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。

直接的方式:

double d = 2.3456789;
String inString = d.toStringAsFixed(2); // '2.35'
double inDouble = double.parse(inString); // 2.35 

使用扩展:

extension Ex on double {
  double toPrecision(int n) => double.parse(toStringAsFixed(n));
}

用法:

void main() {
  double d = 2.3456789;
  double d1 = d.toPrecision(1); // 2.3
  double d2 = d.toPrecision(2); // 2.35
  double d3 = d.toPrecision(3); // 2.345
}

你可以创建一个可重用的函数，接受你想格式化的numberOfDecimal，并利用toStringAsFixed()方法来格式化数字并将其转换回double。

供参考，toStringAsFixed方法不会四舍五入以5结尾的数字(例如:toStringAsFixed四舍五入2.275到2.27而不是2.28)。这是dart toStringAsFixed方法的默认行为(类似于Javascript的toFixed)

作为一种变通方法，我们可以在现有数字的最后一个十进制数后面加上1(例如:将0.0001加到2.275变成2.2751，而2.2751将正确舍入为2.28)

double roundOffToXDecimal(double number, {int numberOfDecimal = 2}) {
  // To prevent number that ends with 5 not round up correctly in Dart (eg: 2.275 round off to 2.27 instead of 2.28)
  String numbersAfterDecimal = number.toString().split('.')[1];
  if (numbersAfterDecimal != '0') {
    int existingNumberOfDecimal = numbersAfterDecimal.length;
    number += 1 / (10 * pow(10, existingNumberOfDecimal));
  }

  return double.parse(number.toStringAsFixed(numberOfDecimal));
}

// Example of usage:
var price = roundOffToXDecimal(2.275, numberOfDecimal: 2)
print(price); // 2.28

如果您想要一个双精度结果，将一个IEEE-754二进制浮点数舍入到特定的十进制位数本身就是有问题的。

In the same way that fractions such as 1/3 can't be exactly represented with a finite number of decimal digits, many (really infinitely many) decimal numbers can't be represented with a finite number of binary digits. For example, the decimal number 0.1 cannot be exactly represented in binary. While you could try to round 0.09999 to 0.1, as a double it would actually be "rounded" to 0.1000000000000000055511151231257827021181583404541015625. Most of the other answers that claim to round doubles with decimal precision actually return the nearest representable double.

你能做的就是使字符串表示形式看起来像一个漂亮的四舍五入的数字，这就是double.toStringAsFixed()所做的。这也是为什么当你打印0.100000000…，如果实现试图打印用户友好的值，则可能会看到0.1。然而，不要被愚弄了:double值实际上永远不会精确到0.1，如果你用这样不精确的值重复计算，就会累积错误。

请注意，以上所有内容都是浮点数工作方式所固有的，并不是Dart所特有的。还看到:

浮点数学坏了吗? 如果你不理解上面的解释，Tom Scott也做了一个很好的视频来解释浮点数是如何工作的。

底线:如果关心十进制精度，就不要使用二进制浮点类型。如果你和钱打交道，这一点尤其重要。

相反，你应该使用:

Integers. For example, if you are dealing with currency, instead of using double dollars = 1.23;, use int cents = 123;. Your calculations then always will be exact, and you can convert to the desired units only when displaying them to the user (and likewise can convert in the opposite direction when reading input from the user). A type designed to represent decimal numbers with arbitrary precision. For example, package:decimal provides a Decimal type. With such a type, some of the other answers (such as multiplying by 100, rounding, and then dividing by 100) then would be appropriate. (But really you should use Decimal.round directly.)

num.toStringAsFixed()轮。它将你的num (n)转换成一个带有你想要的小数点数的字符串(2)，然后在一行漂亮的代码中将它解析回你的num:

n = num.parse(n.toStringAsFixed(2));

使用Dart Extension方法修改@andyw的答案:

extension Precision on double {
    double toPrecision(int fractionDigits) {
        double mod = pow(10, fractionDigits.toDouble());
        return ((this * mod).round().toDouble() / mod);
    }
}

用法:

var latitude = 1.123456;
var latitudeWithFixedPrecision = latitude.toPrecision(3); // Outputs: 1.123