in this extension function on double(which returns a string) i split the converted number to string and i check if the split part of the string is "0" , if so i return the first part only of the split and i neglect this zero

you can modify it according to your needs

extension DoubleExtension on double { String neglectFractionZero() {

    return toString().split(".").last == "0"? toString().split(".").first:toString();
  }
}

edited Feb 22 '22 at 14:43

answered Feb 17 '22 at 09:48

Mohammed Adel

71
3

Remember that Stack Overflow isn't just intended to solve the immediate problem, but also to help future readers find solutions to similar problems, which requires understanding the underlying code. This is especially important for members of our community who are beginners, and not familiar with the syntax. Given that, **can you [edit] your answer to include an explanation of what you're doing** and why you believe it is the best approach? – Jeremy Caney Feb 22 '22 at 00:42

score 0 · Answer 16 · answered Apr 28 '23 at 08:29

The toString() already removes the useless "0" unless it is "1.0". So you just have to remove ".0" if it ends with that.

regex \.0$

for (var e in <double>[
  1,
  10,
  10.0,
  10.00,
  10.1,
  10.10,
  10.01,
  10.010,
]) {
  print(e.toString().replaceAll(RegExp(r'\.0$'), ''));
}

result:

score 0 · Answer 17 · answered Jul 17 '23 at 15:13


String formatWithoutTrailingZeros(double n) {
   return n.toString().replaceFirst(RegExp(r'(?<=\.\d*)(0+$)|(\.0+$)'), '');
}

Explenation:

(?<=\.\d*)(0+$) - Match zeros at the end of a decimal number like 1.040 or 0.00600

(\.0+$) - Match zeros and comma at the end of a decimal number like 1 .00 or 10‎‎‎‎‎‎‎‎‏‏‎ .0

score 0 · Answer 18 · answered Aug 03 '23 at 20:01

void main(List<String> args) {
  print(f('0'));
  print(f('0.0'));
  print(f('12'));
  print(f('12.0'));
  print(f('12.5'));
  print(f('12.5000'));
}

String f(String s) {
  var sepCount = 0;
  return switch (String.fromCharCodes(s.codeUnits.reversed
      .skipWhile((c) => c == 0x30)
      .skipWhile((c) => c == 0x2e && sepCount++ == 0)
      .toList()
      .reversed)) {
    '' => '0',
    final r => r,
  };
}

Output:

score 0 · Answer 19 · answered Aug 05 '23 at 15:59

One way is good, but 25 ways is better.
Another way to remove trailing zeros.

import 'package:parser_combinator/parser/delimited.dart';
import 'package:parser_combinator/parser/digit1.dart';
import 'package:parser_combinator/parser/eof.dart';
import 'package:parser_combinator/parser/not.dart';
import 'package:parser_combinator/parser/preceded.dart';
import 'package:parser_combinator/parser/replace_all.dart';
import 'package:parser_combinator/parser/skip_while1.dart';
import 'package:parser_combinator/parser/tag.dart';
import 'package:parser_combinator/parser/terminated.dart';
import 'package:parser_combinator/parsing.dart';

void main(List<String> args) {
  final list = [
    '',
    '0',
    '0.0',
    '10.0000000000000000000',
    '10.0000000000000000001'
  ];
  for (final element in list) {
    print(convert1(element));
    print(convert2(element));
  }
}

String convert1(String s) {
  const trailingZeros = Terminated(
    Preceded(
      Tag('.'),
      SkipWhile1(_isZero),
    ),
    Eof(),
  );
  const p = ReplaceAll(trailingZeros, _replace);
  return parseString(p.parse, s);
}

String convert2(String s) {
  const trailingZeros = Delimited(
    Tag('.'),
    SkipWhile1(_isZero),
    Not(Digit1()),
  );

  const p = ReplaceAll(trailingZeros, _replace);
  return parseString(p.parse, s);
}

bool _isZero(int c) => c == 0x30;

String _replace(String s) => '';

Output:

0
0
0
0
10
10
10.0000000000000000001
10.0000000000000000001

score -1 · Answer 20 · answered Jul 05 '19 at 11:19

I've came up with improved version of @John.

static String getDisplayPrice(double price) {
    price = price.abs();
    final str = price.toStringAsFixed(price.truncateToDouble() == price ? 0 : 2);
    if (str == '0') return '0';
    if (str.endsWith('.0')) return str.substring(0, str.length - 2);
    if (str.endsWith('0')) return str.substring(0, str.length -1);
    return str;
  }

// 10 -> 10
// 10.0 -> 10
// 10.50 -> 10.5
// 10.05 -> 10.05
// 10.000000000005 -> 10

This is incorrect. When you pass 100, you'll get 10. – benjtupas Mar 14 '20 at 09:31 — benjtupas, Mar 14 '20 at 09:31

score -2 · Answer 21 · answered Mar 14 '19 at 04:39

-2

void main() {
  double x1 = 12.0;
  double x2 = 12.5;
  String s1 = x1.toString().trim();
  String s2 = x2.toString().trim();
  print('s1 is $s1 and s2 is $s2');
  }

try trim method https://api.dartlang.org/stable/2.2.0/dart-core/String/trim.html

answered Mar 14 '19 at 04:39

Rishabh

2,410
4
17
36

How to remove trailing zeros using Dart

21 Answers21

Linked