double value = 2.8032739273;
String formattedValue = value.toStringAsFixed(3);

我在double上做了这个扩展

import 'dart:math';

extension DoubleExtension on double {

  /// rounds the double to a specific decimal place
  double roundedPrecision(int places) {
    double mod = pow(10.0, places) as double;
    return ((this * mod).round().toDouble() / mod);
  }

  /// good for string output because it can remove trailing zeros
  /// and sometimes periods. Or optionally display the exact number of trailing
  /// zeros
  String roundedPrecisionToString(
    int places, {
    bool trailingZeros = false,
  }) {
    double mod = pow(10.0, places) as double;
    double round = ((this * mod).round().toDouble() / mod);
    String doubleToString =
        trailingZeros ? round.toStringAsFixed(places) : round.toString();
    if (!trailingZeros) {
      RegExp trailingZeros = RegExp(r'^[0-9]+.0+$');
      if (trailingZeros.hasMatch(doubleToString)) {
        doubleToString = doubleToString.split('.')[0];
      }
    }
    return doubleToString;
  }

  String toStringNoTrailingZeros() {
    String doubleToString = toString();
    RegExp trailingZeros = RegExp(r'^[0-9]+.0+$');
    if (trailingZeros.hasMatch(doubleToString)) {
      doubleToString = doubleToString.split('.')[0];
    }
    return doubleToString;
  }
}

这是通过的测试。

import 'package:flutter_test/flutter_test.dart';
import 'package:project_name/utils/double_extension.dart';

void main() {
  group("rounded precision", () {
    test("rounding to 0 place results in an int", () {
      double num = 5.1234;
      double num2 = 5.8234;
      expect(num.roundedPrecision(0), 5);
      expect(num2.roundedPrecision(0), 6);
    });
    test("rounding to 1 place rounds correctly to 1 place", () {
      double num = 5.12;
      double num2 = 5.15;
      expect(num.roundedPrecision(1), 5.1);
      expect(num2.roundedPrecision(1), 5.2);
    });
    test(
        "rounding a number to a precision that is more accurate than the origional",
        () {
      double num = 5;
      expect(num.roundedPrecision(5), 5);
    });
  });

  group("rounded precision returns the correct string", () {
    test("rounding to 0 place results in an int", () {
      double num = 5.1234;
      double num2 = 5.8234;
      expect(num.roundedPrecisionToString(0), "5");
      expect(num2.roundedPrecisionToString(0), "6");
    });
    test("rounding to 1 place rounds correct", () {
      double num = 5.12;
      double num2 = 5.15;
      expect(num.roundedPrecisionToString(1), "5.1");
      expect(num2.roundedPrecisionToString(1), "5.2");
    });
    test("rounding to 2 places rounds correct", () {
      double num = 5.123;
      double num2 = 5.156;
      expect(num.roundedPrecisionToString(2), "5.12");
      expect(num2.roundedPrecisionToString(2), "5.16");
    });
    test("cut off all trailing zeros (and periods)", () {
      double num = 5;
      double num2 = 5.03000;
      expect(num.roundedPrecisionToString(5), "5");
      expect(num2.roundedPrecisionToString(5), "5.03");
    });
  });
}

从来没有想过这在Dart是如此复杂，但这是我的解决方案:

double truncateDouble(double val, int decimals) {
    String valString = val.toString();
    int dotIndex = valString.indexOf('.');

    // not enough decimals
    int totalDecimals = valString.length - dotIndex - 1;
    if (totalDecimals < decimals) {
      decimals = totalDecimals;
    }

    valString = valString.substring(0, dotIndex + decimals + 1);

    return double.parse(valString);
  }

var val = truncateDouble(44.999, 2);

您可以调用此函数来获得黑暗(颤振)的精确程度。 Double eval -> Double想要转换的 Int I ->返回的小数点。

double doubleToPrecision(double eval, int i) {
double step1 = eval;//1/3
print(step1); // 0.3333333333333333

String step2 = step1.toStringAsFixed(2);
print(step2); // 0.33

double step3 = double.parse(step2);
print(step3); // 0.33
eval = step3;
return eval; }

如果使用动态类型的数据。你可以使用它。

 typeDecimal(data) => num.parse(data.toString()).toStringAsFixed(2);

如果你不想要任何小数，而结果的小数都是0，这样做是可行的:

String fixedDecimals(double d, int decimals, {bool removeZeroDecimals = true}){
  double mod = pow(10.0, decimals);
  double result = ((d * mod).round().toDouble() / mod);
  if( removeZeroDecimals && result - (result.truncate()) == 0.0 ) decimals = 0;
  return result.toStringAsFixed(decimals);
}

如果输入是9.004并且你想要2个小数，这将简单地输出9而不是9.00。